Question

Given below are the half -cell

$M{n}^{2+}+2^{e-}\to Mn;E^0=-1.18V$

$2(M{n}^{3+}+e^{-}\to M{n}^{2+});E^0=+1.51V$

The $E^0$ for $3M{n}^{2+}\to Mn+2M{n}^{3+}$ will be:

• A. $-2.69$ V; the reaction will not occur
• B. $-2.69$ V; the reaction will occur
• C. $-0.33$ V; the reaction will not occur
• D. $-0.33$ V; the reaction will occur

Answer 1

Answer: (A)

$-2.69$ V; the reaction will not occur

$M{n}^{+2}+2e^{-}\to Mn$ .....(i)

$E^o=-1.18v$

$2M{n}^{+3}+2e6-\to 2M{n}^{+2}{E}^o=+1.51V$

$2M{n}^{+2}\to 2M{n}^{+3}+2e^{-}$

$E^o=-1.51V$ ......(ii)

Add equation (i) and (ii)

$3M{n}^{+2}\to 2M{n}^{+3}+Mn$

$E^o=-(1.18+1.51)$

$E^o=-2.69V$

$\Delta G=-nF{E}^o=+ve$

$-2.69V$ the reaction will not occur

Hence, the correct option is $\text{A}$