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

To calculate the standard Gibbs free energy change (ΔG°) for the reaction, we use the relation between Gibbs free energy and the standard electrode potential (E°) given by the equation: 

ΔG° = -nFE°

Where:

• n = number of moles of electrons exchanged in the balanced equation.
• F = Faraday's constant = 96485 C/mol.
• E° = standard electrode potential = 1.1 V.

For the given reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), the number of moles of electrons exchanged (n) is 2, due to the transfer from Zn⁰ to Zn²⁺ and Cu²⁺ to Cu⁰. Substituting the known values:

ΔG° = -2 × 96485 × 1.1

ΔG° = -212267 J/mol

Finally, to match the options format, we convert this into kJ/mol:

ΔG° = -212.27 J/mol

Thus, the standard Gibbs free energy change for the reaction is approximately -212.27 J/mol, which matches the provided correct answer choice.

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.