MHT CET · Chemistry · Electrochemistry
For the cell reaction,
\(\mathrm{Zn}_{(\mathrm{s})}+2 \mathrm{Ag}_{(\mathrm{aq})}^{+} \longrightarrow \mathrm{Zn}_{(\mathrm{aq})}^{+2}+2 \mathrm{Ag}_{(\mathrm{s})}\)
Cell potential is less than \(\mathrm{E}^{\circ}\) cell by 0.0592 V at 298 K when
- A \(\left[\mathrm{Zn}^{+2}\right]=1 \mathrm{M}\) and \(\left[\mathrm{Ag}^{+}\right]=0.1 \mathrm{M}\)
- B \(\left[\mathrm{Zn}^{+2}\right]=1 \mathrm{M}\) and \(\left[\mathrm{Ag}^{+}\right]=0.01 \mathrm{M}\)
- C \(\left[\mathrm{Zn}^{+2}\right]=0.1 \mathrm{M}\) and \(\left[\mathrm{Ag}^{+}\right]=1 \mathrm{M}\)
- D \(\left[\mathrm{Zn}^{+2}\right]=0.01 \mathrm{M}\) and \(\left[\mathrm{Ag}^{+}\right]=1 \mathrm{M}\)
Answer & Solution
Correct Answer
(A) \(\left[\mathrm{Zn}^{+2}\right]=1 \mathrm{M}\) and \(\left[\mathrm{Ag}^{+}\right]=0.1 \mathrm{M}\)
Step-by-step Solution
Detailed explanation
\( \mathrm{E}_{\text{cell}} = \mathrm{E}^{\circ}_{\text{cell}} - \frac{0.0592}{\mathrm{n}} \log \mathrm{Q} \) \( \mathrm{E}^{\circ}_{\text{cell}} - 0.0592 = \mathrm{E}^{\circ}_{\text{cell}} - \frac{0.0592}{\mathrm{n}} \log \mathrm{Q} \)
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