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If $E^{\circ} ({Cr}^{2 +}, {Cr}^{3 +} ∣ Pt) = - 0.42 V$ and $E^{\circ} ({Cr}^{2 +} ∣ Cr) = - 0.90 V$ then $E^{\circ}$ for the half-cell ${Cr}^{3 +} ∣ Cr$ will be

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a

- 0.48 V

b

0.74 V

c

- 0.74 V

d

- 1.32 V

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detailed solution

Correct option is C

Add $\begin{array}{l} {Cr}^{3 +} + e^{-} \to {Cr}^{2 +} Δ G_{1}^{\circ} = - (1) F (- 0.42 V) \\ {Cr}^{2 +} + 2 e^{-} \to Cr Δ G_{2} = - (2) F (- 0.90 V) \\ \bar{{Cr}^{3 +} + 3 e^{-} \to Cr} \bar{Δ G_{3}^{\circ} = - F (- 0.42 - 2 \times 0.90) V} \end{array}$

Also $Δ G_{3}^{0} = - 3 F {E_{{Cr}^{3 +}}}^{\circ} Cr$ Hence $3 E_{{Cr}^{3 +} ∣ Cr}^{\circ} = - 2.22 V or E^{\circ} = - 0.74 V$

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