Question

The electrode potential of oxidation half-cell is:
(A) Is independent of the concentration of ions in the cell
(B) Decreases with decreased concentration of ions in the cells
(C) Decreases with increased concentration of ions in the cells
(D) None of these

Answer 1

The reaction the occurs at the oxidation half cell can be given as
${M_{(s)}} \to {M^{n + }}_{(aq)} + n{e^ - }$
Nernst’s equation gives the actual potential of the cell depending upon the concentration of the ions in the solution.

Complete step by step solution:
Here, we will find the relation of the electrode potential with the concentration of the ions present in the cell.
- We know that there are two half-cells in the galvanic cell, oxidation half-cell and reduction half-cell. Oxidation reaction occurs at the oxidation half-cell. So, we can describe that in reaction form as
${M_{(s)}} \to {M^{n + }}_{(aq)} + n{e^ - }$
- Now suppose that ${E_{cell}}$ is the potential of the oxidation half-cell. We will use Nernst’s equation to find out the relation of concentration of ions with the potential of oxidation half-cell. So, we can write that according to Nernst equation,
${E_{cell}} = {E^0}_{cell} - \dfrac{{2.303RT}}{{nF}}\log Q$
Putting the values of the constants R, T and F into that equation will give
${E_{cell}} = {E^0}_{cell} - \dfrac{{0.0591}}{n}\log Q$ ……(1)
Now, Q is the equilibrium constant of the oxidation half reaction.
So, equilibrium constant Q = $\dfrac{{{\text{[Product]}}}}{{{\text{[Reactant]}}}} = \dfrac{{[{M^{n + }}]}}{{[M]}}$
Now, concentration of metal M in the solution can be taken as constant. So, equation (1) can be rewritten as
${E_{cell}} = {E^0}_{cell} - \dfrac{{0.0591}}{n}\log [{M^{n + }}]$
We know that as the value of ${M^{n + }}$ increases, the resultant ${E_{cell}}$ value will decrease. Thus, we can conclude that electrode potential of the oxidation half-cell decreases with increase in the concentration of ions in the solution.

Therefore the correct answer of this question is (C).

Note: Remember that change in the concentration of any other ions present in the reaction mixture that does not alter the equilibrium constant of the particular cell reaction does not affect the potential of the cell.