The net ionic equation for 3C 2 H 5 OH(aq)+2K 2 Cr 2 O 7 (aq)+8H 2 SO 4 (aq) → 3CH 3 COOH(aq)+2Cr 2 (SO 4 ) 3 (aq)+2K 2 SO 4 (aq)+11H 2 O(l) for the given reaction has to be written. Concept introduction: Net ionic Equation: The net ionic equation shows only those chemical species that actually undergo chemical change. The ions that do not change during the chemical reaction are called spectator ions. These spectator ions are not shown in net ionic equation.

Question

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Answer 1

Question

Chapter 3, Problem 113QRT

(a)

Interpretation Introduction

Interpretation:

The net ionic equation for $3C2H5OH(aq)+2K2Cr2O7(aq)+8H2SO4(aq)→3CH3COOH(aq)+2Cr2(SO4)3(aq)+2K2SO4(aq)+11H2O(l)$ for the given reaction has to be written.

Concept introduction:

Net ionic Equation:

The net ionic equation shows only those chemical species that actually undergo chemical change. The ions that do not change during the chemical reaction are called spectator ions. These spectator ions are not shown in net ionic equation.

(a)

Expert Solution

Explanation of Solution

The balanced equation for the reaction is given below,

$3 C2H5OH(aq) + 2 K2Cr2O7(aq) + 8 H2SO4(aq) ↓3CH3COOH(aq) + 2 Cr2(SO4)3(aq) + 2 K2SO4(aq) + 11 H2O(l)$

While balancing the equation, the subscripts cannot be altered but coefficients can be changed. The given reaction is already balanced.

The complete ionic equation is given below,

$3 C2H5OH(aq) + 4 K+ + 2 Cr2O72−+ 8 H++ 8HSO4− ↓3 CH3COOH (aq) + 4 Cr3+ + 6 SO42−(aq) + 4 K+ + 2 SO42−(aq) + 11 H2O(l)$

Eliminate the spectator ions. Net ionic equation of the given reaction shown below

$3 C2H5OH(aq) + 2 Cr2O72−+ 8 H++ 8HSO4− ↓3 CH3COOH (aq) + 4 Cr3+ + 6 SO42−(aq) +2 SO42−(aq) + 11 H2O(l)$

The balanced complete net ionic equation is written.

(b)

Interpretation Introduction

Interpretation:

The changing the oxidation numbers in the reaction has to be explained.

Concept introduction:

Oxidation number:

The oxidation number of an element is zero. The oxidation number of a monoatomic ion equals its charge.

The oxidation number is zero for the summation of the oxidation numbers complete atoms in a complete formula.

The charge on the ion is equal to the summation of the oxidation numbers of complete atoms in poly atomic ion.

The oxidation state of alkali metal $(Li, Na, K, Rb, Cs)$ is $+1$ , and alkaline earth metal is $(Be, Mg, Ca, Sr, Ba)$ is $+2$ .

The oxidation state of hydrogen is $+1$ (without bonding with metals), the oxidation state of oxygen is $−2$ .

An oxidizing agent gains the electrons and is reduced in a chemical reaction and it is electron acceptor.

A reducing agent loses electrons and is oxidized in a chemical reaction and it is electron donor.

The oxidation state is called as oxidation number, which describes degree of oxidation (loss of electrons) of an atom in a chemical compound. Theoretically, the oxidation state is positive, negative or zero.

(b)

Expert Solution

Explanation of Solution

The given equation is shown below,

$3 C2H5OH(aq) + 4 K+ + 2 Cr2O72−+ 8 H++ 8HSO4− ↓3 CH3COOH (aq) + 4 Cr3+ + 6 SO42−(aq) + 4 K+ + 2 SO42−(aq) + 11 H2O(l)$

Oxidation number of element in the reactant:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$C2H5OH:2(x) + 6(1) +1(-2)= 02x + 6 - 2 = 0x = −2$

The oxidation number of carbon is $+ 2$ .

Oxidation number of chromium in $Cr2O72−$ is given below,

$Cr2O72−:2(x) + 7(-2) = −22x - 14= −2x = + 6$

The oxidation number of chromium is $+ 6$ .

Oxidation number of sulfur in $HSO4−$ is given below,

$HSO4−:1+ (x) + 4(-2) = −1x - 7= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation number of element in the product:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$CH3COOH:2(x) + 4(1) +2(-2)= 02x + 4 - 4 = 0x = 0$

Oxidation number of carbon is zero.

Oxidation number of sulfur in $HSO4−$ is given below,

$SO42−:(x) + 4(-2) = −2x - 8= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation state of hydrogen is $+ 1$ and oxygen is $−2$ in water.

Oxidation number of carbon is $+ 2$ in the reactant and zero in the product, the oxidation number of chromium is $+ 6$ in the reactant and $+ 3$ in the product. Carbon containing compounds loss the electron therefore it is reducing agent ( $C2H5OH$ ). Chromium containing compound gains the electron therefore it is oxidizing agent ( $CH3COOH$ ).

(c)

Interpretation Introduction

Interpretation:

The oxidized and reduced substance has to be identified.

Concept introduction:

Refer to part (b)

(c)

Expert Solution

Explanation of Solution

The given equation is shown below,

$3 C2H5OH(aq) + 4 K+ + 2 Cr2O72−+ 8 H++ 8HSO4− ↓3 CH3COOH (aq) + 4 Cr3+ + 6 SO42−(aq) + 4 K+ + 2 SO42−(aq) + 11 H2O(l)$

Oxidation number of element in the reactant:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$C2H5OH:2(x) + 6(1) +1(-2)= 02x + 6 - 2 = 0x = −2$

The oxidation number of carbon is $+ 2$ .

Oxidation number of chromium in $Cr2O72−$ is given below,

$Cr2O72−:2(x) + 7(-2) = −22x - 14= −2x = + 6$

The oxidation number of chromium is $+ 6$ .

Oxidation number of sulfur in $HSO4−$ is given below,

$HSO4−:1+ (x) + 4(-2) = −1x - 7= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation number of element in the product:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$CH3COOH:2(x) + 4(1) +2(-2)= 02x + 4 - 4 = 0x = 0$

Oxidation number of carbon is zero.

Oxidation number of sulfur in $HSO4−$ is given below,

$SO42−:(x) + 4(-2) = −2x - 8= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation state of hydrogen is $+ 1$ and oxygen is $−2$ in water.

Oxidation number of carbon is $+ 2$ in the reactant and zero in the product, the oxidation number of chromium is $+ 6$ in the reactant and $+ 3$ in the product. Carbon containing compounds loss the electron therefore it has being oxidized ( $C2H5OH$ ). Chromium containing compound gains the electron therefore it has being reduced ( $K2Cr2O7$ ).

(d)

Interpretation Introduction

Interpretation:

The oxidizing agent and reducing agent has to be identified.

