Question

The standard electrode potential for Daniell cell is 1.1 V. The standard Gibbs free energy for the reaction Zn(s) + Cu$^{2+}$(aq) → Zn$^{2+}$(aq) + Cu(s) is approximately:

• A. -212.27 J/mol
• B. -21.22 J/mol
• C. -212271.4 J/mol
• D. -2.1227 J/mol

Answer 1

The Correct Option is A

Using the formula for Gibbs free energy change, \(\Delta G^{\circ} = -n \times F \times E^{\circ} \\ = -2 \times 96485 \text{ C/mol} \times 1.1 \text{ V} \\ = -212271 \text{ J/mol} \\ \approx -212.27 \text{ kJ/mol}\)$\\$The calculation gives -212.27 J/mol as the result.

Answer 2

Using the formula for Gibbs free energy change, \(\Delta G^{\circ} = -n \times F \times E^{\circ}\), we can calculate the value as follows:

\(\Delta G^{\circ} = -2 \times 96485 \, \text{C/mol} \times 1.1 \, \text{V}\)
\(\Delta G^{\circ} = -212271 \, \text{J/mol}\)
\(\Delta G^{\circ} \approx -212.27 \, \text{kJ/mol}\)

The calculation gives \(\Delta G^{\circ} = -212.27 \, \text{kJ/mol}\) as the result.

This result shows the Gibbs free energy change for the electrochemical reaction, indicating a spontaneous process because the value is negative. The negative sign of \(\Delta G^{\circ}\) reflects that the reaction occurs spontaneously under standard conditions.