Chapter 4, Problem 132MP

(a)

Interpretation Introduction

Interpretation:

The average oxidation states of copper in the materials having formulas ${\text{YBa}}_{\text{2}}{\text{Cu}}_{\text{3}}{\text{O}}_{\text{6}\text{.5}}$ , ${\text{YBa}}_{\text{2}}{\text{Cu}}_{\text{3}}{\text{O}}_7$ and ${\text{YBa}}_{\text{2}}{\text{Cu}}_{\text{3}}{\text{O}}_8$ should be calculated. The result should be interpreted in terms of mixture of $\text{Cu(II) and Cu(III)}$ ions by considering that only ${\text{Y}}^{3+},{\text{ Ba}}^{2+}{\text{ and O}}^{2-}$ are present in addition to copper ions.

Concept Introduction:

Oxidation state is also known as oxidation number. It is defined as the numbers which are assign to the elements in a chemical combination and number represents the electrons which an atom can share, lose or gain to form chemical bonding with an atom of another element.

Therefore, transfer of electrons refers to the oxidation state.

Answer to Problem 132MP

In ${\text{YBa}}_{\text{2}}{\text{Cu}}_{\text{3}}{\text{O}}_{\text{6}\text{.5}}$ , $\text{Cu(II)}$ ion is present.

In ${\text{YBa}}_{\text{2}}{\text{Cu}}_{\text{3}}{\text{O}}_7$ , $\text{Cu(II)}$ ion is present.

In ${\text{YBa}}_{\text{2}}{\text{Cu}}_{\text{3}}{\text{O}}_8$ , $\text{Cu(III)}$ ion is present.

Explanation of Solution

The given material is ${\text{YBa}}_{\text{2}}{\text{Cu}}_{\text{3}}{\text{O}}_{\text{6}\text{.5}}$ .

According to the rules of oxidation states, oxygen is assigned as oxidation state of -2, yttrium is assigned as oxidation state of +3 and barium is assigned as oxidation state of +2. Now, oxidation state of copper should be determined, since overall charge on the molecule is zero.

Therefore, let x be the oxidation state of copper.

  $1\times \left(+3\right)+2\times \left(+2\right)+3x+6.5(-2)=0$

  $+3+4+3x-13=0$

  $7+3x-13=0$

  $3x-6=0$

  $x=\frac{6}{3}=+2$

+6 charge is net charge of per unit cell, which is compensated by three Cu atoms with a net charge of $\frac{6}{3}=+2$

The given material is ${\text{YBa}}_{\text{2}}{\text{Cu}}_{\text{3}}{\text{O}}_7$ .

According to the rules of oxidation states, oxygen is assigned as oxidation state of -2, yttrium is assigned as oxidation state of +3 and barium is assigned as oxidation state of +2. Now, oxidation state of copper should be determined, since overall charge on the molecule is zero.

Therefore, let x be the oxidation state of copper.

  $1\times \left(+3\right)+2\times \left(+2\right)+3x+7(-2)=0$

  $+3+4+3x-14=0$

  $7+3x-14=0$

  $3x-7=0$

  $x=\frac{7}{3}=+2.33\simeq 2$

+7 charge is net charge of per unit cell, which is compensated by three Cu atoms with a net charge of $\frac{7}{3}=+2.33\simeq +2$

The given material is ${\text{YBa}}_{\text{2}}{\text{Cu}}_{\text{3}}{\text{O}}_8$ .

According to the rules of oxidation states, oxygen is assigned as oxidation state of -2, yttrium is assigned as oxidation state of +3 and barium is assigned as oxidation state of +2. Now, oxidation state of copper should be determined, since overall charge on the molecule is zero.

Therefore, let x be the oxidation state of copper.

  $1\times \left(+3\right)+2\times \left(+2\right)+3x+8(-2)=0$

  $+3+4+3x-16=0$

  $7+3x-16=0$

  $3x-9=0$

  $x=\frac{9}{3}=+3$

+9 charge is net charge of per unit cell, which is compensated by three Cu atoms with a net charge of $\frac{9}{3}=+3$

Thus, the average oxidation state of copper in material ${\text{YBa}}_{\text{2}}{\text{Cu}}_{\text{3}}{\text{O}}_{\text{6}\text{.5}}$ is +2 that is $\text{Cu(II)}$ .

