Question

Consider the half-cell reduction reactions :
${Mn}^{2+}+{2e}^{-}\to Mn,E^o=-1.18V$
${Mn}^{2+}\to {2n}^{3+}+e^{-},E^o=-1.51V$
The $E^o$ for the reaction ${3Mn}^{2+}\to {Mn}^0+{2Mn}^{3+}$ and possibility of the forward reaction are respectively:

• A. -4.18 V and yes
• B. +0.33 V and yes
• C. +2.69 V and no
• D. -2.69 V and no

Answer 1

The correct option is D -2.69 V and no

The standard electrode potential of reaction can be calculated as,

$E_0=E_R-E_P$

Where, $E_R$ is the SRP of the reactant, $E_P$is the SRP of the product.

${Mn}^{3+}\to M{n}^{2+}$this reaction takes place with the $E_0-1.51V$

$M{n}^{2+}\to Mn$ this takes place with$E_0-1.18V$

$E_0=-1.51V-1.81V$

$=-2.69V$

and the reaction is will not occurs due to spontaneity.