Question

The SRP for $S^{2-}\left(aq\right)$in CuS/Cu half-cell is:
Given: $K_{sp}\left(CuS\right)={10}^{-35}\text{and }{E}_{c{u}^{2+}/cu}^{\circ }=0.34V$
A. $1.634V$
B. $1.034V$
C. $-0.694V$
D. $0.394V$

Answer 1

Hint:SRP stands for standard reduction potential. In this question we will use the Nernst equation, to find the SRP of the given half-cell. But we know that for the Nernst equation, we will even require the $[C{u}^{2+}]$. This we will calculate using the value of $K_{sp}$ given in the question.

Complete step-by-step answer:As we have already discussed that to find the SRP using Nernst equation, we will have to find out the concentration of copper (II) ions. This we will calculate using the concept of Solubility product.
Let us see the dissociation of the given salt CuS (copper sulfide) in the solution
$CuSC{u}^{2+}+S^{2-}$
$K_{SP}$or solubility product is defined as the product of the concentration of the ions of the salt in its saturated solution at a given temperature raised to the power of number of ions produced by the dissociation of one mole of salt.
Hence, $K_{sp}$for the above dissociation can be written as the product of $[C{u}^{2+}]\mathrm{\&}[S^{2-}]$and since only one mole of both ions are produced, the power to which they will be raised will be 1.
Therefore, $K_{sp}=[C{u}^{2+}][S^{2-}]$
From the above equation, we get that $[C{u}^{2+}]={\displaystyle \frac{K_{sp}}{[S^{2-}]}}$
Now, if we consider the given reaction in the half-cell. In question we are asked to calculate the SRP of CuS/Cu. So, the Reaction will be
$C{u}^{2+}+2e^{-}\to Cu$
We can see that reduction is occurring in the above reaction as the copper (II) ion is gaining two electrons to become copper. Therefore, this half-cell will be termed as reduction half-cell.
Now, to find the SRP we will have to use the Nernst equation.
We know Nernst equation for a reduction half-cell will be written as
$E_{M^{n+}/M}=E_{M^{n+}/M}^{\circ }-{\displaystyle \frac{0.059}{n}}\log {\displaystyle \frac{1}{[M^{n+}]}}$
Here n= number of mole of electrons involved.
Therefore, Nernst equation for the given reduction-half cell can be written as
$E_{C{u}^{2+}/Cu}=E_{{}_{C{u}^{2+}/Cu}}^{\circ }-{\displaystyle \frac{0.059}{n}}\log {\displaystyle \frac{1}{[C{u}^{2+}]}}$ (equation 1)
Substituting the value of $[C{u}^{2+}]={\displaystyle \frac{K_{sp}}{[S^{2-}]}}$in the above equation 1 we get,
$E_{C{u}^{2+}/Cu}=E_{{}_{C{u}^{2+}/Cu}}^{\circ }-{\displaystyle \frac{0.059}{n}}\log {\displaystyle \frac{[S^{2-}]}{K_{sp}}}$
The above equation can be modified as $E_{C{u}^{2+}/Cu}=E_{{}_{C{u}^{2+}/Cu}}^{\circ }+{\displaystyle \frac{0.059}{n}}\log {\displaystyle \frac{K_{sp}}{[S^{2-}]}}$ (equation 2)
Now in the question few values are given to us as;
$\begin{array}{rl}& E_{C{u}^{2+}/Cu}^{\circ }=0.34V\\ & K_{sp}={10}^{-35}\\ & n=2\left(\text{as 2 moles of electrons are involved in the reaction}\right)\end{array}$
Substituting, all these values in the above equation 2, we get
$\begin{array}{rl}& E_{C{u}^{2+}/Cu}=0.34+{\displaystyle \frac{0.059}{2}}\log {\displaystyle \frac{{10}^{-35}}{1}}\\ & \Rightarrow E_{C{u}^{2+}/Cu}=0.34+\left(0.0295\right)(-35\log 10)\\ & \Rightarrow E_{C{u}^{2+}/Cu}=0.34+\left(0.0295\right)(-35)\\ & \Rightarrow E_{C{u}^{2+}/Cu}=0.34+(-1.0325)\\ & \Rightarrow E_{C{u}^{2+}/Cu}=0.34-1.0325\\ & \Rightarrow E_{C{u}^{2+}/Cu}=-0.695\end{array}$
Hence the SRP is -0.695 which is closest to option C.

Hence, the correct option is option C.

Note:We have solved the logarithm in the above equation in the following manner. As per the basic law of logarithm $\log {10}^{-n}$can be written as $-n\times \log 10$. We know that log 10 =1. Therefore, $\log {10}^{-n}=-n\log 10=-n\times 1=-n$. Similarly, in the above calculations $(\log {10}^{-35}=-35)$. We should remember that the standard electrode potential is measured when concentration of electrolyte is 1 M and temperature is 298K. Always be careful while determining the value of n in the Nernst equation.