Chapter 4, Problem 63QAP

Consider the following balanced redox reaction in basic medium.

${\text{3Sn}}^{\text{2}+}\left(aq\right)+{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2}-}\left(aq\right)+{\text{4H}}_{\text{2}}\text{O}\to {\text{3Sn}}^{\text{4}+}\left(aq\right)+{\text{Cr}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)+{\text{8OH}}^{-}\left(aq\right)$

(a) What is the oxidizing agent?

(b) What species has the element that increases its oxidation number?

(c) What species contains the element with the highest oxidation number?

(d) If the reaction were to take place in acidic medium, what species would not be included in the reaction?

Interpretation Introduction

(a)

Interpretation:

The oxidizing agent in the following reaction should be determined.

${\text{3Sn}}^{2+}\text{(aq)}+{\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}({\text{aq)+4H}}_{\text{2}}\text{O}\to 3{\text{Sn}}^{4+}{\text{(aq)+Cr}}_2{\text{O}}_3{\text{(s)+8OH}}^{-}(\text{aq})$

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 half reaction is a component of redox reaction which is either oxidation half reaction or reduction half reaction.

Oxidation is the process in which either loss of electrons, oxidation number increases, or loss of hydrogen atoms takes place. An element is oxidized, when oxidation number increases.

Reduction is the process in which either gain of electrons, oxidation number decreases, or gain of hydrogen atoms takes place. An element is reduced, when oxidation number decreases.

Answer to Problem 63QAP

. ${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}$ is an oxidizing agent.

Explanation of Solution

The given reaction is:

${\text{3Sn}}^{2+}\text{(aq)}+{\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}({\text{aq)+4H}}_{\text{2}}\text{O}\to 3{\text{Sn}}^{4+}{\text{(aq)+Cr}}_2{\text{O}}_3{\text{(s)+8OH}}^{-}(\text{aq})$

Write the oxidation states of each element:

$\text{3}\stackrel{+2}{{\text{Sn}}^{2+}}\text{(aq)}+\stackrel{+6}{{\text{Cr}}_{\text{2}}}\stackrel{-2}{{\text{O}}_7^{2-}}(\text{aq)+4}\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}\to 3\stackrel{+4}{{\text{Sn}}^{4+}}\text{(aq)+}\stackrel{+3}{{\text{Cr}}_2}\stackrel{-2}{{\text{O}}_3}\text{(s)+8}\stackrel{-2}{\text{O}}{\stackrel{+1}{\text{H}}}_{{}^{{}^{-}}}(\text{aq})$

Here, oxidation state of chromium decreases from +6 to +3. Thus, chromium is an oxidizing agent as it is reduced or chromium accepts electrons during the reaction. The species which contains chromium is ${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}$.

Interpretation Introduction

(b)

Interpretation:

The species in the following reaction should be determined which has the element with increasing oxidation number.

${\text{3Sn}}^{2+}\text{(aq)}+{\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}({\text{aq)+4H}}_{\text{2}}\text{O}\to 3{\text{Sn}}^{4+}{\text{(aq)+Cr}}_2{\text{O}}_3{\text{(s)+8OH}}^{-}(\text{aq})$

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 half reaction is a component of redox reaction which is either oxidation half reaction or reduction half reaction.

Oxidation is the process in which either loss of electrons, oxidation number increases, or loss of hydrogen atoms takes place. An element is oxidized, when oxidation number increases.

Reduction is the process in which either gain of electrons, oxidation number decreases, or gain of hydrogen atoms takes place. An element is reduced, when oxidation number decreases.

Answer to Problem 63QAP

The species which contain element that increases oxidation number is ${\text{Sn}}^{2+}$.

Explanation of Solution

The given reaction is:

${\text{3Sn}}^{2+}\text{(aq)}+{\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}({\text{aq)+4H}}_{\text{2}}\text{O}\to 3{\text{Sn}}^{4+}{\text{(aq)+Cr}}_2{\text{O}}_3{\text{(s)+8OH}}^{-}(\text{aq})$

Write the oxidation states of each element:

$\text{3}\stackrel{+2}{{\text{Sn}}^{2+}}\text{(aq)}+\stackrel{+6}{{\text{Cr}}_{\text{2}}}\stackrel{-2}{{\text{O}}_7^{2-}}(\text{aq)+4}\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}\to 3\stackrel{+4}{{\text{Sn}}^{4+}}\text{(aq)+}\stackrel{+3}{{\text{Cr}}_2}\stackrel{-2}{{\text{O}}_3}\text{(s)+8}\stackrel{-2}{\text{O}}{\stackrel{+1}{\text{H}}}_{{}^{{}^{-}}}(\text{aq})$

Here, oxidation state of tin increases from +2 to +4. Thus, tin is a reducing agent as it is oxidized or tin loss electrons during the reaction. The species which contains tin is ${\text{Sn}}^{2+}$.

