For the \(\Delta {G\circ }\) values given here, determine the standard cell potential for the cell.

(a) \(12\;{\rm{kJ}}/{\rm{mol}},{\rm{n}} = 3\)

(b) \( - 45\;{\rm{kJ}}/{\rm{mol}},{\rm{n}} = 1\)

• The electrical work of a cell at standard conditions with\(\Delta {G^{\rm{o}}}\), we get
• \(\begin{aligned}{}\Delta {G^o} = - nFE_{cell}^o\\F = {\rm{ Faraday costant }}(96,500{\rm{C}}/{\rm{mol}})\end{aligned}\)
• 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 standard cell potential from given \(\Delta {G^o}\) for the cell is \( - 0.0415\;{\rm{V}}\)

Given that:

\(\begin{aligned}{}\Delta {G^o} = 12\;{\rm{kJ}}/{\rm{mol}}\\n = 3\end{aligned}\)

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

\(\begin{aligned}{}\Delta {G^o} = - nFE_{cell}^o\\F = {\rm{ Faraday costant }}(96,500{\rm{C}}/{\rm{mol}})\end{aligned}\)

The standard cell potential is calculated as follows:

\(\begin{aligned}{}12\;{\rm{kJ}}/{\rm{ mole }} = - \left( {3{\rm{ mole}}{{\rm{ }}^ - }{\rm{transferred }}} \right)\left( {\frac{{96,500{\rm{C}}}}{{{\rm{ mole}}{{\rm{ }}^ - }}}} \right)\left( {\frac{{1\;{\rm{J}}}}{{1V \times 1{\rm{C}}}}} \right)\left( {\frac{{1\;{\rm{kJ}}}}{{{{10}^3}\;{\rm{J}}}}} \right)\left( {E_{{\rm{cell }}}^o{\rm{ inV }}} \right)\\E_{{\rm{cell }}}^o = - 0.0415\;{\rm{V}}\end{aligned}\)

The standard cell potential from given \(\Delta {G^o}\) for the cell is \( + 0.466\;{\rm{V}}\)

Explanation of Solution

\(\begin{aligned}{}\Delta {G^o} = - 45\;{\rm{kJ}}/{\rm{mol}}\\n = 1\end{aligned}\)

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

\(\begin{aligned}{}\Delta {G^o} = - nFE_{cell}^o\\F = {\rm{ Faraday costant }}(96,500{\rm{C}}/{\rm{mol}})\end{aligned}\)

The standard cell potential is calculated as follows:

\(\begin{aligned}{} - 45\;{\rm{kJ}}/{\rm{ mole }} = - \left( {1{\rm{ mole}}{{\rm{ }}^ - }{\rm{transferred }}} \right)\left( {\frac{{96,500{\rm{C}}}}{{{\rm{ mole}}{{\rm{ }}^ - }}}} \right)\left( {\frac{{1\;{\rm{J}}}}{{1V \times 1{\rm{C}}}}} \right)\left( {\frac{{1\;{\rm{kJ}}}}{{{{10}^3}\;{\rm{J}}}}} \right)\left( {E_{{\rm{cell }}}^o{\rm{ inV }}} \right)\\E_{{\rm{cell }}}^o = + 0.466\;{\rm{V}}\end{aligned}\)