Question

The value of $\Delta G(kJmo{l}^{-1})$ for the given cell is
(take $1F=96500Cmo{l}^{-1}$)

• A. -5.7
• B. 5.7
• C. 11.4
• D. -11.4

Answer 1

Answer: (D)

$\Delta G=-nEF=-\frac{2\times 0.059\times 96500}{1000}kJ=-11.4kJ$