Question

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

The $E^{\ominus }$ for $3M{n}^{2+}\to Mn+2M{n}^{3+}$ will be:
(IIT-JEE 2014)

• A. (a) -0.33 V; the reaction will not occur
• B. (b) -0.33 V; the reaction will occur
• C. (c) -2.69 V; the reaction will not occur
• D. (d) -2.69 V; the reaction will occur

Answer 1

The correct option is C

(c) -2.69 V; the reaction will not occur

$M{n}^{2+}+2e^{\ominus }\to Mn;E^{\ominus }=-1.18V$
$2M{n}^{2+}+\to 2M{n}^{3+}+2e^{\ominus };E^{\ominus }=-1.51V$
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$3M{n}^{2+}\to Mn+2M{n}^{3+}$
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$E_{cell}^{\ominus }=E_{red}^{\ominus }+E_{oxid}^{\ominus }$
= -1.18 + (-1.51) = -2.69 V

Negative EMF reflects non-spontaneous cell reaction.