Question

Consider the following galvanic cells:

For each galvanic cell, give the balanced cell equation and determine. Standard reduction potentials are found in Table 11.1.

Short answer

1. Thebalances cell equation is${\mathrm{Au}}^{3+}+3{\mathrm{Cu}}^{+}\to \mathrm{Au}+3{\mathrm{Cu}}^{2+}$ and the value of ${\mathrm{E}}^0=1.34\mathrm{V}$.
2. The balances cell equation is$\mathrm{Cd}+2{{\mathrm{VO}}_2}^{+}+4{\mathrm{H}}^{+}\to {\mathrm{Cd}}^{+}+2{\mathrm{VO}}^{2+}+2{\mathrm{H}}_2\mathrm{O}$ and thevalue of ${\mathrm{E}}^0=1.40\mathrm{V}$.

Step-by-step solution

Two half reactions

$\begin{array}{l}{\mathrm{Au}}^{3+}+3{\mathrm{e}}^{-}\to \mathrm{Au},{\mathrm{E}}^0=1.50\mathrm{V}\\ {\mathrm{Cu}}^{2+}+{\mathrm{e}}^{-}\to {\mathrm{Cu}}^{+},{\mathrm{E}}^0=0.16\mathrm{V}\end{array}$

To getthe positive value as,

${\mathrm{Cu}}^{+}\to {\mathrm{Cu}}^{2+}+{\mathrm{e}}^{-},{\mathrm{E}}^0=-0.16\mathrm{V}$

Now, the net balanced reaction as,

$\begin{array}{l}{\mathrm{Au}}^{3+}+3{\mathrm{e}}^{-}\to \mathrm{Au},{\mathrm{E}}^0=1.50\mathrm{V}\\ {\mathrm{Cu}}^{+}\to {\mathrm{Cu}}^{2+}+{\mathrm{e}}^{-},{\mathrm{E}}^0=-0.16\mathrm{V}\end{array}$

${\mathrm{Au}}^{3+}+3{\mathrm{Cu}}^{+}\to \mathrm{Au}+3{\mathrm{Cu}}^{2+},{\mathrm{E}}^0=1.50\mathrm{V}-0.16\mathrm{V}=1.34\mathrm{V}$

Two half reactions

$\begin{array}{l}{\mathrm{Cd}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Cd},{\mathrm{E}}^0=-0.40\mathrm{V}\\ {{\mathrm{VO}}_2}^{+}+2{\mathrm{H}}^{+}+{\mathrm{e}}^{-}\to {\mathrm{VO}}^{2+}+{\mathrm{H}}_2\mathrm{O},{\mathrm{E}}^0=1.00\mathrm{V}\end{array}$

The reversed reaction as,

$\begin{array}{l}\mathrm{Cd}\to {\mathrm{Cd}}^{2+}+2{\mathrm{e}}^{-},{\mathrm{E}}^0=0.40\mathrm{V}\\ 2{{\mathrm{VO}}_2}^{+}+4{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\to 2{\mathrm{VO}}^{2+}+2{\mathrm{H}}_2\mathrm{O},{\mathrm{E}}^0=1.00\mathrm{V}\end{array}$

$\mathrm{Cd}+2{{\mathrm{VO}}_2}^{+}+4{\mathrm{H}}^{+}\to {\mathrm{Cd}}^{+}+2{\mathrm{VO}}^{2+}+2{\mathrm{H}}_2\mathrm{O},{\mathrm{E}}^0=0.40\mathrm{V}+1.00\mathrm{V}=1.40\mathrm{V}$