For the following reaction
\[
2 \text{Mn}^{3+} (\text{aq}) + 3 \text{N}_2^{2+} (\text{aq}) \rightarrow 3 \text{Mn}^{2+} (\text{aq}) + 2 \text{N}_2 (\text{s})
\]
The standard Gibbs free energy change (\(\Delta G^\circ\)) of the reaction is
Show Hint
Use the formula \(\Delta G^\circ = -nFE_{\text{cell}}\) to relate Gibbs free energy change to the cell potential.
We are given that the cell potential \( E_{\text{cell}} = 2 \, \text{V} \), and we need to find the standard Gibbs free energy change (\(\Delta G^\circ\)).
The relation between the standard Gibbs free energy change and the cell potential is:
\[
\Delta G^\circ = -nFE_{\text{cell}}
\]
where:
- \(n\) is the number of electrons transferred in the reaction (which is 6 electrons based on the balanced equation),
- \(F\) is the Faraday constant (\(F = 96,485 \, \text{C/mol}\)),
- \(E_{\text{cell}}\) is the cell potential in volts.
Substituting the values:
\[
\Delta G^\circ = -6 \times 96,485 \times 2 = -1,158,000 \, \text{J/mol} = -12 \, \text{F}
\]
Thus, the standard Gibbs free energy change is \(-12 \, \text{F}\).
