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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.


Learning Progress: Step 29 of 37 in this series