The average oxidation state of copper in material ${\text{YBa}}_{\text{2}}{\text{Cu}}_{\text{3}}{\text{O}}_7$ is nearly equals to+2 that is $\text{Cu(II)}$ .

The average oxidation state of copper in material ${\text{YBa}}_{\text{2}}{\text{Cu}}_{\text{3}}{\text{O}}_8$ is +3 that is $\text{Cu(III)}$ .

(b)

Interpretation Introduction

Interpretation:

The equations should be balanced which are involved in copper analysis.

Concept Introduction:

The chemical reaction in which both oxidation and reduction process takes place is known as redox reaction. In this reaction, transfer of electrons takes place among the elements.

A reduction or an oxidation reaction is known as half reaction.

Balance all atoms including oxygen and hydrogen atoms are carried out by addition of water molecule (to balance oxygen) and hydrogen ion (to balance hydrogen) in the half reactions. Number of electrons and charge should be balanced after that makes the number of electrons equal in both oxidation and reduction reactions by multiplying with an integer. The last step is to add both half reactions.

Answer to Problem 132MP

The formula of superconductor is ${\text{YBa}}_{\text{2}}{\text{Cu}}_{\text{3}}{\text{O}}_7$ .

Oxidation state of copper is +2.

Explanation of Solution

Given information:

A sample of one superconductor treated directly with ${\text{I}}^{-}$ as shown as:

  ${\text{Cu}}^{\text{2+}}\text{(aq)}+{\text{I}}^{-}\text{(aq)}\to \text{CuI(s)}+{\text{I}}_3^{-}\text{(aq)}$

  ${\text{Cu}}^{\text{3+}}\text{(aq)}+{\text{I}}^{-}\text{(aq})\to \text{CuI(s)}+{\text{I}}_3^{-}\text{(aq)}$

A sample of second superconductor is dissolved in acid which converts all copper to $\text{Cu(II)}$ .

  ${\text{Cu}}^{\text{2+}}\text{(aq)}+{\text{I}}^{-}\text{(aq)}\to \text{CuI(s)}+{\text{I}}_3^{-}\text{(aq)}$

Now, two reactions which are involved in copper analysis is given as:

  ${\text{Cu}}^{\text{2+}}\text{(aq)}+{\text{I}}^{-}\text{(aq)}\to \text{CuI(s)}+{\text{I}}_3^{-}\text{(aq)}$

  ${\text{Cu}}^{\text{3+}}\text{(aq)}+{\text{I}}^{-}\text{(aq})\to \text{CuI(s)}+{\text{I}}_3^{-}\text{(aq)}$

For first reaction:

  ${\text{Cu}}^{\text{2+}}\text{(aq)}+{\text{I}}^{-}\text{(aq)}\to \text{CuI(s)}+{\text{I}}_3^{-}\text{(aq)}$

The above reaction is separated (oxidation-reduction reaction) as:

  ${\text{Cu}}^{\text{2+}}\text{(aq)}\to \text{CuI(s) (Reduction)}$

  ${\text{I}}^{-}(\text{aq)}\to {\text{I}}_3^{-}\text{(aq})\text{ (Oxidation)}$

Balance all the atoms other than hydrogen and oxygen.

  ${\text{Cu}}^{\text{2+}}{\text{(aq)+I}}^{-}\to \text{CuI(s) (Reduction)}$

  ${\text{3I}}^{-}(\text{aq)}\to {\text{I}}_3^{-}\text{(aq})\text{ (Oxidation)}$

Balance oxygen atoms.

  ${\text{Cu}}^{\text{2+}}{\text{(aq)+I}}^{-}\to \text{CuI(s) (Reduction)}$

  ${\text{3I}}^{-}(\text{aq)}\to {\text{I}}_3^{-}\text{(aq})\text{ (Oxidation)}$

Balance hydrogen atoms.

  ${\text{Cu}}^{\text{2+}}{\text{(aq)+I}}^{-}\to \text{CuI(s) (Reduction)}$

  ${\text{3I}}^{-}(\text{aq)}\to {\text{I}}_3^{-}\text{(aq})\text{ (Oxidation)}$

Balance charge and number of electrons.

