Question

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

• A. \(-6 \, \text{F}\)
• B. \(-12 \, \text{F}\)
• C. \(-3 \, \text{F}\)
• D. \(-9 \, \text{F}\)

Hint:

Use the formula \(\Delta G^\circ = -nFE_{\text{cell}}\) to relate Gibbs free energy change to the cell potential.

Answer 1

The Correct Option is B

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 \(-12 \, \text{F}\).