Question

EMF of a cell having the following cell reaction at 298 K is 0.059 V.
Zn(s) + 2H\(^+\)(aq) $\rightarrow$ Zn\(^{2+}\) + H\(_2\)(g)
What is the value of \(\Delta G\)?

• A. –5.7 kJ
• B. –11.4 kJ
• C. –8.3 kJ
• D. –14.1 kJ

Hint:

Always convert EMF into Gibbs free energy using \(\Delta G = -nFE\).

Answer 1

The Correct Option is B

Step 1: Write the relation between \(\Delta G\) and EMF.
\[ \Delta G = -nFE \]
Step 2: Identify the values.
For the given reaction, number of electrons transferred \(n = 2\).
Faraday constant \(F = 96500\ \text{C mol}^{-1}\).
EMF \(E = 0.059\ \text{V}\).
Step 3: Substitute the values.
\[ \Delta G = -2 \times 96500 \times 0.059 \]
\[ \Delta G = -11387\ \text{J} \]
Step 4: Convert joules to kilojoules.
\[ \Delta G = -11.4\ \text{kJ} \]
Step 5: Conclusion.
The value of Gibbs free energy change is –11.4 kJ.