Chapter 14, Problem 38E

Sum each of the following oxidation and reduction reactions to obtain the overall reaction:

$\begin{array}{l}\text{a}\text{. Ni}\to {\text{Ni}}^{\text{2+}}\text{+2}{e}^{\text{-}}\text{(oxidation);}\\ {\text{2Ag}}^{\text{+}}\text{+2}{e}^{\text{-}}\to \text{Ag(reduction)}\\ \text{b}{\text{. C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH+4OH}}^{\text{-}}\to {\text{CH}}_{\text{3}}{\text{COOH+3H}}_{\text{2}}\text{O+4}{e}^{\text{-}}\text{(oxidation);}\\ {\text{O}}_{\text{2}}{\text{+2H}}_{\text{2}}\text{O+4}{e}^{\text{-}}\to {\text{4OH}}^{\text{-}}\text{(reduction)}\end{array}$

Interpretation Introduction

Interpretation:

The overall cell reaction for the given half-cell is to be determined.

Concept Introduction:

Oxidation is the addition of an electronegative element or the removal of an electropositive element in a chemical reaction.

Reduction is the addition of an electropositive element or the removal of an electronegative element in a chemical reaction.

The chemical reaction in which both oxidation and reduction takes place simultaneously is called a redox reaction.

Electrolysis is the process in which a redox reaction takes place at the electrodes. The electrode that is charged positively is called the anode. The electrode that is charged negatively is called the cathode. During electrolysis, reduction takes place at the cathode and oxidation takes place at the anode.

Answer to Problem 38E

Solution: a)

$2{\text{Ag}}^{+}+\text{Ni}\to 2\text{Ag}\text{+}{\text{Ni}}^{2+}$ b)

${\text{C}}_2{\text{H}}_5\text{OH}+{\text{O}}_2\to {\text{CH}}_{\text{3}}\text{COOH}+{\text{H}}_{\text{2}}\text{O}$

Explanation of Solution

a)

$\begin{array}{l}\text{Ni}\to {\text{Ni}}^{2+}+2e^{-}(\text{oxidation})\\ \text{and}\\ 2{\text{Ag}}^{+}+2e^{-}\to 2\text{Ag (reduction)}\end{array}$.

$\text{Ag}$ half-cell reaction: This half-cell reaction will occur as reduction at the cathode.

$2{\text{Ag}}^{+}+2e^{-}\to 2\text{Ag}$.

$\text{Ni}$ half-cell reaction: This half-cell reaction will occur as oxidation at the anode.

$\text{Ni}\to {\text{Ni}}^{2+}+2e^{-}$

Adding the two half reactions gives the overall cell reaction as follows:

$\begin{array}{l}2{\text{Ag}}^{+}+\text{ 2}{e}^{-}\to 2\text{Ag}\\ \text{Ni}\to {\text{Ni}}^{2+}+\text{ 2}{e}^{-}\\ \overline{2{\text{Ag}}^{+}+\text{Ni}\to {\text{Ni}}^{2+}\text{+}2\text{Ag}}\end{array}$

Therefore, the overall cell reaction is as follows:

$2{\text{Ag}}^{+}+\text{Ni}\to 2\text{Ag}\text{+}{\text{Ni}}^{2+}$

b)

$\begin{array}{l}{\text{C}}_2{\text{H}}_5\text{OH}\text{+}4{\text{OH}}^{-}\to {\text{CH}}_{\text{3}}\text{COOH}+3{\text{H}}_{\text{2}}\text{O}\text{+}4e^{-}(\text{oxidation})\\ \text{and}\\ {\text{O}}_2+2{\text{H}}_2\text{O}+4e^{-}\to 4{\text{OH}}^{-}\text{ (reduction)}\end{array}$.

${\text{H}}_2\text{O}$ half-cell reaction: This half-cell reaction will occur as reduction at the cathode.

$2{\text{H}}_2\text{O}+4e^{-}+{\text{O}}_2\to 4{\text{OH}}^{-}$

${\text{C}}_2{\text{H}}_5\text{OH}$ half-cell reaction: This half-cell reaction will occur as oxidation at the anode.

${\text{C}}_2{\text{H}}_5\text{OH}\text{+}4{\text{OH}}^{-}\to {\text{CH}}_{\text{3}}\text{COOH}+3{\text{H}}_{\text{2}}\text{O}\text{+}4e^{-}$

Adding the two half-cell reactions gives the overall cell reaction as follows:

$\begin{array}{l}{\text{2H}}_2\text{O}+\text{ 4}{e}^{-}+{\text{O}}_2\to 4{\text{OH}}^{-}\\ {\text{C}}_2{\text{H}}_5\text{OH}\text{+}4{\text{OH}}^{-}\to {\text{CH}}_{\text{3}}\text{COOH}+4e^{-}+3{\text{H}}_2\text{O}\\ \overline{{\text{C}}_2{\text{H}}_5\text{OH}+{\text{O}}_2\to {\text{CH}}_{\text{3}}\text{COOH}+{\text{H}}_2\text{O}}\end{array}$

Therefore, the overall cell reaction is as follows:

${\text{C}}_2{\text{H}}_5\text{OH}+{\text{O}}_2\to {\text{CH}}_{\text{3}}\text{COOH}+{\text{H}}_{\text{2}}\text{O}$