Question

Given below are the half-cell reactions:
$M{n}^{2+}+2e^{-}\to Mn;E^{\circ }=-1.18V$
$2(M{n}^{3+}+e^{-}\to M{n}^{2+});E^{\circ }=+1.51V$

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

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

Answer 1

Answer: (C)

$-2.69V$; the reaction will not occur

$M{n}^{2+}+2e^{-}\to Mn{E}_{M{n}^{2+}/Mn}^{\circ }=-1.18V$

$\frac{2M{n}^{2+}\to 2M{n}^{3+}+2e^{-}}{3M{n}^{2+}\to 2M{n}^{3+}+Mn};E_{M{n}^{3+}/M{n}^{2+}}^{\circ }=+1.51V$

$E_{Redoxprocess}^{\circ }=E_{M{n}^{2+}/Mn}^{\circ }-E_{M{n}^{3+}/M{n}^{2+}}^{\circ }=-1.81-1.51$

$=-2.69V$

Negative value shows that the process will not occur.