Concept introduction:

Refer to part (b)

(d)

Expert Solution

Explanation of Solution

The given equation is shown below,

$3 C2H5OH(aq) + 4 K+ + 2 Cr2O72−+ 8 H++ 8HSO4− ↓3 CH3COOH (aq) + 4 Cr3+ + 6 SO42−(aq) + 4 K+ + 2 SO42−(aq) + 11 H2O(l)$

Oxidation number of element in the reactant:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$C2H5OH:2(x) + 6(1) +1(-2)= 02x + 6 - 2 = 0x = −2$

The oxidation number of carbon is $+ 2$ .

Oxidation number of chromium in $Cr2O72−$ is given below,

$Cr2O72−:2(x) + 7(-2) = −22x - 14= −2x = + 6$

The oxidation number of chromium is $+ 6$ .

Oxidation number of sulfur in $HSO4−$ is given below,

$HSO4−:1+ (x) + 4(-2) = −1x - 7= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation number of element in the product:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$CH3COOH:2(x) + 4(1) +2(-2)= 02x + 4 - 4 = 0x = 0$

Oxidation number of carbon is zero.

Oxidation number of sulfur in $HSO4−$ is given below,

$SO42−:(x) + 4(-2) = −2x - 8= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation state of hydrogen is $+ 1$ and oxygen is $−2$ in water.

Oxidation number of carbon is $+ 2$ in the reactant and zero in the product, the oxidation number of chromium is $+ 6$ in the reactant and $+ 3$ in the product. Carbon containing compounds loss the electron therefore it is reducing agent ( $C2H5OH$ ). Chromium containing compound gains the electron therefore it is oxidizing agent ( $CH3COOH$ ).

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Answer 2

Question

Chapter 3, Problem 113QRT

(a)

Interpretation Introduction

Interpretation:

The net ionic equation for $3C2H5OH(aq)+2K2Cr2O7(aq)+8H2SO4(aq)→3CH3COOH(aq)+2Cr2(SO4)3(aq)+2K2SO4(aq)+11H2O(l)$ for the given reaction has to be written.

Concept introduction:

Net ionic Equation:

The net ionic equation shows only those chemical species that actually undergo chemical change. The ions that do not change during the chemical reaction are called spectator ions. These spectator ions are not shown in net ionic equation.

(a)

Expert Solution

Explanation of Solution

The balanced equation for the reaction is given below,

$3 C2H5OH(aq) + 2 K2Cr2O7(aq) + 8 H2SO4(aq) ↓3CH3COOH(aq) + 2 Cr2(SO4)3(aq) + 2 K2SO4(aq) + 11 H2O(l)$

While balancing the equation, the subscripts cannot be altered but coefficients can be changed. The given reaction is already balanced.

The complete ionic equation is given below,

$3 C2H5OH(aq) + 4 K+ + 2 Cr2O72−+ 8 H++ 8HSO4− ↓3 CH3COOH (aq) + 4 Cr3+ + 6 SO42−(aq) + 4 K+ + 2 SO42−(aq) + 11 H2O(l)$

Eliminate the spectator ions. Net ionic equation of the given reaction shown below

$3 C2H5OH(aq) + 2 Cr2O72−+ 8 H++ 8HSO4− ↓3 CH3COOH (aq) + 4 Cr3+ + 6 SO42−(aq) +2 SO42−(aq) + 11 H2O(l)$

The balanced complete net ionic equation is written.

(b)

Interpretation Introduction

Interpretation:

The changing the oxidation numbers in the reaction has to be explained.

Concept introduction:

Oxidation number:

The oxidation number of an element is zero. The oxidation number of a monoatomic ion equals its charge.

The oxidation number is zero for the summation of the oxidation numbers complete atoms in a complete formula.

The charge on the ion is equal to the summation of the oxidation numbers of complete atoms in poly atomic ion.

The oxidation state of alkali metal $(Li, Na, K, Rb, Cs)$ is $+1$ , and alkaline earth metal is $(Be, Mg, Ca, Sr, Ba)$ is $+2$ .

The oxidation state of hydrogen is $+1$ (without bonding with metals), the oxidation state of oxygen is $−2$ .

An oxidizing agent gains the electrons and is reduced in a chemical reaction and it is electron acceptor.

A reducing agent loses electrons and is oxidized in a chemical reaction and it is electron donor.

The oxidation state is called as oxidation number, which describes degree of oxidation (loss of electrons) of an atom in a chemical compound. Theoretically, the oxidation state is positive, negative or zero.

(b)

Expert Solution

Explanation of Solution

The given equation is shown below,

$3 C2H5OH(aq) + 4 K+ + 2 Cr2O72−+ 8 H++ 8HSO4− ↓3 CH3COOH (aq) + 4 Cr3+ + 6 SO42−(aq) + 4 K+ + 2 SO42−(aq) + 11 H2O(l)$

Oxidation number of element in the reactant:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$C2H5OH:2(x) + 6(1) +1(-2)= 02x + 6 - 2 = 0x = −2$

The oxidation number of carbon is $+ 2$ .

Oxidation number of chromium in $Cr2O72−$ is given below,

$Cr2O72−:2(x) + 7(-2) = −22x - 14= −2x = + 6$

The oxidation number of chromium is $+ 6$ .

Oxidation number of sulfur in $HSO4−$ is given below,

$HSO4−:1+ (x) + 4(-2) = −1x - 7= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation number of element in the product:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$CH3COOH:2(x) + 4(1) +2(-2)= 02x + 4 - 4 = 0x = 0$

Oxidation number of carbon is zero.

Oxidation number of sulfur in $HSO4−$ is given below,

$SO42−:(x) + 4(-2) = −2x - 8= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation state of hydrogen is $+ 1$ and oxygen is $−2$ in water.

Oxidation number of carbon is $+ 2$ in the reactant and zero in the product, the oxidation number of chromium is $+ 6$ in the reactant and $+ 3$ in the product. Carbon containing compounds loss the electron therefore it is reducing agent ( $C2H5OH$ ). Chromium containing compound gains the electron therefore it is oxidizing agent ( $CH3COOH$ ).

(c)

Interpretation Introduction

Interpretation:

The oxidized and reduced substance has to be identified.

Concept introduction:

Refer to part (b)

(c)

Expert Solution

Explanation of Solution

The given equation is shown below,

$3 C2H5OH(aq) + 4 K+ + 2 Cr2O72−+ 8 H++ 8HSO4− ↓3 CH3COOH (aq) + 4 Cr3+ + 6 SO42−(aq) + 4 K+ + 2 SO42−(aq) + 11 H2O(l)$

Oxidation number of element in the reactant:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$C2H5OH:2(x) + 6(1) +1(-2)= 02x + 6 - 2 = 0x = −2$

The oxidation number of carbon is $+ 2$ .

Oxidation number of chromium in $Cr2O72−$ is given below,

$Cr2O72−:2(x) + 7(-2) = −22x - 14= −2x = + 6$

The oxidation number of chromium is $+ 6$ .

Oxidation number of sulfur in $HSO4−$ is given below,

$HSO4−:1+ (x) + 4(-2) = −1x - 7= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation number of element in the product:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$CH3COOH:2(x) + 4(1) +2(-2)= 02x + 4 - 4 = 0x = 0$

Oxidation number of carbon is zero.

