Identify the type of cell

The cell is $M(s)\,|\,M^+(0.05\,\text{M})\,||\,M^+(1\,\text{M})\,|\,M(s)$

This is a concentration cell, where both electrodes are the same metal and the emf arises only due to difference in ion concentration.

Find the direction of spontaneous reaction

For the half-reaction $M^+ + e^- \rightarrow M(s)$

By Nernst equation, $E = E^\circ + \frac{0.0591}{1}\log [M^+]$

So, higher $[M^+]$ gives higher reduction potential.

Therefore:

• Right electrode: $[M^+] = 1\,\text{M}$ has higher reduction potential, so it acts as cathode.
• Left electrode: $[M^+] = 0.05\,\text{M}$ acts as anode.

Hence the spontaneous cell reaction is from left to right, so $E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} > 0$

Check the given magnitude

For a concentration cell, $E_{\text{cell}} = \frac{0.0591}{1}\log\left(\frac{[M^+]_{\text{cathode}}}{[M^+]_{\text{anode}}}\right)$

$E_{\text{cell}} = 0.0591\log\left(\frac{1}{0.05}\right)$

$= 0.0591\log(20)$

Since $\log 20 \approx 1.301$, $E_{\text{cell}} \approx 0.0591 \times 1.301 \approx 0.077\,\text{V}$

which is about $70\,\text{mV}$, consistent with the question. So indeed, $E_{\text{cell}} > 0$

Determine sign of $\Delta G$

We use $\Delta G = -nFE_{\text{cell}}$

Since $E_{\text{cell}} > 0$ for spontaneous operation, $\Delta G < 0$

Determine sign of $\Delta G^\circ$

For a concentration cell, both electrodes are identical, so $E^\circ_{\text{cell}} = 0$

Therefore, $\Delta G^\circ = -nF E^\circ_{\text{cell}} = 0$

So any option claiming $\Delta G^\circ > 0$ is false.

Evaluate options

• A: $E_{\text{cell}} < 0$ ; $\Delta G > 0$
  False.
• B: $E_{\text{cell}} > 0$ ; $\Delta G < 0$
  True.
• C: $E_{\text{cell}} < 0$ ; $\Delta G^\circ > 0$
  False.
• D: $E_{\text{cell}} > 0$ ; $\Delta G^\circ > 0$
  False because $\Delta G^\circ = 0$.

Final answer

The correct option is $\boxed{\text{B}}$