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  4. To find the standard potential of M3+/M electrode, the following cell is constituted: Pt/M/M3+ (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 298 K. The standard potential of half reaction M3+ + 3e- → M at 298 K will be: (Given E ominusAg+ /Ag at 298 K = 0.80 Volt)

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Q. To find the standard potential of $M^{3+}/M$ electrode, 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 $298\, K$. The standard potential of half reaction $M^{3+} + 3e^{-} \to M$ at 298 K will be :
(Given $E^{\ominus}_{Ag^{+} /Ag}$ at 298 K = 0.80 Volt)

JEE MainJEE Main 2017Electrochemistry

Solution:

$0.421=E^{\circ}-\frac{0.059}{3}log \frac{0.001}{\left(0.01\right)^{3}}$

$E^{\circ}=0.421+\frac{0.059}{3}log\left(10^{3}\right)$

$E^{\circ}=0.480=0.8-E^{\circ}_{M^{+3}/M}$

$E^{\circ}_{M^{+3}/M}=0.32$