AP EAMCET · Chemistry · Electrochemistry
The standard Gibbs' energy change for Daniel cell reaction is
\[
\begin{aligned}
& \mathrm{Zn}(s)+\mathrm{Cu}^{2+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Cu}(s) \\
& E_{\text {cell }}^{\circ}=1.1 \mathrm{~V} \text {. }
\end{aligned}
\]
- A \(-212.3 \mathrm{~kJ}\)
- B \(106.15 \mathrm{~kJ}\)
- C \(+212.3 \mathrm{~kJ}\)
- D \(100 \mathrm{~kJ}\)
Answer & Solution
Correct Answer
(A) \(-212.3 \mathrm{~kJ}\)
Step-by-step Solution
Detailed explanation
Given, \[ \begin{aligned} n, E_{\text {cell }}^{\circ} & =1.1 \mathrm{~V} \\ n & =2 \\ F & =96500 \mathrm{C} / \mathrm{mol} \\ \Delta G & =-n F E^{\circ} \\ & =-2 \times 96500 \times 1.1 \\ & =-212300 \mathrm{~J} \\ & =-212.3 \mathrm{~kJ} \end{aligned} \]
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