Given $$E^\circ_{Cr^{3+}/Cr} = -0.72\ V,\ E^\circ_{Fe^{2+}/Fe} = -0.42\ V$$. The potential for the cell $$Cr|Cr^{3+}(0.1M)\ ||\ Fe^{2+}(0.01M)|Fe$$ is

For the cell $$Cr|Cr^{3+}(0.1\text{ M})\ ||\ Fe^{2+}(0.01\text{ M})|Fe$$ first identify the half-reactions and their standard reduction potentials:

Cathode candidate $$Fe^{2+}+2e^- \rightarrow Fe,\qquad E^\circ_{Fe^{2+}/Fe}=-0.42\text{ V}$$
Anode candidate  $$Cr^{3+}+3e^- \rightarrow Cr,\qquad E^\circ_{Cr^{3+}/Cr}=-0.72\text{ V}$$

The less negative potential (-0.42 V) is higher, so iron undergoes reduction (cathode) and chromium undergoes oxidation (anode).

Standard cell potential:

$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = (-0.42) - (-0.72) = +0.30\text{ V}$$

Write balanced overall reaction by equalising electrons (LCM = 6):

Oxidation: $$2Cr \rightarrow 2Cr^{3+} + 6e^⁻$$

Reduction: $$3Fe^{2+} + 6e^- \rightarrow 3Fe$$

Overall: $$2Cr + 3Fe^{2+} \rightarrow 2Cr^{3+} + 3Fe$$

The reaction quotient $$Q$$ involves only ions in solution (solids are excluded):

$$Q=\frac{\left[Cr^{3+}\right]^2}{\left[Fe^{2+}\right]^3} =\frac{(0.10)^2}{(0.010)^3} =\frac{0.01}{1.0\times10^{-6}} =10^{4}$$

Total electrons transferred $$n=6$$. Apply the Nernst equation at 298 K:

$$E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n}\log Q = 0.30 - \frac{0.0591}{6}\log(10^{4})$$

$$\log(10^{4}) = 4,\qquad \frac{0.0591}{6}=0.00985$$

$$E_{\text{cell}} = 0.30 - (0.00985)(4)=0.30-0.0394 \approx 0.26\text{ V}$$

Therefore the cell potential is 0.26 V.

Option A which is: 0.26 V