Question

Calculate the equilibrium constant of the reaction:
$Cd_{(aq)}^{2 + } + Z{n_{(s)}} \to Zn_{(aq)}^{2 + } + C{d_{(s)}}$
If $E_{C{d^{2 + }}/Cd}^o = - 0.403V$ and $E_{Z{n^{2 + }}/Zn}^o = - 0.763V$
A . $K = 1.45 \times {10^{12}}$
B. $K = 4.25 \times {10^{14}}$
C. $K = 1.45 \times {10^{18}}$
D. $K = 14.4 \times {10^{11}}$

Answer 1

Nernst equation at equilibrium gets modified, since the rate of forward and backward reaction is equal and hence no net potential is developed. Substitute the value for equilibrium constant as it is the ratio of concentration to concentration of products and use this modified equation to find the value of equilibrium constant.

Formula Used:
1. $E_{cell}^o = E_{C{d^{2 + }}/Cd}^o - E_{Z{n^{2 + }}/Zn}^o$
2. $E_{cell}^o = \dfrac{{0.059}}{n}{K_c}$
3. $Anti\log [\log {K_c}] = {K_c}$
Where, $E_{cell}^o$ = Standard electrode potential
$E_{C{d^{2 + }}/Cd}^o$ = Electrode potential of cadmium
$E_{Z{n^{2 + }}/Zn}^o$ = Electrode potential of zinc
$n$ = Transfer of electrons takes place in the reaction
${K_c}$ = Equilibrium constant of the reaction

Complete step by step answer:
In the above reaction, cadmium ion $C{d^{2 + }}$ undergoes reduction in the presence of zinc $Zn$ to give $Cd$ and $Z{n^{2 + }}$ .
The Complete reaction is:
$Cd_{(aq)}^{2 + } + Z{n_{(s)}} \to Zn_{(aq)}^{2 + } + C{d_{(s)}}$
The value of equilibrium constant will be:
${K_c} = \dfrac{{[Cd][Z{n^{2 + }}]}}{{[C{d^{2 + }}][Zn]}}$
Remember this value, as we will need to substitute in the upcoming equations
Let us write the general formula for Nernst reaction at standard conditions
${E_{cell}} = E_{cell}^o + \dfrac{{0.059}}{n}\log \dfrac{{{\text{concentration of products}}}}{{{\text{concentration of reactants}}}}$
where, ${E_{cell}} = {\text{ Electrode potential}}$
At equilibrium, ${E_{cell}} = 0$ since the rate of forward reaction will be equal to the rate of backward reaction , hence no net current will flow.
$\log \dfrac{{{\text{concentration of product}}}}{{{\text{concentration of reactant}}}} = \log \dfrac{{[Cd][Z{n^{2 + }}]}}{{[Zn][C{d^{2 + }}]}}$
we determined the value of this ratio as the equilibrium constant:
hence, we can state it as:
${K_c} = \dfrac{{[Cd][Z{n^{2 + }}]}}{{[C{d^{2 + }}][Zn]}}$
Substituting all the above information in the Nernst equation we get,
$E_{cell}^o = \dfrac{{0.059}}{n}{K_c}$
We know that,
$E_{cell}^o = E_{C{d^{2 + }}/Cd}^o - E_{Z{n^{2 + }}/Zn}^o$
and from the question we have the information as, $E_{C{d^{2 + }}/Cd}^o = - 0.403V$ and $E_{Z{n^{2 + }}/Zn}^o = - 0.763V$
hence substituting these values we get,
$E_{cell}^o = [ - 0.403 - ( - 0.763)]V \Rightarrow 0.360V$
To find out the value of number of electrons,
Consider the half cell reactions,
$C{d^{2 + }} + 2{e^ - } \to Cd$
And also,
$Zn \to Z{n^{2 + }} + 2{e^ - }$
Hence the value of number of transfer electrons $(n) = 2$
Substituting the value for standard electrode potential and number of electrons in the Nernst equation, we get this equation.
$0.360 = \dfrac{{0.059}}{2}\log {K_c}$
On further Solving this equation we will get,
$\log {K_c} = 12.20$
to find out the value of equilibrium constant, we will have to take antilog,
So, by applying antilog both the sides:
$\Rightarrow Anti\log [\log {K_c}] = {K_c}$
On substituting the values we will get:
$Anti\log [12.20] = {K_c}$
So, the value of equilibrium constant will be:
${K_c} = 1.45 \times {10^{12}}$

So, the correct answer is Option A.

Note: Please note the general Nernst equation for any reaction is
${E_{cell}} = E_{cell}^o + \dfrac{{2.303RT}}{{nF}}\log \dfrac{{{\text{concentration of products}}}}{{{\text{concentration of reactants}}}}$
Where, $R$ = Gas constant
$T$ = Temperature
$F$ = Faraday constant
$n$ = Number of transferred electrons
But, we took the equation as:
${E_{cell}} = E_{cell}^o + \dfrac{{0.059}}{n}\log \dfrac{{{\text{concentration of products}}}}{{{\text{concentration of reactants}}}}$
This is because standard electrode potential was given to us, at STP values and hence standard temperature value was substituted $T = 298K$ .