Question

Given below are the half-cell reactions:
\[M{n^{ + 2}} + 2{e^ - }\xrightarrow{{}}Mn\] ; ${E^0} = - 1.18V$
\[2(M{n^{ + 3}} + {e^ - }\xrightarrow{{}}M{n^{2 + }})\] ; ${E^0} = + 1.51V$
The ${E^0}$ for $3M{n^{ + 2}}\xrightarrow{{}}Mn + 2M{n^{ + {3^{}}}}$ will be
(A) -0.33V; the reaction will not occur
(B) -0.33V; the reaction will occur
(C) -2.69; the reaction will not occur
(D) -2.69; the reaction will occur

Answer 1

Hint: To solve this question we should be aware of the standard potential of a reaction which is denoted ${E^0}$ and also know when a reaction is said to be spontaneous or nonspontaneous. Always be careful with the sign while solving as the sign plays an important role in these kinds of questions.

Complete step by step answer:
Standard potential of reaction is the measure of potential between electrodes. It is denoted by ${E^0}$. It is given by:
\[{E^0} = E_{OXIDISED}^0 - E_{REDUCED}^0\]………..equation 1
Firstly, let me write down the half reaction given in the question proceed further:
\[M{n^{ + 2}} + 2{e^ - }\xrightarrow{{}}Mn\] ……………..Reaction1
\[2(M{n^{ + 3}} + {e^ - }\xrightarrow{{}}M{n^{2 + }})\]…………...Reaction 2
$E_1^0$ be standard potential of reaction 1. So, $E_1^0 = - 1.18V$ (given)………equation 2
$E_2^0$be standard potential of reaction 2. So, $E_2^0 = + 1.51V$ (given)……...equation 3
Reaction 1 is oxidation reaction that is Mn is oxidized and in reaction 2 Mn is reduced
Thus, substituting equation 2 and 3 in 1.
\[{E^0} = - 1.18 - ( + 1.51)\]
\[{E^0} = - 2.69\]
Therefore, ${E^0}$ value is -2.69.
From thermodynamics we know that when the Gibbs energy value is positive, then the reaction will be non-spontaneous as it absorbs energy (heat) from the surrounding. When the Gibbs energy value is negative then, the reaction will be spontaneous.
\[\Delta {G^0} = - nF{E^0}\] ……………….equation 4
From this it is evident that if the ${E^0}$ value is negative then, $\Delta {G^0}$ value will be positive and vice-versa.
Hence, it is a non-spontaneous reaction.
So, the correct answer is “Option C”.

Note: When the ${E^0}$ value is negative and $\Delta {G^0}$ value is positive both account for non-spontaneous reactions. When the ${E^0}$ value is positive and $\Delta {G^0}$ value is negative will account for spontaneous reaction. So, it is very important to be careful with the sign while solving. The sign plays an important role in these kinds of questions.