Question

: Consider the given electrochemical data
${M^{3 + }} + 3{e^ - } \to M$ ${E^{\circ}} = - 0.74V$
${M^{3 + }} + {e^ - } \to {{\mathbf{M}}^{2 + }}$ ${E^{\circ}} = - 0.4V$
Then the ${E^{\circ}}$ for the reaction
${M^{2 + }} + 2{e^ - } \to M$
A) -0.91V
B) +0.91V
C) -0.3V
D) 1.82V

Answer 1

Hint: We can use Gibbs free energy $\left( {\Delta G} \right)$. It is used to measure the maximum amount of work down in a thermodynamic system when the temperature and pressure is kept constant.

Complete step by step solution:
In question we have given two reactions,
${M^{3 + }} + 3{e^ - } \to M$ --(1)
${M^{3 + }} + {e^ - } \to {{\text{M}}^{2 + }}$ --(2)
To get this reaction from (1) and (2)
${M^{2 + }} + 2{e^ - } \to M$ --(3)
Now, we need form the reaction (1) and (2) in somehow to reaction (3)
When we observe we can see that in reaction (3) there is no ${M^{3 + }}$ so we should remove that. And in reaction (1) the product is as same as the reaction (3) which we require. So we can keep the reactions (1) as it is. Now in reaction (2) the ${{\text{M}}^{2 + }}$ is the product, but we need it as the reactant. So we can reverse the reaction (2). Since we reversed the reaction, the ${E^{\circ}}$ will change into its opposite sign. i.e., ${E^{\circ}} = + 0.4V$
Now we get the reaction (2) as
${{\text{M}}^{2 + }} \to {M^{3 + }} + {e^ - }$ --(4)
Now we can add reaction (4) and reaction (1)
Then we get,
${M^{3 + }} + 3{e^ - } + {M^{2 + }} \to M + {M^{3 + }} + {e^ - }$
Reducing into simple forms, we get
${M^{2 + }} + 2{e^ - } \to M$ --(5)
Since we have added the reaction then, the Gibbs free energy will also be some.
If Gibbs free energy of reactions (1), (4) and (5) are respectively.
Therefore,
$\vartriangle G_3^{\circ} = \vartriangle G_1^{\circ} + \vartriangle G_2^{\circ}$ --(6)
Now Gibbs free energy $(\Delta {G^{\circ}}) = - nF{E^{\circ}}$
where $n$ is the number of electrons involved in cell reaction
             $F$ is Faraday’s constant
$\Rightarrow$ From the equation of Gibbs free energy, we can substitute that in equation (4), then we get,
$ - {n_3}FE_3^{\circ} = - {n_1}FE_1^{\circ} + - {n_2}FE_2^{\circ}$
Also, from question it is given that,
$E_1^{\circ} = - 0.74V$ and $E_2^{\circ} = + 0.4$
And from the equation (1), (4) and (5) we understood that number of electrons involved is.
${n_3} = 2,\;{n_2} = 1\;and\;{n_1} = 3$
Substituting the values,
$ - 2 \times {\text{F}} \times {\text{E}}_3^{\circ} = 3 \times {\text{F}} \times \left( { - 0.74} \right) + \left( { - 1 \times {\text{F}} \times 0.4} \right)$
Since $F$ is common, we can cancel $F$ and multiplying, we get
$ - 2E_3^{\circ} = 2.22 - 0.4$
$ \Rightarrow E_3^{\circ} = - \dfrac{{1.82}}{2} = - 0.91V$
The ${E^{\circ}}$ for the reaction will be $ - 0.91V$

Therefore option (a) is correct

Note:
We cannot simply add the ${E^{\circ}}$ of the reaction. Since this is not an anode or cathode reaction. But we can use the energy since they will be conserved, that is why we used Gibbs free energy for finding the ${E^{\circ}}$ of the reaction.