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To find the standard potential of $M^{3 +} / M$ electrodes the following cell is constituted. $Pt M / M^{3 +} (0.001 mol L^{- 1}) / {Ag}^{+} (0.01 mol L^{- 1}) / Ag$ the emf of the cell is found to be 0.421 volt at 298K. the standard potential of half reaction $M^{3 +} + 3 e^{-} \to M$ at 298K will be $G i v e n E_{A g^{+} / A g}^{0} a t 298 K = 0.80 V o l t$ )

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a

0.32 Volt

b

1.28 Volt

c

0.66 Volt

d

0.38 Volt

answer is B.

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Detailed Solution

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$E_{M^{3 +} / M}^{\circ}$ $E_{M^{3 +} / M}^{\circ}$ Cell reaction
M+ 3 Ag+ (aq) $\to$ = 0.32 volt
E = E^o - Question Image log
0.421= E⁰ - $1.73 ≅ 2$ log $\frac{0.001}{{(0.01)}^{3}}$
E⁰ =0.48
$E^{\circ} = E_{A g^{+} / A g}^{\circ} - E_{M^{3 +} / M}^{\circ}$
= 0.8v – 0.48v = 0.32 volt

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To find the standard potential of M3+/M electrodes the following cell is constituted. Pt M/M3+(0.001 mol L−1)/Ag+(0.01mol L−1)/Ag the emf of the cell is found to be 0.421 volt at 298K. the standard potential of half reaction M3++3e−→M at 298K will be Given EAg+/Ag0at 298K=0.80 Volt )