  ${\text{Cu}}^{\text{2+}}{\text{(aq)+I}}^{-}+e^{-}\to \text{CuI(s) (Reduction)}$ (1)

  ${\text{3I}}^{-}(\text{aq)}\to {\text{I}}_3^{-}\text{(aq})\text{ +2}{e}^{-}\text{ (Oxidation)}$ (2)

Multiply (1) by 2.

  ${\text{2Cu}}^{\text{2+}}{\text{(aq)+2I}}^{-}+2e^{-}\to 2\text{CuI(s) (Reduction)}$

  ${\text{3I}}^{-}(\text{aq)}\to {\text{I}}_3^{-}\text{(aq})\text{ +2}{e}^{-}\text{ (Oxidation)}$

Add both equations:

  ${\text{2Cu}}^{\text{2+}}{\text{(aq)+2I}}^{-}+2e^{-}+{\text{3I}}^{-}(\text{aq)}\to 2{\text{CuI(s)+I}}_3^{-}\text{(aq})\text{ +2}{e}^{-}$

The balanced equation is written as:

  ${\text{2Cu}}^{\text{2+}}{\text{(aq)+5I}}^{-}(\text{aq)}\to 2{\text{CuI(s)+I}}_3^{-}\text{(aq})$

For second reaction:

  ${\text{Cu}}^{\text{3+}}\text{(aq)}+{\text{I}}^{-}\text{(aq})\to \text{CuI(s)}+{\text{I}}_3^{-}\text{(aq)}$

The above reaction is separated (oxidation-reduction reaction) as:

  ${\text{Cu}}^{\text{3+}}\text{(aq)}\to \text{CuI(s) (Reduction)}$

  ${\text{I}}^{-}(\text{aq)}\to {\text{I}}_3^{-}\text{(aq})\text{ (Oxidation)}$

Balance all the atoms other than hydrogen and oxygen.

  ${\text{Cu}}^{\text{3+}}{\text{(aq)+I}}^{-}\to \text{CuI(s) (Reduction)}$

  ${\text{3I}}^{-}(\text{aq)}\to {\text{I}}_3^{-}\text{(aq})\text{ (Oxidation)}$

Balance oxygen atoms.

  ${\text{Cu}}^{\text{3+}}{\text{(aq)+I}}^{-}\to \text{CuI(s) (Reduction)}$

  ${\text{3I}}^{-}(\text{aq)}\to {\text{I}}_3^{-}\text{(aq})\text{ (Oxidation)}$

Balance hydrogen atoms.

  ${\text{Cu}}^{\text{3+}}{\text{(aq)+I}}^{-}\to \text{CuI(s) (Reduction)}$

  ${\text{3I}}^{-}(\text{aq)}\to {\text{I}}_3^{-}\text{(aq})\text{ (Oxidation)}$

Balance charge and number of electrons.

  ${\text{Cu}}^{\text{3+}}{\text{(aq)+I}}^{-}+2e^{-}\to \text{CuI(s) (Reduction)}$ (1)

  ${\text{3I}}^{-}(\text{aq)}\to {\text{I}}_3^{-}\text{(aq})\text{ +2}{e}^{-}\text{ (Oxidation)}$ (2)

Add both equations:

  ${\text{Cu}}^{\text{3+}}{\text{(aq)+I}}^{-}+2e^{-}+{\text{3I}}^{-}(\text{aq)}\to {\text{CuI(s)+I}}_3^{-}\text{(aq})\text{ +2}{e}^{-}$

The balanced equation is written as:

  ${\text{Cu}}^{\text{3+}}{\text{(aq)+4I}}^{-}(\text{aq)}\to {\text{CuI(s)+I}}_3^{-}\text{(aq})$

(c)

Interpretation Introduction

Interpretation:

The formula of the superconductor sample should be determined along with the average oxidation state of copper in this material.

Concept Introduction:

Mole is SI unit which is used to measure the quantity of the substance. It is the quantity of a substance which contains same number of atoms as present in accurately 12.00 g of carbon-12 is known as mole.

Number of moles of a compound is defined as the ratio of given mass of the compound to the molar or molecular mass of the compound.

