If the free-energy change of the following voltaic cell
is -46.3 kJ, what is the standard potential of the cell?

Question

If the free-energy change of the following voltaic cell  
is -46.3 kJ, what is the standard potential of the cell?

Quizplus · Accepted Answer

The standard potential of the cell can be calculated using the formula:
ΔG° = -nFE°
where ΔG° is the free energy change, n is the number of electrons transferred, F is Faraday's constant (96,485 C/mol), and E° is the standard potential of the cell.

Substituting the given values, we get:
ΔG° = -nFE°
-1046.3 kJ/mol = (-11 mol)(96,485 C/mol)(E°)
E° = -1046.3 kJ/mol / (-11 mol)(96,485 C/mol)
E° = 0.10024 V

Since the reduction half-reaction taking place at the cathode involves F1F1F1S1 ion, the standard potential of the cell will be positive. Therefore, the correct option is (C) (F1F1F1S1F1F1F100.240 V).

If the Free-Energy Change of the Following Voltaic Cell