Interpretation Introduction

(c)

Interpretation:

The species in the following reaction should be determined which has the element with highest oxidation number.

${\text{3Sn}}^{2+}\text{(aq)}+{\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}({\text{aq)+4H}}_{\text{2}}\text{O}\to 3{\text{Sn}}^{4+}{\text{(aq)+Cr}}_2{\text{O}}_3{\text{(s)+8OH}}^{-}(\text{aq})$

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 half reaction is a component of redox reaction which is either oxidation half reaction or reduction half reaction.

Oxidation is the process in which either loss of electrons, oxidation number increases, or loss of hydrogen atoms takes place. An element is oxidized, when oxidation number increases.

Reduction is the process in which either gain of electrons, oxidation number decreases, or gain of hydrogen atoms takes place. An element is reduced, when oxidation number decreases.

Answer to Problem 63QAP

The species which contains chromium having highest oxidation state or number is ${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}$.

Explanation of Solution

The given reaction is:

${\text{3Sn}}^{2+}\text{(aq)}+{\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}({\text{aq)+4H}}_{\text{2}}\text{O}\to 3{\text{Sn}}^{4+}{\text{(aq)+Cr}}_2{\text{O}}_3{\text{(s)+8OH}}^{-}(\text{aq})$

Write the oxidation states of each element:

$\text{3}\stackrel{+2}{{\text{Sn}}^{2+}}\text{(aq)}+\stackrel{+6}{{\text{Cr}}_{\text{2}}}\stackrel{-2}{{\text{O}}_7^{2-}}(\text{aq)+4}\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}\to 3\stackrel{+4}{{\text{Sn}}^{4+}}\text{(aq)+}\stackrel{+3}{{\text{Cr}}_2}\stackrel{-2}{{\text{O}}_3}\text{(s)+8}\stackrel{-2}{\text{O}}{\stackrel{+1}{\text{H}}}_{{}^{{}^{-}}}(\text{aq})$

Here, oxidation state or number of chromium is highest that is +6 in the given redox reaction. Thus, the species which contains chromium having highest oxidation state or number is ${\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}$.

Interpretation Introduction

(d)

Interpretation:

The species which would not include in the reaction should be determined when the following reaction occurs in acidic medium.

${\text{3Sn}}^{2+}\text{(aq)}+{\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}({\text{aq)+4H}}_{\text{2}}\text{O}\to 3{\text{Sn}}^{4+}{\text{(aq)+Cr}}_2{\text{O}}_3{\text{(s)+8OH}}^{-}(\text{aq})$

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 half reaction is a component of redox reaction which is either oxidation half reaction or reduction half reaction.

Oxidation is the process in which either loss of electrons, oxidation number increases, or loss of hydrogen atoms takes place. An element is oxidized, when oxidation number increases.

Reduction is the process in which either gain of electrons, oxidation number decreases, or gain of hydrogen atoms takes place. An element is reduced, when oxidation number decreases.

Answer to Problem 63QAP

Hydroxide ion is not present in the reaction as the reaction occurs in acidic medium

Explanation of Solution

The given reaction is:

${\text{3Sn}}^{2+}\text{(aq)}+{\text{Cr}}_{\text{2}}{\text{O}}_7^{2-}({\text{aq)+4H}}_{\text{2}}\text{O}\to 3{\text{Sn}}^{4+}{\text{(aq)+Cr}}_2{\text{O}}_3{\text{(s)+8OH}}^{-}(\text{aq})$

Write the oxidation states of each element:

$\text{3}\stackrel{+2}{{\text{Sn}}^{2+}}\text{(aq)}+\stackrel{+6}{{\text{Cr}}_{\text{2}}}\stackrel{-2}{{\text{O}}_7^{2-}}(\text{aq)+4}\stackrel{+1}{{\text{H}}_{\text{2}}}\stackrel{-2}{\text{O}}\to 3\stackrel{+4}{{\text{Sn}}^{4+}}\text{(aq)+}\stackrel{+3}{{\text{Cr}}_2}\stackrel{-2}{{\text{O}}_3}\text{(s)+8}\stackrel{-2}{\text{O}}{\stackrel{+1}{\text{H}}}_{{}^{{}^{-}}}(\text{aq})$

Divide the above redox reaction into two half reactions that is oxidation half reaction and reduction half reaction.

Oxidation Half reaction:

$\stackrel{+2}{{\text{Sn}}^{2+}}\text{(aq)}\to \stackrel{+4}{{\text{Sn}}^{4+}}\text{(aq)}$

Reduction half reaction:

$\stackrel{+6}{{\text{Cr}}_{\text{2}}}\stackrel{-2}{{\text{O}}_7^{2-}}(\text{aq)}\to \stackrel{+3}{{\text{Cr}}_2}\stackrel{-2}{{\text{O}}_3}\text{(s)}$

Balance the elements except oxygen. No adjustment needed

Now, add number of electrons (required) to that side of reaction where reduction occurs.