The mathematical expression is given by:

Number of moles = $\frac{\text{ mass of the compound}}{\text{molar mass of the compound}}$

Molarity is defined as the ratio of number of moles to the volume of solution in L.

The mathematical expression is:

  $\text{Molarity =}\frac{\text{number of moles}}{\text{volume of solution in L}}$

Answer to Problem 132MP

Molarity of ${\text{Na}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{3}}$ = $\text{0}\text{.0468 M}$

Explanation of Solution

Given information:

In step i,

Volume of ${\mathrm{Na}}_2{S}_2{O}_3$ = 37.7 mL

Molarity of ${\mathrm{Na}}_2{S}_2{O}_3$ = 0.1000 M

Mass of sample = 562.5 mg

In step ii,

Volume of ${\mathrm{Na}}_2{S}_2{O}_3$ = 22.57 mL

Molarity of ${\mathrm{Na}}_2{S}_2{O}_3$ = 0.1000 M

Mass of sample = 504.2 mg

The mathematical expression for calculating molarity is:

  $\text{Molarity =}\frac{\text{number of moles}}{\text{volume of solution in L}}$

Rearrange the above formula in terms of number of moles:

  $\text{Number of moles}=\text{Molarity }\times \text{volume of solution in L}$

For step (i)

Convert the given volume in mL to L.

Since, 1L=1000 mL

Thus, volume in L = $37.77\text{ mL}\times \frac{1\text{ L}}{1000\text{ mL}}$

= $0.03777\text{ L}$

Number of moles of ${\mathrm{Na}}_2{S}_2{O}_3$ = $0.1000\text{ mole/L }\times \text{0}\text{.03777 L}$

= $0.003777\text{ mole}$

For step (ii)

Convert the given volume in mL to L.

Since, 1L=1000 mL

Thus, volume in L = $22.57\text{ mL}\times \frac{1\text{ L}}{1000\text{ mL}}$

= $0.02257\text{ L}$

Number of moles of ${\mathrm{Na}}_2{S}_2{O}_3$ = $0.1000\text{ mole/L }\times \text{0}\text{.02257 L}$

= $0.002257\text{ mole}$

Now, the balanced reaction between ${\text{S}}_{\text{2}}{\text{O}}_3^{2-}$ and ${\text{I}}_3^{-}$ is:

  ${\text{I}}_3^{-}({\text{aq)+2S}}_2{\text{O}}_3^{2-}(\text{aq)}\to {\text{S}}_4{\text{O}}_6^{2-}(\text{aq)}+3{\mathrm{I}}^{-}(\text{aq)}$

Now, for step (i)

Number of moles of ${\text{I}}_3^{-}$ = $\frac{0.003777\text{ mole}}{2}=0.0018885\text{ mole}$

Now, for step (ii)

Number of moles of ${\text{I}}_3^{-}$ = $\frac{0.002257\text{ mole}}{2}=0.0011285\text{ mole}$

Balanced equation for step (i):

  ${\text{2Cu}}^{\text{2+}}{\text{(aq)+5I}}^{-}(\text{aq)}\to 2{\text{CuI(s)+I}}_3^{-}\text{(aq})$

  ${\text{Cu}}^{\text{3+}}{\text{(aq)+4I}}^{-}(\text{aq)}\to {\text{CuI(s)+I}}_3^{-}\text{(aq})$

Number of moles of ${\text{Cu}}^{\text{2+}}$ = $2(0.0018885\text{ mole})=0.003777\text{ mole}$

Number of moles of ${\text{Cu}}^{\text{3+}}$ = $0.0018885\text{ mole}$

Total number of moles = $0.003777\text{ mole}+0.001885\text{ mole}$

= $0.005662\text{ mole}$

Since, 1 mole of sample has three moles of copper.

