To find the standard potential of M^{3+}/M el | vedclass

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(B) The cell reaction is: $M(s) + 3Ag^{+}(aq) 	o M^{3+}(aq) + 3Ag(s)$.The number of electrons transferred is $n = 3$.Using the Nernst equation: $E_{c

Vedclass · Accepted Answer

$0.32$

Vedclass · Answer

(B) The cell reaction is: $M(s) + 3Ag^{+}(aq) 	o M^{3+}(aq) + 3Ag(s)$.The number of electrons transferred is $n = 3$.Using the Nernst equation: $E_{cell} = E^{o}_{cell} - \frac{0.0591}{n} \log \frac{[M^{3+}]}{[Ag^{+}]^{3}}$.Substituting the given values: $0.421 = E^{o}_{cell} - \frac{0.0591}{3} \log \frac{0.001}{(0.01)^{3}}$.$0.421 = E^{o}_{cell} - 0.0197 \log \frac{10^{-3}}{10^{-6}} = E^{o}_{cell} - 0.0197 \log(10^{3})$.$0.421 = E^{o}_{cell} - 0.0197 	imes 3 = E^{o}_{cell} - 0.0591$.$E^{o}_{cell} = 0.421 + 0.0591 = 0.4801 \ V \approx 0.48 \ V$.Since $E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode} = E^{o}_{Ag^{+}/Ag} - E^{o}_{M^{3+}/M}$.$0.48 = 0.80 - E^{o}_{M^{3+}/M}$.$E^{o}_{M^{3+}/M} = 0.80 - 0.48 = 0.32 \ V$.

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