Question

Calculate the equilibrium constant for the reaction $Fe + CuSO_4 \rightleftharpoons FeSO_4 + Cu$ at \[{25^o}C\].
Given \[{E^0}(Fe/Fe^{2+}) = 0.44V,{E^0}(Cu/Cu^{2+}) = - 0.337V\]

Answer 1

Hint: We need to use Nernst Equation to get Equilibrium constant K.
We know, \[{E_{cell}} = 0\] and Q=K at equilibrium, now to proceed ahead, we need to find Standard EMF using below formula:
\[{E^0}_{cell} = {E^0}_{Cathode} - {E^0}_{Anode}\]
Then use Nernst equation formula, \[{E_{cell}} = {E^0}_{cell} - \dfrac{{0.059}}{n}\log Q\]

Complete step by step answer:
Firstly calculate EMF of cell in standard condition using below formula: \[{E^0}_{cell} = {E^0}_{Cathode} - {E^0}_{Anode}\]
In any cell, Reduction happens at Cathode, and oxidation happens at anode.
So in the given question Standard reduction potential values are given, for iron it is positive and for copper it is negative.
Thus Iron will be reduced as Standard reduction value is high, and copper will be oxidised.
So, using \[{E^0}_{cell} = {E^0}_{Cathode} - {E^0}_{Anode}\]
${E^0}_{cathode} = 0.44V$
${E^0}_{anode} = - 0.337V$
Substitute cathode and anode values to calculate standard EMF of cell:
$ {E^0}_{cell} = 0.44 - ( - 0.337) $
$= 0.777V$
At equilibrium, \[{E_{cell}} = 0\] and Q=K.
To get, number of electrons transferred, we can see from Oxidation Half reaction and reduction half reaction.
$F{e^{2 + }} + 2{e^ - } \to Fe $
$Cu \to C{u^{2 + }} + 2{e^ - } $

Thus, n=2.
Now use Nernst equation at \[{25^o}C\] ,
\[{E_{cell}} = {E^0}_{cell} - \dfrac{{0.059}}{n}\log Q\]
Substituting all values from above, we get
\[0 = 0.777 - \dfrac{{0.059}}{2}\log K\]
Take 0.777 on the other side of the equal sign, and then on both sides, the negative sign will get canceled out.
\[0.777 = \dfrac{{0.059}}{2}\log K\]
Now, multiply 2 with 0.777 and divide by 0.059, so on one side we will get log K,
\[\dfrac{{1.554}}{{0.059}} = \log K\]
Calculating the numerical side, we get value of log K
\[26.34 = \log K\]
Now, taking antilog on both sides, and rearranging we get
\[anti\log (\log (K)) = anti\log (26.34)\]
Now log and antilog cancel each other,
\[K = anti\log (26.34)\]
Now, directly we can look in log tables and get approximate value:
\[\therefore K = 2*{10^{26}}\]

Note:
Be careful in finding antilog values from the log table. Q changes to K at equilibrium and the cell comes to rest at equilibrium, means it does not function or no redox reaction takes place, or potential becomes 0 V. So the EMF for cell at equilibrium becomes 0 V.