Question

To find the standard potential of $M^{3+}/M$ electrode, the following cell is constituted : $Pt/M/M^{3+}(0.001mol{L}^{-1})/A{g}^{+}(0.01mol{L}^{-1})/Ag$
The emf of the cell is found to be $0.421$ volt at $298K$. The standard potential of half reaction $M^{3+}+3e^{-}\to M$ at 298 K will be :
(Given $E_{A{g}^{+}/Ag}^{\ominus }$ at 298 K = 0.80 Volt)

• A. $0.38Volt$
• B. $0.32Volt$
• C. $1.28Volt$
• D. $0.66Volt$

Answer 1

Answer: (B)

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

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

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

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