Oxidation number of sulfur in $HSO4−$ is given below,

$SO42−:(x) + 4(-2) = −2x - 8= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation state of hydrogen is $+ 1$ and oxygen is $−2$ in water.

Oxidation number of carbon is $+ 2$ in the reactant and zero in the product, the oxidation number of chromium is $+ 6$ in the reactant and $+ 3$ in the product. Carbon containing compounds loss the electron therefore it has being oxidized ( $C2H5OH$ ). Chromium containing compound gains the electron therefore it has being reduced ( $K2Cr2O7$ ).

(d)

Interpretation Introduction

Interpretation:

The oxidizing agent and reducing agent has to be identified.

Concept introduction:

Refer to part (b)

(d)

Expert Solution

Explanation of Solution

The given equation is shown below,

$3 C2H5OH(aq) + 4 K+ + 2 Cr2O72−+ 8 H++ 8HSO4− ↓3 CH3COOH (aq) + 4 Cr3+ + 6 SO42−(aq) + 4 K+ + 2 SO42−(aq) + 11 H2O(l)$

Oxidation number of element in the reactant:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$C2H5OH:2(x) + 6(1) +1(-2)= 02x + 6 - 2 = 0x = −2$

The oxidation number of carbon is $+ 2$ .

Oxidation number of chromium in $Cr2O72−$ is given below,

$Cr2O72−:2(x) + 7(-2) = −22x - 14= −2x = + 6$

The oxidation number of chromium is $+ 6$ .

Oxidation number of sulfur in $HSO4−$ is given below,

$HSO4−:1+ (x) + 4(-2) = −1x - 7= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation number of element in the product:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$CH3COOH:2(x) + 4(1) +2(-2)= 02x + 4 - 4 = 0x = 0$

Oxidation number of carbon is zero.

Oxidation number of sulfur in $HSO4−$ is given below,

$SO42−:(x) + 4(-2) = −2x - 8= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation state of hydrogen is $+ 1$ and oxygen is $−2$ in water.

Oxidation number of carbon is $+ 2$ in the reactant and zero in the product, the oxidation number of chromium is $+ 6$ in the reactant and $+ 3$ in the product. Carbon containing compounds loss the electron therefore it is reducing agent ( $C2H5OH$ ). Chromium containing compound gains the electron therefore it is oxidizing agent ( $CH3COOH$ ).

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Answer 3

Question

Chapter 3, Problem 113QRT

(a)

Interpretation Introduction

Interpretation:

The net ionic equation for $3C2H5OH(aq)+2K2Cr2O7(aq)+8H2SO4(aq)→3CH3COOH(aq)+2Cr2(SO4)3(aq)+2K2SO4(aq)+11H2O(l)$ for the given reaction has to be written.

Concept introduction:

Net ionic Equation:

The net ionic equation shows only those chemical species that actually undergo chemical change. The ions that do not change during the chemical reaction are called spectator ions. These spectator ions are not shown in net ionic equation.

(a)

Expert Solution

Explanation of Solution

The balanced equation for the reaction is given below,

$3 C2H5OH(aq) + 2 K2Cr2O7(aq) + 8 H2SO4(aq) ↓3CH3COOH(aq) + 2 Cr2(SO4)3(aq) + 2 K2SO4(aq) + 11 H2O(l)$

While balancing the equation, the subscripts cannot be altered but coefficients can be changed. The given reaction is already balanced.

The complete ionic equation is given below,

$3 C2H5OH(aq) + 4 K+ + 2 Cr2O72−+ 8 H++ 8HSO4− ↓3 CH3COOH (aq) + 4 Cr3+ + 6 SO42−(aq) + 4 K+ + 2 SO42−(aq) + 11 H2O(l)$

Eliminate the spectator ions. Net ionic equation of the given reaction shown below

$3 C2H5OH(aq) + 2 Cr2O72−+ 8 H++ 8HSO4− ↓3 CH3COOH (aq) + 4 Cr3+ + 6 SO42−(aq) +2 SO42−(aq) + 11 H2O(l)$

The balanced complete net ionic equation is written.

(b)

Interpretation Introduction

Interpretation:

The changing the oxidation numbers in the reaction has to be explained.

Concept introduction:

Oxidation number:

The oxidation number of an element is zero. The oxidation number of a monoatomic ion equals its charge.

The oxidation number is zero for the summation of the oxidation numbers complete atoms in a complete formula.

The charge on the ion is equal to the summation of the oxidation numbers of complete atoms in poly atomic ion.

The oxidation state of alkali metal $(Li, Na, K, Rb, Cs)$ is $+1$ , and alkaline earth metal is $(Be, Mg, Ca, Sr, Ba)$ is $+2$ .

The oxidation state of hydrogen is $+1$ (without bonding with metals), the oxidation state of oxygen is $−2$ .

An oxidizing agent gains the electrons and is reduced in a chemical reaction and it is electron acceptor.

A reducing agent loses electrons and is oxidized in a chemical reaction and it is electron donor.

The oxidation state is called as oxidation number, which describes degree of oxidation (loss of electrons) of an atom in a chemical compound. Theoretically, the oxidation state is positive, negative or zero.

(b)

Expert Solution

Explanation of Solution

The given equation is shown below,

$3 C2H5OH(aq) + 4 K+ + 2 Cr2O72−+ 8 H++ 8HSO4− ↓3 CH3COOH (aq) + 4 Cr3+ + 6 SO42−(aq) + 4 K+ + 2 SO42−(aq) + 11 H2O(l)$

Oxidation number of element in the reactant:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$C2H5OH:2(x) + 6(1) +1(-2)= 02x + 6 - 2 = 0x = −2$

The oxidation number of carbon is $+ 2$ .

Oxidation number of chromium in $Cr2O72−$ is given below,

$Cr2O72−:2(x) + 7(-2) = −22x - 14= −2x = + 6$

The oxidation number of chromium is $+ 6$ .

Oxidation number of sulfur in $HSO4−$ is given below,

$HSO4−:1+ (x) + 4(-2) = −1x - 7= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation number of element in the product:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$CH3COOH:2(x) + 4(1) +2(-2)= 02x + 4 - 4 = 0x = 0$

Oxidation number of carbon is zero.

Oxidation number of sulfur in $HSO4−$ is given below,

$SO42−:(x) + 4(-2) = −2x - 8= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation state of hydrogen is $+ 1$ and oxygen is $−2$ in water.

Oxidation number of carbon is $+ 2$ in the reactant and zero in the product, the oxidation number of chromium is $+ 6$ in the reactant and $+ 3$ in the product. Carbon containing compounds loss the electron therefore it is reducing agent ( $C2H5OH$ ). Chromium containing compound gains the electron therefore it is oxidizing agent ( $CH3COOH$ ).

(c)

Interpretation Introduction

Interpretation:

The oxidized and reduced substance has to be identified.

