For the standard cell potentials given here, determine the \(\Delta {G\circ }\) for the cell in \({\rm{kJ}}\).

(a) \(0.000\;{\rm{V}},{\rm{n}} = 2\)

(b) \( + 0.434\;{\rm{V}},{\rm{n}} = 2\)

(c) \( - 2.439\;{\rm{V}},{\rm{n}} = 1\)

• The electrical work of a cell at standard conditions with\(\Delta {G^{\rm{o}}}\), we get
• \(\begin{aligned}{l}\Delta {G^o} = - nFE_{cell}^o\\F = {\rm{ Faraday costant }}(96,500{\rm{C}}/{\rm{mol}})\end{aligned}\)
• For positive value of standard cell potential, the reaction is spontaneous and it is non-spontaneous for negative value of the standard cell potential.

The \(\Delta {G^o}\) for the cell is \(0.00\;{\rm{kJ}}\).

Given that:

\(\begin{aligned}{l}E_{{\rm{cell }}}^{\rm{o}} = 0.00\;{\rm{V}}\\n = 2\end{aligned}\)

The electrical work of a cell at standard conditions with \(\Delta {G\circ }\) is calculated as follows:

\(\Delta {G^{\rm{o}}} = - nFE_{{\rm{cell}}}^{\rm{o}}\)

\(F = \)Faraday constant \((96,500{\rm{C}}/{\rm{mol}})\)

\(n = 2\).

The Gibbs free energy change when this quantity of reaction is converted is

\(\Delta {G^{\rm{o}}} = - \left( {\frac{{2{\rm{ mol }}{e^ - }}}{{{\rm{transferred }}}}} \right)\left( {\frac{{96,500{\rm{C}}}}{{{\rm{ mol }}{e^ - }}}} \right)\left( {\frac{{1\;{\rm{J}}}}{{1\;{\rm{V}} \times 1{\rm{C}}}}} \right)\left( {\frac{{1\;{\rm{kJ}}}}{{{{10}^3}\;{\rm{J}}}}} \right)(0.00\;{\rm{V}})\)

\(\Delta {G^{\rm{o}}} = 0.00\;{\rm{kJ}}\)

The \(\Delta {G^o}\) for the cell is \( - 83.7\;{\rm{kJ}}\).

Given that:

\(\begin{aligned}{l}E_{{\rm{cell }}}^{\rm{o}} = + 0.434\;{\rm{V}}\\n = 2\end{aligned}\)

The electrical work of a cell at standard conditions with \(\Delta {G\circ }\) is calculated as follows:

\(\Delta {G^{\rm{o}}} = - nFE_{{\rm{cell }}}^{\rm{o}}\)

\(F = \)Faraday constant \((96,500{\rm{C}}/{\rm{mol}})\)

\(n = 2\). The Gibbs free energy change when this quantity of reaction is converted is

\(\begin{aligned}{l}\Delta {G^{\rm{o}}} = - \left( {\frac{{2{\rm{ mole }}{e^ - }}}{{{\rm{transferred }}}}} \right)\left( {\frac{{96,500{\rm{C}}}}{{{\rm{ mole }}{e^ - }}}} \right)\left( {\frac{{1\;{\rm{J}}}}{{1\;{\rm{V}} \times 1{\rm{C}}}}} \right)\left( {\frac{{1\;{\rm{kJ}}}}{{{{10}^3}\;{\rm{J}}}}} \right)( + 0.434\;{\rm{V}})\Delta {G^{\rm{o}}}\\ = - 83.7\;{\rm{kJ}}\end{aligned}\)

The \(\Delta {G^o}\) for the cell is \( + 235.3\;{\rm{kJ}}\).

Given that:

\(\begin{aligned}{l}E_{{\rm{cell }}}^{\rm{o}} = - 2.439\;{\rm{V}}\\n = 1\end{aligned}\)

The electrical work of a cell at standard conditions with \(\Delta {G^{\rm{o}}}\) is calculated as follows:

\(\Delta {G^{\rm{o}}} = - nFE_{{\rm{cell }}}^{\rm{o}}\)

\(F = \)Faraday constant \((96,500{\rm{C}}/{\rm{mol}})\)

\(n = 1.\)The Gibbs free energy change when this quantity of reaction is converted is

\(\Delta {G^{\rm{o}}} = - \left( {\frac{{1\;{\rm{mole}}{{\rm{e}}^ - }}}{{{\rm{transferred }}}}} \right)\left( {\frac{{96,500{\rm{C}}}}{{{\rm{mol}}{{\rm{e}}^ - }}}} \right)\left( {\frac{{1\;{\rm{J}}}}{{1\;{\rm{V}} \times 1{\rm{C}}}}} \right)\left( {\frac{{1\;{\rm{kJ}}}}{{{{10}^3}\;{\rm{J}}}}} \right)( - 2.439\;{\rm{V}})\)

\(\Delta {G^{\rm{o}}} = + 235.3\;{\rm{kJ}}\)