Oxidation Half reaction:

$\stackrel{+2}{{\text{Sn}}^{2+}}\text{(aq)}\to \stackrel{+4}{{\text{Sn}}^{4+}}{\text{(aq)+2e}}^{-}$

Reduction half reaction:

$\stackrel{+6}{{\text{Cr}}_{\text{2}}}\stackrel{-2}{{\text{O}}_7^{2-}}({\text{aq)+6e}}^{-}\to \stackrel{+3}{{\text{Cr}}_2}\stackrel{-2}{{\text{O}}_3}\text{(s)}$

Balance the charge by adding ${\text{H}}^{\text{+}}$ in both reactions.

Oxidation Half reaction:

$\stackrel{+2}{{\text{Sn}}^{2+}}\text{(aq)}\to \stackrel{+4}{{\text{Sn}}^{4+}}{\text{(aq)+2e}}^{-}$

Reduction half reaction:

$\stackrel{+6}{{\text{Cr}}_{\text{2}}}\stackrel{-2}{{\text{O}}_7^{2-}}({\text{aq)+6e}}^{-}+{\text{8H}}^{\text{+}}\to \stackrel{+3}{{\text{Cr}}_2}\stackrel{-2}{{\text{O}}_3}\text{(s)}$

Balance the hydrogen and oxygen atoms by adding water molecules.

Oxidation Half reaction:

$\stackrel{+2}{{\text{Sn}}^{2+}}\text{(aq)}\to \stackrel{+4}{{\text{Sn}}^{4+}}{\text{(aq)+2e}}^{-}$

Reduction half reaction:

$\stackrel{+6}{{\text{Cr}}_{\text{2}}}\stackrel{-2}{{\text{O}}_7^{2-}}({\text{aq)+6e}}^{-}+{\text{8H}}^{\text{+}}\to \stackrel{+3}{{\text{Cr}}_2}\stackrel{-2}{{\text{O}}_3}{\text{(s)+4H}}_2\text{O}$

Multiply the oxidation half reaction with 3 to balance the number of electrons in both reactions.

Oxidation Half reaction:

$3\stackrel{+2}{{\text{Sn}}^{2+}}\text{(aq)}\to \stackrel{+4}{{\text{3Sn}}^{4+}}{\text{(aq)+6e}}^{-}$

Reduction half reaction:

$\stackrel{+6}{{\text{Cr}}_{\text{2}}}\stackrel{-2}{{\text{O}}_7^{2-}}({\text{aq)+6e}}^{-}+{\text{8H}}^{\text{+}}\to \stackrel{+3}{{\text{Cr}}_2}\stackrel{-2}{{\text{O}}_3}{\text{(s)+4H}}_2\text{O}$

Now, add both reactions.

$3\stackrel{+2}{{\text{Sn}}^{2+}}\text{(aq)}\to \stackrel{+4}{{\text{3Sn}}^{4+}}{\text{(aq)+6e}}^{-}$

$\stackrel{+6}{{\text{Cr}}_{\text{2}}}\stackrel{-2}{{\text{O}}_7^{2-}}({\text{aq)+6e}}^{-}+{\text{8H}}^{\text{+}}\to \stackrel{+3}{{\text{Cr}}_2}\stackrel{-2}{{\text{O}}_3}{\text{(s)+4H}}_2\text{O}$

$\overline{3\stackrel{}{{\text{Sn}}^{2+}}\text{(aq)}+\stackrel{}{{\text{Cr}}_{\text{2}}}\stackrel{}{{\text{O}}_7^{2-}}({\text{aq)+6e}}^{-}+{\text{8H}}^{\text{+}}\to \stackrel{}{{\text{3Sn}}^{4+}}{\text{(aq)+6e}}^{-}+\stackrel{}{{\text{Cr}}_2}\stackrel{}{{\text{O}}_3}{\text{(s)+4H}}_2\text{O}}$

Cancel out the common terms in opposite sides. Thus, balanced chemical equation is given by:

$3\stackrel{}{{\text{Sn}}^{2+}}\text{(aq)}+\stackrel{}{{\text{Cr}}_{\text{2}}}\stackrel{}{{\text{O}}_7^{2-}}(\text{aq)}+{\text{8H}}^{\text{+}}\to \stackrel{}{{\text{3Sn}}^{4+}}\text{(aq)}+\stackrel{}{{\text{Cr}}_2}\stackrel{}{{\text{O}}_3}{\text{(s)+4H}}_2\text{O}$

According to above reaction and the given reaction, hydroxide ion is not present in the reaction as the reaction occurs in acidic medium whereas the given reaction occurs in basic medium.