MHT CET · Chemistry · Electrochemistry
A reaction, \(\mathrm{Ni}_{(\mathrm{s})}+\mathrm{Cu}_{(\mathrm{IM})}^{++} \rightarrow \mathrm{Ni}_{(\mathrm{IM} 2}^{++}+\mathrm{Cu}_{(\mathrm{s})}\) occurs in a cell. Calculate \(\mathrm{E}_{\mathrm{cell}}^6\) if \(\mathrm{E}_{\mathrm{Cu}}^0=0.337 \mathrm{~V}\) and \(\mathrm{E}_{\mathrm{Ni}}^{\mathrm{o}}=-0.257 \mathrm{~V}\)
- A \(0.594 \mathrm{~V}\)
- B \(-0.594 \mathrm{~V}\)
- C \(-0.08 \mathrm{~V}\)
- D \(0.08 \mathrm{~V}\)
Answer & Solution
Correct Answer
(A) \(0.594 \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
The standard cell potential is given by
\(\begin{aligned}
\mathrm{E}_{\text {cell }}^o & =\mathrm{E}_{\text {cathode }}^{\circ}-\mathrm{E}_{\text {anode }}^o \\
\mathrm{E}_{\text {cell }}^{\circ} & =\mathrm{E}_{\mathrm{Cu}}^{\circ}-\mathrm{E}_{\mathrm{Ni}}^{\circ} \\
& =(0.337 \mathrm{~V})-(-0.257 \mathrm{~V}) \\
& =0.337 \mathrm{~V}+0.257 \mathrm{~V} \\
& =0.594 \mathrm{~V}
\end{aligned}\)
\(\begin{aligned}
\mathrm{E}_{\text {cell }}^o & =\mathrm{E}_{\text {cathode }}^{\circ}-\mathrm{E}_{\text {anode }}^o \\
\mathrm{E}_{\text {cell }}^{\circ} & =\mathrm{E}_{\mathrm{Cu}}^{\circ}-\mathrm{E}_{\mathrm{Ni}}^{\circ} \\
& =(0.337 \mathrm{~V})-(-0.257 \mathrm{~V}) \\
& =0.337 \mathrm{~V}+0.257 \mathrm{~V} \\
& =0.594 \mathrm{~V}
\end{aligned}\)
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