Question

If \[O{x_{_1}} + a{e^ - } \rightleftharpoons \operatorname{Re} {d_1}{\text{ }}{{\text{E}}^ \circ }_1\]
\[O{x_{_2}} + b{e^ - } \rightleftharpoons \operatorname{Re} {d_2}{\text{ }}{{\text{E}}^ \circ }_2\]
The potential at the equivalence point is:
(A) \[E = \dfrac{{a{E_1}^ \circ + b{E_2}^ \circ }}{{(a + b)}}\]
(B) \[E = \dfrac{{a{E_1}^ \circ - b{E_2}^ \circ }}{{(a + b)}}\]
(C) \[E = \dfrac{{b{E_1}^ \circ - a{E_2}^ \circ }}{{(a + b)}}\]
(D) \[E = \dfrac{{a{E_1}^ \circ - b{E_2}^ \circ }}{{(a - b)}}\]

Answer 1

Hint: Here, Ox stands for oxidizing agent and Red stands for reducing agent. Free energy of the reaction, number of electrons involved in the reaction and the electrode potential is related by the following formula.
\[\Delta {G^ \circ } = - nF{E^ \circ }_{cell}\]


Complete step by step solution:
Here, two reactions are given. Two different oxidizing agents are undergoing reduction reaction to give a reducing agent. They require a specific number of electrons to undergo reaction, written as ‘a’ and ‘b’.

The standard electrode potentials of both the reactions are also given here. So, we are going to find the potential of the cell at the equivalence point.
- Here, the number of electrons involved in the reaction and the electrode potential is given in the question. So, formula we can use here is \[\Delta {G^ \circ } = - nF{E^ \circ }_{cell}\]…………..(1)

- Now, we will have to relate both the reactions. We can say that net Gibbs free energy of this reaction will be equal to the sum of the Gibbs free energy of both these reactions.
So, we can write that,
\[\Delta {G^ \circ }_{net} = \Delta {G^ \circ }_1 + \Delta {G^ \circ }_2\]……(2)
We can also write the equation (1) for \[\Delta {G^ \circ }_{net}\], \[\Delta {G^ \circ }_1\] and \[\Delta {G^ \circ }_2\] as below.
\[\Delta {G^ \circ }_{net} = - nF{E^ \circ }_{cell}\]
But here total electrons involved in both the reactions are (a+b). So,
 \[\Delta {G^ \circ }_{net} = - (a + b)F{E^ \circ }\]
\[\Delta {G^ \circ }_1 = - aF{E^ \circ }\]
\[\Delta {G^ \circ }_2 = - bF{E^ \circ }\]
Now put all these values in equation (2), we will get,
\[ - (a + b)F{E^ \circ }_{cell} = - aF{E^ \circ }_1 + ( - bF{E^ \circ }_2)\]
We can cancel out (-F) from both sides, the equation will be
\[(a + b){E^ \circ }_{cell} = a{E^ \circ }_1 + b{E^ \circ }_2\]
\[{E^ \circ }_{cell} = \dfrac{{a{E^ \circ }_1 + b{E^ \circ }_2}}{{(a + b)}}\]

Hence we can say that this is the value of the potential of the resultant cell made from the two reactions given in the question.


Note: Do not forget the minus sign in the formula to find Gibbs free energy which is shown in equation(1). Make sure that you do not need to add minus sign to the number of electrons involved in the reaction as the electron is negatively charged.