Question

The standard reduction potential for $C{u^{2 + }}/Cu$ is +0.34V. What will be the reduction potential at pH=14? [Given ${K_{sp}}$ of $Cu{(OH)_2}$ is $1 \times {10^{ - 19}}$]
(A) 2.2V
(B) 3.4V
(C) -0.22V
(D) -2.2V

Answer 1

The Nernst’s equation gives the relation between the concentration of the ions in the solutions and the potential which can be given as
${E_{cell}} = {E^0} - \dfrac{{0.0591}}{n}\log \dfrac{{[{\text{Product]}}}}{{{\text{[Reactant]}}}}$
pH and pOH are related by the equation pH + pOH =14.

Complete step by step solution:
We are given the solubility product of $Cu{(OH)_2}$ alongside the standard reduction potential of the reduction half-cell. We can find the resultant potential of the given cell if we find the value of the concentration of the copper ions present in the solution. We will obtain the concentration of copper ions in the solution by the use of solubility products.
- Solubility product is the product of the concentration of the ions present in the solution upon its dissolution. So, we can write that
$Cu{(OH)_2} \to C{u^{2 + }} + 2O{H^ - }$
We are given that pH of the solution is 14. We know that the sum of pH and pOH of the solution is always 14. So, we can write that
pOH = 14 – pH = 14 – 14 = 0
We also know that
$pOH = - \log [O{H^ - }]$
But we obtained that pOH=0 . So,
$0 = - \log [O{H^ - }]$
$[O{H^ - }] = 1M$
Now, we can say that
${\text{Solubility product = [C}}{{\text{u}}^{2 + }}{\text{][O}}{{\text{H}}^ - }{]^2}$
Putting the available values in the above equation, we get
$1 \times {10^{ - 19}} = [C{u^{2 + }}]{[1]^2}$
$[C{u^{2 + }}] = {10^{ - 19}}M$
Now, we obtained that concentration of copper ions present in the reaction. Now, we will write the reaction occurring in the reduction half-cell.
$C{u^{2 + }}_{(aq)} + 2{e^ - } \to C{u_{(s)}}$
Now, we will use Nernst’s equation here. It can be written as
${E_{cell}} = {E^0} - \dfrac{{0.0591}}{n}\log \dfrac{{[{\text{Product]}}}}{{{\text{[Reactant]}}}}$
Here, n is the number of electrons involved in the reaction.
Putting the available values into the above equation, we get
${E_{cell}} = 0.34 - \dfrac{{0.0591}}{2}\log \left[ {\dfrac{1}{{{{10}^{ - 19}}}}} \right]$
${E_{cell}} = 0.34 - \dfrac{{0.0591}}{2} \times 19$
${E_{cell}} = - 0.22V$
Thus, we obtained that the reduction potential of the given cell at pH=14 will be -0.22V.

So, the correct answer to this question is (C).

Note: Remember that in Nernst's equation, we took the concentration of the Cu metal as unity because only the concentration of the ions will affect the potential of the cell. Do not forget that as two hydroxide ions are produced from the dissociation of $Cu{(OH)_2}$, its solubility product will be equal to $[C{u^{2 + }}]{[O{H^ - }]^2}$ .