Concept introduction:

Refer to part (b)

(c)

Expert Solution

Explanation of Solution

The given equation is shown below,

$3 C2H5OH(aq) + 4 K+ + 2 Cr2O72−+ 8 H++ 8HSO4− ↓3 CH3COOH (aq) + 4 Cr3+ + 6 SO42−(aq) + 4 K+ + 2 SO42−(aq) + 11 H2O(l)$

Oxidation number of element in the reactant:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$C2H5OH:2(x) + 6(1) +1(-2)= 02x + 6 - 2 = 0x = −2$

The oxidation number of carbon is $+ 2$ .

Oxidation number of chromium in $Cr2O72−$ is given below,

$Cr2O72−:2(x) + 7(-2) = −22x - 14= −2x = + 6$

The oxidation number of chromium is $+ 6$ .

Oxidation number of sulfur in $HSO4−$ is given below,

$HSO4−:1+ (x) + 4(-2) = −1x - 7= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation number of element in the product:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$CH3COOH:2(x) + 4(1) +2(-2)= 02x + 4 - 4 = 0x = 0$

Oxidation number of carbon is zero.

Oxidation number of sulfur in $HSO4−$ is given below,

$SO42−:(x) + 4(-2) = −2x - 8= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation state of hydrogen is $+ 1$ and oxygen is $−2$ in water.

Oxidation number of carbon is $+ 2$ in the reactant and zero in the product, the oxidation number of chromium is $+ 6$ in the reactant and $+ 3$ in the product. Carbon containing compounds loss the electron therefore it has being oxidized ( $C2H5OH$ ). Chromium containing compound gains the electron therefore it has being reduced ( $K2Cr2O7$ ).

(d)

Interpretation Introduction

Interpretation:

The oxidizing agent and reducing agent has to be identified.

Concept introduction:

Refer to part (b)

(d)

Expert Solution

Explanation of Solution

The given equation is shown below,

$3 C2H5OH(aq) + 4 K+ + 2 Cr2O72−+ 8 H++ 8HSO4− ↓3 CH3COOH (aq) + 4 Cr3+ + 6 SO42−(aq) + 4 K+ + 2 SO42−(aq) + 11 H2O(l)$

Oxidation number of element in the reactant:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$C2H5OH:2(x) + 6(1) +1(-2)= 02x + 6 - 2 = 0x = −2$

The oxidation number of carbon is $+ 2$ .

Oxidation number of chromium in $Cr2O72−$ is given below,

$Cr2O72−:2(x) + 7(-2) = −22x - 14= −2x = + 6$

The oxidation number of chromium is $+ 6$ .

Oxidation number of sulfur in $HSO4−$ is given below,

$HSO4−:1+ (x) + 4(-2) = −1x - 7= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation number of element in the product:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$CH3COOH:2(x) + 4(1) +2(-2)= 02x + 4 - 4 = 0x = 0$

Oxidation number of carbon is zero.

Oxidation number of sulfur in $HSO4−$ is given below,

$SO42−:(x) + 4(-2) = −2x - 8= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation state of hydrogen is $+ 1$ and oxygen is $−2$ in water.

Oxidation number of carbon is $+ 2$ in the reactant and zero in the product, the oxidation number of chromium is $+ 6$ in the reactant and $+ 3$ in the product. Carbon containing compounds loss the electron therefore it is reducing agent ( $C2H5OH$ ). Chromium containing compound gains the electron therefore it is oxidizing agent ( $CH3COOH$ ).

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Answer 4

Question

Chapter 3, Problem 113QRT

(a)

Interpretation Introduction

Interpretation:

The net ionic equation for $3C2H5OH(aq)+2K2Cr2O7(aq)+8H2SO4(aq)→3CH3COOH(aq)+2Cr2(SO4)3(aq)+2K2SO4(aq)+11H2O(l)$ for the given reaction has to be written.

Concept introduction:

Net ionic Equation:

The net ionic equation shows only those chemical species that actually undergo chemical change. The ions that do not change during the chemical reaction are called spectator ions. These spectator ions are not shown in net ionic equation.

(a)

Expert Solution

Explanation of Solution

The balanced equation for the reaction is given below,

$3 C2H5OH(aq) + 2 K2Cr2O7(aq) + 8 H2SO4(aq) ↓3CH3COOH(aq) + 2 Cr2(SO4)3(aq) + 2 K2SO4(aq) + 11 H2O(l)$

While balancing the equation, the subscripts cannot be altered but coefficients can be changed. The given reaction is already balanced.

The complete ionic equation is given below,

$3 C2H5OH(aq) + 4 K+ + 2 Cr2O72−+ 8 H++ 8HSO4− ↓3 CH3COOH (aq) + 4 Cr3+ + 6 SO42−(aq) + 4 K+ + 2 SO42−(aq) + 11 H2O(l)$

Eliminate the spectator ions. Net ionic equation of the given reaction shown below

$3 C2H5OH(aq) + 2 Cr2O72−+ 8 H++ 8HSO4− ↓3 CH3COOH (aq) + 4 Cr3+ + 6 SO42−(aq) +2 SO42−(aq) + 11 H2O(l)$

The balanced complete net ionic equation is written.

(b)

Interpretation Introduction

Interpretation:

The changing the oxidation numbers in the reaction has to be explained.

Concept introduction:

Oxidation number:

The oxidation number of an element is zero. The oxidation number of a monoatomic ion equals its charge.

The oxidation number is zero for the summation of the oxidation numbers complete atoms in a complete formula.

The charge on the ion is equal to the summation of the oxidation numbers of complete atoms in poly atomic ion.

The oxidation state of alkali metal $(Li, Na, K, Rb, Cs)$ is $+1$ , and alkaline earth metal is $(Be, Mg, Ca, Sr, Ba)$ is $+2$ .

The oxidation state of hydrogen is $+1$ (without bonding with metals), the oxidation state of oxygen is $−2$ .

An oxidizing agent gains the electrons and is reduced in a chemical reaction and it is electron acceptor.

A reducing agent loses electrons and is oxidized in a chemical reaction and it is electron donor.

The oxidation state is called as oxidation number, which describes degree of oxidation (loss of electrons) of an atom in a chemical compound. Theoretically, the oxidation state is positive, negative or zero.

(b)

Expert Solution

Explanation of Solution

The given equation is shown below,

$3 C2H5OH(aq) + 4 K+ + 2 Cr2O72−+ 8 H++ 8HSO4− ↓3 CH3COOH (aq) + 4 Cr3+ + 6 SO42−(aq) + 4 K+ + 2 SO42−(aq) + 11 H2O(l)$

Oxidation number of element in the reactant:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$C2H5OH:2(x) + 6(1) +1(-2)= 02x + 6 - 2 = 0x = −2$

The oxidation number of carbon is $+ 2$ .

Oxidation number of chromium in $Cr2O72−$ is given below,

$Cr2O72−:2(x) + 7(-2) = −22x - 14= −2x = + 6$

The oxidation number of chromium is $+ 6$ .

Oxidation number of sulfur in $HSO4−$ is given below,

$HSO4−:1+ (x) + 4(-2) = −1x - 7= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation number of element in the product:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$CH3COOH:2(x) + 4(1) +2(-2)= 02x + 4 - 4 = 0x = 0$

Oxidation number of carbon is zero.

