Given E∘$$Cr3+$$∣$$Cr=$$−$$0.72V$$ and E∘$$Fe2+$$∣$$Fe=$$−$$0.42V$$ The potential of the cell $$CrCr3+(0.1M)$$∥$$Fe2+(0.01M)Fe$$ is

The cell potential is given by Ecell $$=ER$$−EL where Right $$half-cell$$ $$Fe2++2e$$−→Fe $$ER=ER$$∘−$$RT2Fln$$⁡$$1Fe2+/c$$∘ $$=$$−$$0.42V+0.059V2log$$⁡(0.01) $$=$$−$$0.42V$$−$$0.06V=$$−$$0.48V$$ Left $$half-cell$$ $$Cr3++3e$$−→$$CrEL=EL$$∘−$$RT3Fln$$⁡$$1Cr3+/c$$∘$$=$$−$$0.72+0.059V3log$$⁡(0.1) $$=$$−$$0.72V$$−$$0.02V=$$−$$0.74VHence$$, $$Ecell=$$−$$0.48V$$−$$($$−$$0.74V)=0.26V$$

Questions

Given $E^{\circ} ({Cr}^{3 +} ∣ Cr) = - 0.72 V$ and $E^{\circ} ({Fe}^{2 +} ∣ Fe) = - 0.42 V$ The potential of the cell $Cr |{Cr}^{3 +} (0.1 M) ∥ {Fe}^{2 +} (0.01 M)| Fe$ is

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0.26 V

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detailed solution

Correct option is B

The cell potential is given by $E_{cell} = E_{R} - E_{L}$ where

Right half-cell ${Fe}^{2 +} + 2 e^{-} \to Fe$

$E_{R} = E_{R}^{\circ} - \frac{R T}{2 F} \ln \frac{1}{[{Fe}^{2 +}] / c^{\circ}} = - 0.42 V + (\frac{0.059 V}{2}) \log (0.01) = - 0.42 V - 0.06 V = - 0.48 V$

Left half-cell ${Cr}^{3 +} + 3 e^{-} \to Cr$

$E_{L} = E_{L}^{\circ} - \frac{R T}{3 F} \ln \frac{1}{[{Cr}^{3 +}] / c^{\circ}} = - 0.72 + (\frac{0.059 V}{3}) \log (0.1) = - 0.72 V - 0.02 V = - 0.74 V$