JEE PYQ: Electrochemistry Question 37
Question 37 - 2019 (12 Jan Shift 2)
The standard electrode potential $E^0$ and its temperature coefficient $\left(\frac{dE^0}{dT}\right)$ for a cell are 2 V and $-5 \times 10^{-4}$ VK$^{-1}$ at 300 K respectively. The cell reaction is: $\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}$ The standard reaction enthalpy ($\Delta_rH^0$) at 300 K in kJ mol$^{-1}$ is: [Use $R = 8$ JK$^{-1}$ mol$^{-1}$ and $F = 96,000$ C mol$^{-1}$]
(1) -412.8
(2) -384.0
(3) 192.0
(4) 206.4
Show Answer
Answer: (1) -412.8
Solution
$\Delta_rH^0 = -nF\left(E^0 - T\frac{dE^0}{dT}\right) = -2 \times 96000(2 + 0.15) = -412.8$ kJ/mol.
BETA