Oxidation number of sulfur in $HSO4−$ is given below,

$SO42−:(x) + 4(-2) = −2x - 8= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation state of hydrogen is $+ 1$ and oxygen is $−2$ in water.

Oxidation number of carbon is $+ 2$ in the reactant and zero in the product, the oxidation number of chromium is $+ 6$ in the reactant and $+ 3$ in the product. Carbon containing compounds loss the electron therefore it is reducing agent ( $C2H5OH$ ). Chromium containing compound gains the electron therefore it is oxidizing agent ( $CH3COOH$ ).

(c)

Interpretation Introduction

Interpretation:

The oxidized and reduced substance has to be identified.

Concept introduction:

Refer to part (b)

(c)

Expert Solution

Explanation of Solution

The given equation is shown below,

$3 C2H5OH(aq) + 4 K+ + 2 Cr2O72−+ 8 H++ 8HSO4− ↓3 CH3COOH (aq) + 4 Cr3+ + 6 SO42−(aq) + 4 K+ + 2 SO42−(aq) + 11 H2O(l)$

Oxidation number of element in the reactant:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$C2H5OH:2(x) + 6(1) +1(-2)= 02x + 6 - 2 = 0x = −2$

The oxidation number of carbon is $+ 2$ .

Oxidation number of chromium in $Cr2O72−$ is given below,

$Cr2O72−:2(x) + 7(-2) = −22x - 14= −2x = + 6$

The oxidation number of chromium is $+ 6$ .

Oxidation number of sulfur in $HSO4−$ is given below,

$HSO4−:1+ (x) + 4(-2) = −1x - 7= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation number of element in the product:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$CH3COOH:2(x) + 4(1) +2(-2)= 02x + 4 - 4 = 0x = 0$

Oxidation number of carbon is zero.

Oxidation number of sulfur in $HSO4−$ is given below,

$SO42−:(x) + 4(-2) = −2x - 8= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation state of hydrogen is $+ 1$ and oxygen is $−2$ in water.

Oxidation number of carbon is $+ 2$ in the reactant and zero in the product, the oxidation number of chromium is $+ 6$ in the reactant and $+ 3$ in the product. Carbon containing compounds loss the electron therefore it has being oxidized ( $C2H5OH$ ). Chromium containing compound gains the electron therefore it has being reduced ( $K2Cr2O7$ ).

(d)

Interpretation Introduction

Interpretation:

The oxidizing agent and reducing agent has to be identified.

Concept introduction:

Refer to part (b)

(d)

Expert Solution

Explanation of Solution

The given equation is shown below,

$3 C2H5OH(aq) + 4 K+ + 2 Cr2O72−+ 8 H++ 8HSO4− ↓3 CH3COOH (aq) + 4 Cr3+ + 6 SO42−(aq) + 4 K+ + 2 SO42−(aq) + 11 H2O(l)$

Oxidation number of element in the reactant:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$C2H5OH:2(x) + 6(1) +1(-2)= 02x + 6 - 2 = 0x = −2$

The oxidation number of carbon is $+ 2$ .

Oxidation number of chromium in $Cr2O72−$ is given below,

$Cr2O72−:2(x) + 7(-2) = −22x - 14= −2x = + 6$

The oxidation number of chromium is $+ 6$ .

Oxidation number of sulfur in $HSO4−$ is given below,

$HSO4−:1+ (x) + 4(-2) = −1x - 7= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation number of element in the product:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$CH3COOH:2(x) + 4(1) +2(-2)= 02x + 4 - 4 = 0x = 0$

Oxidation number of carbon is zero.

Oxidation number of sulfur in $HSO4−$ is given below,

$SO42−:(x) + 4(-2) = −2x - 8= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation state of hydrogen is $+ 1$ and oxygen is $−2$ in water.

Oxidation number of carbon is $+ 2$ in the reactant and zero in the product, the oxidation number of chromium is $+ 6$ in the reactant and $+ 3$ in the product. Carbon containing compounds loss the electron therefore it is reducing agent ( $C2H5OH$ ). Chromium containing compound gains the electron therefore it is oxidizing agent ( $CH3COOH$ ).

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Answer 5

Chapter 3, Problem 113QRT

(a)

Interpretation Introduction

Interpretation:

The net ionic equation for $3C2H5OH(aq)+2K2Cr2O7(aq)+8H2SO4(aq)→3CH3COOH(aq)+2Cr2(SO4)3(aq)+2K2SO4(aq)+11H2O(l)$ for the given reaction has to be written.

Concept introduction:

Net ionic Equation:

The net ionic equation shows only those chemical species that actually undergo chemical change. The ions that do not change during the chemical reaction are called spectator ions. These spectator ions are not shown in net ionic equation.

(a)

Expert Solution

Explanation of Solution

The balanced equation for the reaction is given below,

$3 C2H5OH(aq) + 2 K2Cr2O7(aq) + 8 H2SO4(aq) ↓3CH3COOH(aq) + 2 Cr2(SO4)3(aq) + 2 K2SO4(aq) + 11 H2O(l)$

While balancing the equation, the subscripts cannot be altered but coefficients can be changed. The given reaction is already balanced.

The complete ionic equation is given below,

$3 C2H5OH(aq) + 4 K+ + 2 Cr2O72−+ 8 H++ 8HSO4− ↓3 CH3COOH (aq) + 4 Cr3+ + 6 SO42−(aq) + 4 K+ + 2 SO42−(aq) + 11 H2O(l)$

Eliminate the spectator ions. Net ionic equation of the given reaction shown below

$3 C2H5OH(aq) + 2 Cr2O72−+ 8 H++ 8HSO4− ↓3 CH3COOH (aq) + 4 Cr3+ + 6 SO42−(aq) +2 SO42−(aq) + 11 H2O(l)$

The balanced complete net ionic equation is written.

(b)

Interpretation Introduction

Interpretation:

The changing the oxidation numbers in the reaction has to be explained.

Concept introduction:

Oxidation number:

The oxidation number of an element is zero. The oxidation number of a monoatomic ion equals its charge.

The oxidation number is zero for the summation of the oxidation numbers complete atoms in a complete formula.

The charge on the ion is equal to the summation of the oxidation numbers of complete atoms in poly atomic ion.

The oxidation state of alkali metal $(Li, Na, K, Rb, Cs)$ is $+1$ , and alkaline earth metal is $(Be, Mg, Ca, Sr, Ba)$ is $+2$ .

The oxidation state of hydrogen is $+1$ (without bonding with metals), the oxidation state of oxygen is $−2$ .

An oxidizing agent gains the electrons and is reduced in a chemical reaction and it is electron acceptor.

A reducing agent loses electrons and is oxidized in a chemical reaction and it is electron donor.

The oxidation state is called as oxidation number, which describes degree of oxidation (loss of electrons) of an atom in a chemical compound. Theoretically, the oxidation state is positive, negative or zero.

