MHT CET · Chemistry · Electrochemistry
Calculate the \(\mathrm{E}_{\text {cell }}^o\) for \(\mathrm{Zn}_{(\mathrm{s})}\left|\mathrm{Zn}_{(\mathrm{IM})}^{+}\right|\left|\mathrm{Cd}_{(\mathrm{IM})}^{+}\right| \mathrm{Cd}_{(\mathrm{s})}\) at \(25^{\circ} \mathrm{C}\left[\mathrm{E}_{\mathrm{Zn}}^{\circ}=-0.763 \mathrm{~V} ; \mathrm{E}_{\mathrm{Cd}}^{\circ}=-0.403 \mathrm{~V}\right]\)
- A \(0.36 \mathrm{~V}\)
- B \(1.17 \mathrm{~V}\)
- C \(-0.36 \mathrm{~V}\)
- D \(-1.17 \mathrm{~V}\)
Answer & Solution
Correct Answer
(A) \(0.36 \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
For the given cell reaction, anode is \(\mathrm{Zn}\) and cathode is \(\mathrm{Cd}\).
\(\begin{aligned}
\mathrm{E}_{\text {cell }}^o & =\mathrm{E}_{\text {cathode }}^{\circ}-\mathrm{E}_{\text {anode }}^0 \\
& =-0.403-(-0.763) \\
& =0.36 \mathrm{~V}
\end{aligned}\)
\(\begin{aligned}
\mathrm{E}_{\text {cell }}^o & =\mathrm{E}_{\text {cathode }}^{\circ}-\mathrm{E}_{\text {anode }}^0 \\
& =-0.403-(-0.763) \\
& =0.36 \mathrm{~V}
\end{aligned}\)
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