TS EAMCET · Chemistry · Electrochemistry
Calculate the emf of the cell \(\mathrm{Cu}(s)\left|\mathrm{Cu}^{2+}(a q)\right|\left|\mathrm{Ag}^{+}(a q)\right| \mathrm{Ag}(s)\) Given \(E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^0=+0.34 \mathrm{~V}, E_{\mathrm{Ag}^{+} / \mathrm{Ag}}^0=0.80 \mathrm{~V}\)
- A \(+0.46 \mathrm{~V}\)
- B \(+1.14 \mathrm{~V}\)
- C \(+0.57 \mathrm{~V}\)
- D \(-0.46 \mathrm{~V}\)
Answer & Solution
Correct Answer
(A) \(+0.46 \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} E_{\text {cell }}^{\circ} & =E_{\text {red (cathode) }}^{\circ}-E_{\text {oxi (anode) }}^{\circ} \\ & =E_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{\circ}-E_{\mathrm{Cu}^{+} / \mathrm{Cu}^{2+}}^0 \\ & =0.80-(+0.34)=+0.46 \mathrm{~V}\end{aligned}\)
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