(b)

Expert Solution

Explanation of Solution

The given equation is shown below,

$3 C2H5OH(aq) + 4 K+ + 2 Cr2O72−+ 8 H++ 8HSO4− ↓3 CH3COOH (aq) + 4 Cr3+ + 6 SO42−(aq) + 4 K+ + 2 SO42−(aq) + 11 H2O(l)$

Oxidation number of element in the reactant:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$C2H5OH:2(x) + 6(1) +1(-2)= 02x + 6 - 2 = 0x = −2$

The oxidation number of carbon is $+ 2$ .

Oxidation number of chromium in $Cr2O72−$ is given below,

$Cr2O72−:2(x) + 7(-2) = −22x - 14= −2x = + 6$

The oxidation number of chromium is $+ 6$ .

Oxidation number of sulfur in $HSO4−$ is given below,

$HSO4−:1+ (x) + 4(-2) = −1x - 7= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation number of element in the product:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$CH3COOH:2(x) + 4(1) +2(-2)= 02x + 4 - 4 = 0x = 0$

Oxidation number of carbon is zero.

Oxidation number of sulfur in $HSO4−$ is given below,

$SO42−:(x) + 4(-2) = −2x - 8= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation state of hydrogen is $+ 1$ and oxygen is $−2$ in water.

Oxidation number of carbon is $+ 2$ in the reactant and zero in the product, the oxidation number of chromium is $+ 6$ in the reactant and $+ 3$ in the product. Carbon containing compounds loss the electron therefore it is reducing agent ( $C2H5OH$ ). Chromium containing compound gains the electron therefore it is oxidizing agent ( $CH3COOH$ ).

(c)

Interpretation Introduction

Interpretation:

The oxidized and reduced substance has to be identified.

Concept introduction:

Refer to part (b)

(c)

Expert Solution

Explanation of Solution

The given equation is shown below,

$3 C2H5OH(aq) + 4 K+ + 2 Cr2O72−+ 8 H++ 8HSO4− ↓3 CH3COOH (aq) + 4 Cr3+ + 6 SO42−(aq) + 4 K+ + 2 SO42−(aq) + 11 H2O(l)$

Oxidation number of element in the reactant:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$C2H5OH:2(x) + 6(1) +1(-2)= 02x + 6 - 2 = 0x = −2$

The oxidation number of carbon is $+ 2$ .

Oxidation number of chromium in $Cr2O72−$ is given below,

$Cr2O72−:2(x) + 7(-2) = −22x - 14= −2x = + 6$

The oxidation number of chromium is $+ 6$ .

Oxidation number of sulfur in $HSO4−$ is given below,

$HSO4−:1+ (x) + 4(-2) = −1x - 7= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation number of element in the product:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$CH3COOH:2(x) + 4(1) +2(-2)= 02x + 4 - 4 = 0x = 0$

Oxidation number of carbon is zero.

Oxidation number of sulfur in $HSO4−$ is given below,

$SO42−:(x) + 4(-2) = −2x - 8= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation state of hydrogen is $+ 1$ and oxygen is $−2$ in water.

Oxidation number of carbon is $+ 2$ in the reactant and zero in the product, the oxidation number of chromium is $+ 6$ in the reactant and $+ 3$ in the product. Carbon containing compounds loss the electron therefore it has being oxidized ( $C2H5OH$ ). Chromium containing compound gains the electron therefore it has being reduced ( $K2Cr2O7$ ).

(d)

Interpretation Introduction

Interpretation:

The oxidizing agent and reducing agent has to be identified.

Concept introduction:

Refer to part (b)

(d)

Expert Solution

Explanation of Solution

The given equation is shown below,

$3 C2H5OH(aq) + 4 K+ + 2 Cr2O72−+ 8 H++ 8HSO4− ↓3 CH3COOH (aq) + 4 Cr3+ + 6 SO42−(aq) + 4 K+ + 2 SO42−(aq) + 11 H2O(l)$

Oxidation number of element in the reactant:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$C2H5OH:2(x) + 6(1) +1(-2)= 02x + 6 - 2 = 0x = −2$

The oxidation number of carbon is $+ 2$ .

Oxidation number of chromium in $Cr2O72−$ is given below,

$Cr2O72−:2(x) + 7(-2) = −22x - 14= −2x = + 6$

The oxidation number of chromium is $+ 6$ .

Oxidation number of sulfur in $HSO4−$ is given below,

$HSO4−:1+ (x) + 4(-2) = −1x - 7= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation number of element in the product:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$CH3COOH:2(x) + 4(1) +2(-2)= 02x + 4 - 4 = 0x = 0$

Oxidation number of carbon is zero.

Oxidation number of sulfur in $HSO4−$ is given below,

$SO42−:(x) + 4(-2) = −2x - 8= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation state of hydrogen is $+ 1$ and oxygen is $−2$ in water.

Oxidation number of carbon is $+ 2$ in the reactant and zero in the product, the oxidation number of chromium is $+ 6$ in the reactant and $+ 3$ in the product. Carbon containing compounds loss the electron therefore it is reducing agent ( $C2H5OH$ ). Chromium containing compound gains the electron therefore it is oxidizing agent ( $CH3COOH$ ).

Answer 6

Chapter 3, Problem 113QRT

(a)

Interpretation Introduction

Interpretation:

The net ionic equation for $3C2H5OH(aq)+2K2Cr2O7(aq)+8H2SO4(aq)→3CH3COOH(aq)+2Cr2(SO4)3(aq)+2K2SO4(aq)+11H2O(l)$ for the given reaction has to be written.

Concept introduction:

Net ionic Equation:

The net ionic equation shows only those chemical species that actually undergo chemical change. The ions that do not change during the chemical reaction are called spectator ions. These spectator ions are not shown in net ionic equation.

(a)

Expert Solution

Explanation of Solution

The balanced equation for the reaction is given below,

$3 C2H5OH(aq) + 2 K2Cr2O7(aq) + 8 H2SO4(aq) ↓3CH3COOH(aq) + 2 Cr2(SO4)3(aq) + 2 K2SO4(aq) + 11 H2O(l)$

While balancing the equation, the subscripts cannot be altered but coefficients can be changed. The given reaction is already balanced.

The complete ionic equation is given below,

$3 C2H5OH(aq) + 4 K+ + 2 Cr2O72−+ 8 H++ 8HSO4− ↓3 CH3COOH (aq) + 4 Cr3+ + 6 SO42−(aq) + 4 K+ + 2 SO42−(aq) + 11 H2O(l)$

Eliminate the spectator ions. Net ionic equation of the given reaction shown below

$3 C2H5OH(aq) + 2 Cr2O72−+ 8 H++ 8HSO4− ↓3 CH3COOH (aq) + 4 Cr3+ + 6 SO42−(aq) +2 SO42−(aq) + 11 H2O(l)$

The balanced complete net ionic equation is written.

Answer 7

Chapter 3, Problem 113QRT

(a)

Interpretation Introduction

Interpretation:

The net ionic equation for $3C2H5OH(aq)+2K2Cr2O7(aq)+8H2SO4(aq)→3CH3COOH(aq)+2Cr2(SO4)3(aq)+2K2SO4(aq)+11H2O(l)$ for the given reaction has to be written.

