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Question: Answered & Verified by Expert
If for the cell reaction, $\mathrm{Zn}+\mathrm{Cu}^{2+} \rightleftharpoons \mathrm{Cu}+\mathrm{Zn}^{2+}$ Entropy change $\Delta \mathrm{S}^{\circ}$ is $96.5 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$, then temperature coefficient of the emf of a cell is
ChemistryThermodynamics (C)VITEEEVITEEE 2013
Options:
  • A $5 \times 10^{-4} \mathrm{VK}^{-1}$
  • B $1 \times 10^{-3} \mathrm{VK}^{-1}$
  • C $2 \times 10^{-3} \mathrm{VK}^{-1}$
  • D $9.65 \times 10^{-4} \mathrm{VK}^{-1}$
Solution:
2156 Upvotes Verified Answer
The correct answer is: $5 \times 10^{-4} \mathrm{VK}^{-1}$
$\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{nFT}\left(\frac{\mathrm{dE}}{\mathrm{dT}}\right)_{\mathrm{P}}$
and $\Delta \mathrm{G}=\Delta \mathrm{T}-\mathrm{T} \Delta \mathrm{S}$
$\therefore \frac{\Delta \mathrm{S}}{\mathrm{nF}}=\left(\frac{\mathrm{dE}}{\mathrm{dT}}\right)_{\mathrm{P}}$
$\therefore \frac{96.5}{2 \times 96500}=\left(\frac{\mathrm{dE}}{\mathrm{dT}}\right)_{\mathrm{P}}$
$\therefore\left(\frac{\mathrm{dE}_{\mathrm{cell}}}{\mathrm{dT}}\right)_{\mathrm{P}}=\frac{1 \times 10^{-3}}{2}=5 \times 10^{-4} \mathrm{VK}^{-1}$

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