Question

The reduction potential at pH= 14 for the $C{u^{2 + }}/Cu$
 couples is :
[Given, $E_{C{u^{2 + }}/Cu}^0 = 0.34V$; $K_{s p} C u(O H)_{2}=1 \times 10^{-19}$ ]
A. 0.34V
B. -0.34V
C. 0.22V
D. -0.22V

Answer 1

Hint: Using Nernst equation of reduction potential we will find out the reduction potential of the given chemical equation.

Complete step by step answer:-
Nernst equation: $E=E^{\circ}-\dfrac{k T}{F} \ln \frac{[\operatorname{Re} d]}{[O x]}$
Where
E= Reduction potential of the electrochemical process
E$^\circ$= Standard half reduction potential
k= Constant(R/n)
T= temperature in kelvins
F= Faraday's constant
Let's solve by step by step:
Redox potential: it is also known as reduction and oxidation potential.Redox potential is a measure of the tendency of a chemical species to acquire electrons from or lose electrons to an electrode and thereby be reduced or oxidized respectively.
As we know the given pH=14,
$\therefore$ pOH=0 and we can see [$H^+$] = 1 M
Now $\left[\mathrm{Cu}^{2+}\right]\left[\mathrm{OH}^{-}\right]^{2}=K_{w}=1 \times 10^{-19}$
$\left[\mathrm{Cu}^{2+}\right]=1 \times 10^{-19} \mathrm{M}$
For the half-reaction will be,
$Cu^{2+}+2 e \rightarrow Cu$
Like how the convergence of a hydrogen particle decides the corrosiveness or pH of a fluid arrangement, the inclination of electron moves between compound animal groups and a terminal decides the redox capability of an anode couple. Like pH, redox potential speaks to how effectively electrons are moved to or from species in the arrangement. Redox potential portrays the capacity under the particular state of a compound animal to lose or pick up electrons rather than the measure of electrons accessible for oxidation or decrease.
Then
$=0.34-20.0591 \log \dfrac{1}{10^{-19}}$
$=-0.22 V$

Hence the final answer will be option D that is -0.22V.

Note:-
Students generally make mistakes in determining the oxidation element and pure element. They mostly mis-interpret the sign taken in the Nernst Equation and calculate the wrong answer.