Concept introduction:

Net ionic Equation:

The net ionic equation shows only those chemical species that actually undergo chemical change. The ions that do not change during the chemical reaction are called spectator ions. These spectator ions are not shown in net ionic equation.

Answer 8

(a)

Expert Solution

Explanation of Solution

The balanced equation for the reaction is given below,

$3 C2H5OH(aq) + 2 K2Cr2O7(aq) + 8 H2SO4(aq) ↓3CH3COOH(aq) + 2 Cr2(SO4)3(aq) + 2 K2SO4(aq) + 11 H2O(l)$

While balancing the equation, the subscripts cannot be altered but coefficients can be changed. The given reaction is already balanced.

The complete ionic equation is given below,

$3 C2H5OH(aq) + 4 K+ + 2 Cr2O72−+ 8 H++ 8HSO4− ↓3 CH3COOH (aq) + 4 Cr3+ + 6 SO42−(aq) + 4 K+ + 2 SO42−(aq) + 11 H2O(l)$

Eliminate the spectator ions. Net ionic equation of the given reaction shown below

$3 C2H5OH(aq) + 2 Cr2O72−+ 8 H++ 8HSO4− ↓3 CH3COOH (aq) + 4 Cr3+ + 6 SO42−(aq) +2 SO42−(aq) + 11 H2O(l)$

The balanced complete net ionic equation is written.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

Interpretation:

The changing the oxidation numbers in the reaction has to be explained.

Concept introduction:

Oxidation number:

The oxidation number of an element is zero. The oxidation number of a monoatomic ion equals its charge.

The oxidation number is zero for the summation of the oxidation numbers complete atoms in a complete formula.

The charge on the ion is equal to the summation of the oxidation numbers of complete atoms in poly atomic ion.

The oxidation state of alkali metal $(Li, Na, K, Rb, Cs)$ is $+1$ , and alkaline earth metal is $(Be, Mg, Ca, Sr, Ba)$ is $+2$ .

The oxidation state of hydrogen is $+1$ (without bonding with metals), the oxidation state of oxygen is $−2$ .

An oxidizing agent gains the electrons and is reduced in a chemical reaction and it is electron acceptor.

A reducing agent loses electrons and is oxidized in a chemical reaction and it is electron donor.

The oxidation state is called as oxidation number, which describes degree of oxidation (loss of electrons) of an atom in a chemical compound. Theoretically, the oxidation state is positive, negative or zero.

(b)

Expert Solution

Explanation of Solution

The given equation is shown below,

$3 C2H5OH(aq) + 4 K+ + 2 Cr2O72−+ 8 H++ 8HSO4− ↓3 CH3COOH (aq) + 4 Cr3+ + 6 SO42−(aq) + 4 K+ + 2 SO42−(aq) + 11 H2O(l)$

Oxidation number of element in the reactant:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$C2H5OH:2(x) + 6(1) +1(-2)= 02x + 6 - 2 = 0x = −2$

The oxidation number of carbon is $+ 2$ .

Oxidation number of chromium in $Cr2O72−$ is given below,

$Cr2O72−:2(x) + 7(-2) = −22x - 14= −2x = + 6$

The oxidation number of chromium is $+ 6$ .

Oxidation number of sulfur in $HSO4−$ is given below,

$HSO4−:1+ (x) + 4(-2) = −1x - 7= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation number of element in the product:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$CH3COOH:2(x) + 4(1) +2(-2)= 02x + 4 - 4 = 0x = 0$

Oxidation number of carbon is zero.

Oxidation number of sulfur in $HSO4−$ is given below,

$SO42−:(x) + 4(-2) = −2x - 8= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation state of hydrogen is $+ 1$ and oxygen is $−2$ in water.

Oxidation number of carbon is $+ 2$ in the reactant and zero in the product, the oxidation number of chromium is $+ 6$ in the reactant and $+ 3$ in the product. Carbon containing compounds loss the electron therefore it is reducing agent ( $C2H5OH$ ). Chromium containing compound gains the electron therefore it is oxidizing agent ( $CH3COOH$ ).

Answer 13

(b)

Interpretation Introduction

Interpretation:

The changing the oxidation numbers in the reaction has to be explained.

Concept introduction:

Oxidation number:

The oxidation number of an element is zero. The oxidation number of a monoatomic ion equals its charge.

The oxidation number is zero for the summation of the oxidation numbers complete atoms in a complete formula.

The charge on the ion is equal to the summation of the oxidation numbers of complete atoms in poly atomic ion.

The oxidation state of alkali metal $(Li, Na, K, Rb, Cs)$ is $+1$ , and alkaline earth metal is $(Be, Mg, Ca, Sr, Ba)$ is $+2$ .

The oxidation state of hydrogen is $+1$ (without bonding with metals), the oxidation state of oxygen is $−2$ .

An oxidizing agent gains the electrons and is reduced in a chemical reaction and it is electron acceptor.

A reducing agent loses electrons and is oxidized in a chemical reaction and it is electron donor.

The oxidation state is called as oxidation number, which describes degree of oxidation (loss of electrons) of an atom in a chemical compound. Theoretically, the oxidation state is positive, negative or zero.

Answer 14

(b)

Expert Solution

Explanation of Solution

The given equation is shown below,

$3 C2H5OH(aq) + 4 K+ + 2 Cr2O72−+ 8 H++ 8HSO4− ↓3 CH3COOH (aq) + 4 Cr3+ + 6 SO42−(aq) + 4 K+ + 2 SO42−(aq) + 11 H2O(l)$

Oxidation number of element in the reactant:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$C2H5OH:2(x) + 6(1) +1(-2)= 02x + 6 - 2 = 0x = −2$

The oxidation number of carbon is $+ 2$ .

Oxidation number of chromium in $Cr2O72−$ is given below,

$Cr2O72−:2(x) + 7(-2) = −22x - 14= −2x = + 6$

The oxidation number of chromium is $+ 6$ .

Oxidation number of sulfur in $HSO4−$ is given below,

$HSO4−:1+ (x) + 4(-2) = −1x - 7= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation number of element in the product:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$CH3COOH:2(x) + 4(1) +2(-2)= 02x + 4 - 4 = 0x = 0$

Oxidation number of carbon is zero.

Oxidation number of sulfur in $HSO4−$ is given below,

$SO42−:(x) + 4(-2) = −2x - 8= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation state of hydrogen is $+ 1$ and oxygen is $−2$ in water.

Oxidation number of carbon is $+ 2$ in the reactant and zero in the product, the oxidation number of chromium is $+ 6$ in the reactant and $+ 3$ in the product. Carbon containing compounds loss the electron therefore it is reducing agent ( $C2H5OH$ ). Chromium containing compound gains the electron therefore it is oxidizing agent ( $CH3COOH$ ).

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

Interpretation Introduction

Interpretation:

The oxidized and reduced substance has to be identified.