Thus,

Number of moles of copper in sample = $\frac{0.005662\text{ mole}}{3}$

= $0.001888\text{ mole}$

Molar mass of ${\text{YBa}}_{\text{2}}{\text{Cu}}_{\text{3}}{\text{O}}_x$

= $1\times 88.9\text{ g/mole + 2}\times \text{137}\text{.327 g/mole +3}\times \text{63}\text{.546 g/mole + }x\times 16\text{ g/mole}$

= $\text{554}\text{.192 g/mole + 16 }x\text{ g/mole}$

Number of moles = $\frac{\text{mass}}{\text{Molar mass}}$

Put the values,

  $0.001888\text{ mole}$ = $\frac{562.5\text{ mg}\times \frac{1\text{ g}}{1000\text{ mg}}}{\text{554}\text{.192 g/mole + 16 }x\text{ g/mole}}$

Now, for step (ii)

  ${\text{2Cu}}^{\text{2+}}{\text{(aq)+5I}}^{-}(\text{aq)}\to 2{\text{CuI(s)+I}}_3^{-}\text{(aq})$

Number of moles of copper= $0.0011285\text{ mole}\times \text{2= 0}\text{.002257 mole}$

1 mole of sample contains 3 moles of copper.

Thus, number of moles of copper in sample = $\frac{\text{0}\text{.002257 mole}}{3}$

= $0.000752\text{ mole}$

Number of moles = $\frac{\text{mass}}{\text{Molar mass}}$

Mass of sample = 504.2 mg

Put the values,

  $0.000752\text{ mole}=\frac{\text{504}\text{.2 mg}\times \frac{1\text{ g}}{1000\text{ mg}}}{\text{Molar mass}}$

  $\text{Molar mass}=\frac{0.5042\text{ g}}{0.000752\text{ mole}}$

  $\text{Molar mass}=670.48\text{ g/mole}$

Now,

For step (ii)

  ${\text{2Cu}}^{\text{2+}}{\text{(aq)+5I}}^{-}(\text{aq)}\to 2{\text{CuI(s)+I}}_3^{-}\text{(aq})$

  ${\text{I}}_3^{-}({\text{aq)+2S}}_2{\text{O}}_3^{2-}(\text{aq)}\to {\text{S}}_4{\text{O}}_6^{2-}(\text{aq)}+3{\mathrm{I}}^{-}(\text{aq)}$

Add both equations,

  ${\text{2Cu}}^{\text{2+}}{\text{(aq)+5I}}^{-}({\text{aq)+I}}_3^{-}({\text{aq)+2S}}_2{\text{O}}_3^{2-}(\text{aq)}\to 2{\text{CuI(s)+I}}_3^{-}\text{(aq})+{\text{S}}_4{\text{O}}_6^{2-}(\text{aq)}+3{\mathrm{I}}^{-}(\text{aq)}$

The balance combined equation is written as:

  ${\text{2Cu}}^{\text{2+}}{\text{(aq)+2I}}^{-}({\text{aq)+2S}}_2{\text{O}}_3^{2-}(\text{aq)}\to 2\text{CuI(s)}+{\text{S}}_4{\text{O}}_6^{2-}(\text{aq)}$

Now,

Equate value of molar mass from step (ii) and molar mass from step (i)

  $\text{554}\text{.192 g/mole + 16 }x\text{ g/mole = }670.48\text{ g/mole}$

  $\text{16 }x\text{ g/mole = }670.48\text{ g/mole}-\text{554}\text{.192 g/mole}$

  $\text{16 }x\text{ g/mole = 116}\text{.288 g/mole}$

  $x\text{ = }\frac{\text{116}\text{.288 g/mole}}{\text{16 g/mole}}$

  $x\text{ = 7}\text{.23}$

  $x\simeq \text{ 7}$

Thus, the formula of superconductor is ${\text{YBa}}_{\text{2}}{\text{Cu}}_{\text{3}}{\text{O}}_7$ .

Now,

According to the rules of oxidation states, oxygen is assigned as oxidation state of -2, yttrium is assigned as oxidation state of +3 and barium is assigned as oxidation state of +2. Now, oxidation state of copper should be determined, since overall charge on the molecule is zero.

Therefore, let x be the oxidation state of copper.

  $1\times \left(+3\right)+2\times \left(+2\right)+3x+7(-2)=0$

  $+3+4+3x-14=0$

  $7+3x-14=0$

  $3x-7=0$

  $x=\frac{7}{3}=+2.33\simeq 2$

+7 charge is net charge of per unit cell, which is compensated by three Cu atoms with a net charge of $\frac{7}{3}=+2.33\simeq 2$