MHT CET · Chemistry · Electrochemistry
If \(\mathrm{E}^{\circ}\) cell for \(\mathrm{Cd}_{(\mathrm{s})}\left|\mathrm{Cd}_{(\mathrm{IM})}^{2+} \square \mathrm{Ag}_{(\mathrm{IM})}^{+}\right| \mathrm{Ag}_{(\mathrm{s})}\) is 1.2 V . What is the emf of the cell at \(25^{\circ} \mathrm{C}\) ?
- A -1.2 V
- B 2.4 V
- C -2.4 V
- D 1.2 V
Answer & Solution
Correct Answer
(D) 1.2 V
Step-by-step Solution
Detailed explanation
The electrode reaction is:
\(\mathrm{Cd}+2 \mathrm{Ag}^{+} \longrightarrow \mathrm{Cd}^{2+}+2 \mathrm{Ag}\)
For this cell reaction, \(\mathrm{n}=2\).
Using Nernst equation at 298 K :
\(\mathrm{E}_{\text {cell }} =\mathrm{E}_{\text {cell }}^o-\frac{0.0592}{2} \log _{10} \frac{\left[\mathrm{Cd}^{2+}\right]}{\left.\mathrm{Ag}^{+}\right]^2} \)
\( =1.2-\frac{0.0592}{2} \log _{10} \frac{1}{1} \)
\( \therefore \mathrm{E}_{\text {cell }} =1.2-0=1.2 \mathrm{~V} \quad\left(\because \log _{10} 1=0\right)\)
\(\mathrm{Cd}+2 \mathrm{Ag}^{+} \longrightarrow \mathrm{Cd}^{2+}+2 \mathrm{Ag}\)
For this cell reaction, \(\mathrm{n}=2\).
Using Nernst equation at 298 K :
\(\mathrm{E}_{\text {cell }} =\mathrm{E}_{\text {cell }}^o-\frac{0.0592}{2} \log _{10} \frac{\left[\mathrm{Cd}^{2+}\right]}{\left.\mathrm{Ag}^{+}\right]^2} \)
\( =1.2-\frac{0.0592}{2} \log _{10} \frac{1}{1} \)
\( \therefore \mathrm{E}_{\text {cell }} =1.2-0=1.2 \mathrm{~V} \quad\left(\because \log _{10} 1=0\right)\)
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