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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

  1. A \(\left[\mathrm{Zn}^{+2}\right]=1 \mathrm{M}\) and \(\left[\mathrm{Ag}^{+}\right]=0.1 \mathrm{M}\)
  2. B \(\left[\mathrm{Zn}^{+2}\right]=1 \mathrm{M}\) and \(\left[\mathrm{Ag}^{+}\right]=0.01 \mathrm{M}\)
  3. C \(\left[\mathrm{Zn}^{+2}\right]=0.1 \mathrm{M}\) and \(\left[\mathrm{Ag}^{+}\right]=1 \mathrm{M}\)
  4. D \(\left[\mathrm{Zn}^{+2}\right]=0.01 \mathrm{M}\) and \(\left[\mathrm{Ag}^{+}\right]=1 \mathrm{M}\)
Verified Solution

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

\( \mathrm{E}_{\text{cell}} = \mathrm{E}^{\circ}_{\text{cell}} - \frac{0.0592}{\mathrm{n}} \log \mathrm{Q} \) \( \mathrm{E}^{\circ}_{\text{cell}} - 0.0592 = \mathrm{E}^{\circ}_{\text{cell}} - \frac{0.0592}{\mathrm{n}} \log \mathrm{Q} \)