If E∘M+∣M=1.7V and  E∘M3+∣M=1.4V then E∘M3+,M+∣Ptwill be about

# If ${E}^{\circ }\left({\mathrm{M}}^{+}\mid \mathrm{M}\right)=1.7\mathrm{V}$ and  ${E}^{\circ }\left({\mathrm{M}}^{3+}\mid \mathrm{M}\right)=1.4\mathrm{V}$ then ${E}^{\circ }\left({\mathrm{M}}^{3+},{\mathrm{M}}^{+}\mid \mathrm{Pt}\right)$will be about

1. A

0.3 V

2. B

0.95 V

3. C

0.48 V

4. D

1.25 V

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### Solution:

We have

Thus ${\mathrm{M}}^{3+}+2{\mathrm{e}}^{-}\to {\mathrm{M}}^{+}$;    $\mathrm{\Delta }{G}^{\circ }=-2F\left({E}^{\circ }{\mathrm{M}}^{3+},{\mathrm{M}}^{+}\mid \mathrm{Pt}\right)$

Equation (iii) is obtained by subtracting Eq. (i) from Eq. (ii). Hence

$-2F\left({E}_{{\mathrm{M}}^{3+}\cdot {\mathrm{M}}^{+}\mid \mathrm{Pt}}^{\circ }\right)=-3F\left(1.4\mathrm{V}\right)+F\left(1.7\mathrm{V}\right)$ or ${E}_{{\mathrm{M}}^{3+},{\mathrm{M}}^{+}{\mathrm{Pt}}^{\circ }}=1.25\mathrm{V}$

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