JEE PYQ: Electrochemistry Question 29
Question 29 - 2019 (09 Apr Shift 1)
The standard Gibbs energy for the given cell reaction in kJ mol$^{-1}$ at 298 K is: $\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)$, $E^0 = 2$ V (Faraday’s constant, F = 96000 C mol$^{-1}$)
(1) -384
(2) 384
(3) 192
(4) -192
Show Answer
Answer: (1) -384
Solution
$\Delta G^0 = -nFE^0 = -2 \times 96000 \times 2 = -384$ kJ/mol.