MHT CET · Chemistry · Electrochemistry
\(\text {Calculate } \mathrm{E}_{\text {cell }}^{\circ} \text { for } \mathrm{Cd}_{(\mathrm{s})}\left|\mathrm{Cd}_{(\mathrm{IM})}^{++}\right|\left|\mathrm{Ag}_{(\mathrm{MM})}^{+}\right| \mathrm{Ag}_{(\mathrm{s})}.\) \(\ {\left[\mathrm{E}_{\mathrm{Cd}}^{\circ}=-0.403 \mathrm{~V} ; \mathrm{E}_{\mathrm{Ag}}^{\circ}=0.799 \mathrm{~V}\right]}\)
- A \(1.202 V\)
- B \(-1.202 V\)
- C \(0.396 V\)
- D \(-0.396 V\)
Answer & Solution
Correct Answer
(A) \(1.202 V\)
Step-by-step Solution
Detailed explanation
For the given cell, anode is \(\mathrm{Cd}\) and cathode is Ag.
\(\begin{aligned}
\mathrm{E}_{\text {cell }}^{\circ} & =\mathrm{E}_{\text {cathode }}^{\circ}-\mathrm{E}_{\text {anode }}^{\circ} \\
& =0.799-(-0.403) \\
& =1.202 \mathrm{~V}
\end{aligned}\)
\(\begin{aligned}
\mathrm{E}_{\text {cell }}^{\circ} & =\mathrm{E}_{\text {cathode }}^{\circ}-\mathrm{E}_{\text {anode }}^{\circ} \\
& =0.799-(-0.403) \\
& =1.202 \mathrm{~V}
\end{aligned}\)
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