Question

A cell has been set up as shown in the following diagram and E$^0$ has been measured as 1.00V at 25°C. Calculate $\Delta$G$^0$ for the reaction.

• A. -386 kJ
• B. -193 kJ
• C. 1.00 kJ
• D. 193 kJ

Hint:

Remember, the cell potential is directly related to the change in Gibbs free energy.

Answer 1

The Correct Option is B

The relationship between Gibbs free energy ($\Delta$G$^0$) and cell potential (E$^0$) is given by: \[ \Delta G^0 = -nFE^0 \] where: - $n$ = number of moles of electrons transferred, - $F$ = Faraday's constant (96,500 C/mol), - $E^0$ = cell potential (1.00 V), Substituting values: \[ \Delta G^0 = -n \times 96,500 \times 1.00 \] Using $n = 2$ for the reaction, we get: \[ \Delta G^0 = -193,000 \, \text{J} = -193 \, \text{kJ} \]