Concept introduction:

Refer to part (b)

(c)

Expert Solution

Explanation of Solution

The given equation is shown below,

$3 C2H5OH(aq) + 4 K+ + 2 Cr2O72−+ 8 H++ 8HSO4− ↓3 CH3COOH (aq) + 4 Cr3+ + 6 SO42−(aq) + 4 K+ + 2 SO42−(aq) + 11 H2O(l)$

Oxidation number of element in the reactant:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$C2H5OH:2(x) + 6(1) +1(-2)= 02x + 6 - 2 = 0x = −2$

The oxidation number of carbon is $+ 2$ .

Oxidation number of chromium in $Cr2O72−$ is given below,

$Cr2O72−:2(x) + 7(-2) = −22x - 14= −2x = + 6$

The oxidation number of chromium is $+ 6$ .

Oxidation number of sulfur in $HSO4−$ is given below,

$HSO4−:1+ (x) + 4(-2) = −1x - 7= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation number of element in the product:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$CH3COOH:2(x) + 4(1) +2(-2)= 02x + 4 - 4 = 0x = 0$

Oxidation number of carbon is zero.

Oxidation number of sulfur in $HSO4−$ is given below,

$SO42−:(x) + 4(-2) = −2x - 8= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation state of hydrogen is $+ 1$ and oxygen is $−2$ in water.

Oxidation number of carbon is $+ 2$ in the reactant and zero in the product, the oxidation number of chromium is $+ 6$ in the reactant and $+ 3$ in the product. Carbon containing compounds loss the electron therefore it has being oxidized ( $C2H5OH$ ). Chromium containing compound gains the electron therefore it has being reduced ( $K2Cr2O7$ ).

Answer 19

(c)

Interpretation Introduction

Interpretation:

The oxidized and reduced substance has to be identified.

Concept introduction:

Refer to part (b)

Answer 20

(c)

Expert Solution

Explanation of Solution

The given equation is shown below,

$3 C2H5OH(aq) + 4 K+ + 2 Cr2O72−+ 8 H++ 8HSO4− ↓3 CH3COOH (aq) + 4 Cr3+ + 6 SO42−(aq) + 4 K+ + 2 SO42−(aq) + 11 H2O(l)$

Oxidation number of element in the reactant:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$C2H5OH:2(x) + 6(1) +1(-2)= 02x + 6 - 2 = 0x = −2$

The oxidation number of carbon is $+ 2$ .

Oxidation number of chromium in $Cr2O72−$ is given below,

$Cr2O72−:2(x) + 7(-2) = −22x - 14= −2x = + 6$

The oxidation number of chromium is $+ 6$ .

Oxidation number of sulfur in $HSO4−$ is given below,

$HSO4−:1+ (x) + 4(-2) = −1x - 7= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation number of element in the product:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$CH3COOH:2(x) + 4(1) +2(-2)= 02x + 4 - 4 = 0x = 0$

Oxidation number of carbon is zero.

Oxidation number of sulfur in $HSO4−$ is given below,

$SO42−:(x) + 4(-2) = −2x - 8= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation state of hydrogen is $+ 1$ and oxygen is $−2$ in water.

Oxidation number of carbon is $+ 2$ in the reactant and zero in the product, the oxidation number of chromium is $+ 6$ in the reactant and $+ 3$ in the product. Carbon containing compounds loss the electron therefore it has being oxidized ( $C2H5OH$ ). Chromium containing compound gains the electron therefore it has being reduced ( $K2Cr2O7$ ).

Answer 21

(c)

Expert Solution

Answer 22

(c)

Expert Solution

Answer 23

Expert Solution

Answer 24

(d)

Interpretation Introduction

Interpretation:

The oxidizing agent and reducing agent has to be identified.

Concept introduction:

Refer to part (b)

(d)

Expert Solution

Explanation of Solution

The given equation is shown below,

$3 C2H5OH(aq) + 4 K+ + 2 Cr2O72−+ 8 H++ 8HSO4− ↓3 CH3COOH (aq) + 4 Cr3+ + 6 SO42−(aq) + 4 K+ + 2 SO42−(aq) + 11 H2O(l)$

Oxidation number of element in the reactant:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$C2H5OH:2(x) + 6(1) +1(-2)= 02x + 6 - 2 = 0x = −2$

The oxidation number of carbon is $+ 2$ .

Oxidation number of chromium in $Cr2O72−$ is given below,

$Cr2O72−:2(x) + 7(-2) = −22x - 14= −2x = + 6$

The oxidation number of chromium is $+ 6$ .

Oxidation number of sulfur in $HSO4−$ is given below,

$HSO4−:1+ (x) + 4(-2) = −1x - 7= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation number of element in the product:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$CH3COOH:2(x) + 4(1) +2(-2)= 02x + 4 - 4 = 0x = 0$

Oxidation number of carbon is zero.

Oxidation number of sulfur in $HSO4−$ is given below,

$SO42−:(x) + 4(-2) = −2x - 8= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation state of hydrogen is $+ 1$ and oxygen is $−2$ in water.

Oxidation number of carbon is $+ 2$ in the reactant and zero in the product, the oxidation number of chromium is $+ 6$ in the reactant and $+ 3$ in the product. Carbon containing compounds loss the electron therefore it is reducing agent ( $C2H5OH$ ). Chromium containing compound gains the electron therefore it is oxidizing agent ( $CH3COOH$ ).

Answer 25

(d)

Interpretation Introduction

Interpretation:

The oxidizing agent and reducing agent has to be identified.

Concept introduction:

Refer to part (b)

Answer 26

(d)

Expert Solution

Explanation of Solution

The given equation is shown below,

$3 C2H5OH(aq) + 4 K+ + 2 Cr2O72−+ 8 H++ 8HSO4− ↓3 CH3COOH (aq) + 4 Cr3+ + 6 SO42−(aq) + 4 K+ + 2 SO42−(aq) + 11 H2O(l)$

Oxidation number of element in the reactant:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$C2H5OH:2(x) + 6(1) +1(-2)= 02x + 6 - 2 = 0x = −2$

The oxidation number of carbon is $+ 2$ .

Oxidation number of chromium in $Cr2O72−$ is given below,

$Cr2O72−:2(x) + 7(-2) = −22x - 14= −2x = + 6$

The oxidation number of chromium is $+ 6$ .

Oxidation number of sulfur in $HSO4−$ is given below,

$HSO4−:1+ (x) + 4(-2) = −1x - 7= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation number of element in the product:

Oxidation number of potassium is $+ 1$ in the reactant, the oxidation number of carbon in ethanol is given below,

$CH3COOH:2(x) + 4(1) +2(-2)= 02x + 4 - 4 = 0x = 0$

Oxidation number of carbon is zero.

Oxidation number of sulfur in $HSO4−$ is given below,

$SO42−:(x) + 4(-2) = −2x - 8= −1x = + 6$

The oxidation number of sulfur is $+ 6$ .

Oxidation state of hydrogen is $+ 1$ and oxygen is $−2$ in water.

Oxidation number of carbon is $+ 2$ in the reactant and zero in the product, the oxidation number of chromium is $+ 6$ in the reactant and $+ 3$ in the product. Carbon containing compounds loss the electron therefore it is reducing agent ( $C2H5OH$ ). Chromium containing compound gains the electron therefore it is oxidizing agent ( $CH3COOH$ ).