Question

Calculate the value of ${E_{cell}}$ at $298K$ for the following cell:
$Al/A{l^{3 + }}(0.01M)||S{n^{2 + }}(0.015M)/Sn$
${E^ \circ }_{A{l^{3 + }}/Al} = - 1.66V$ and ${E^ \circ }_{S{n^{2 + }}/Sn} = - 0.14V$

Answer 1

Hint:Here we are given with the ${E^ \circ }_{cell}$ of the reaction but we have to find the ${E_{cell}}$. So for the relation between ${E^ \circ }_{cell}$ and ${E_{cell}}$ we have an equation which is known as Nernst equation. First we will find the value of ${E^ \circ }_{cell}$ from the given data and then we have to apply Nernst equation to find ${E_{cell}}$ of the reaction.

Complete step by step answer:
First of all let us talk about how to find ${E^ \circ }_{cell}$ of the reaction when you know the values of ${E^ \circ }$ of its components i.e. the ions or atoms which are reduced or oxidised in the reaction.
Reduced ions or atoms are that in which there is decrease in the positive charge on the ion i.e. gain of electrons.
Oxidised ions or atoms are that in which there is an increase in the positive charge on the ion i.e. loss of electrons.
Now if in the reaction we know which is oxidised and which is reduced atom and is we also know the values of ${E^ \circ }$ of its components i.e. the ions or atoms. Then we can calculate the value of ${E^ \circ }_{cell}$ of the reaction as: \[\;{E^ \circ }_{cell} = \;{E^ \circ }_R - \;{E^ \circ }_L\] where \[{E^ \circ }_R\] is the ${E^ \circ }$ of the atom which is at right side and \[{E^ \circ }_L\] is the \[{E^ \circ }\] of the atom which is at left side. Here in the question the atom on the right side is $Sn$ and the atom on left side is $Al$. So the ${E^ \circ }_{cell}$ of the reaction will be ${E^ \circ }_{cell} = {E^ \circ }_R - {E^ \circ }_L$, where ${E^ \circ }_R = - 0.14V$ and ${E^ \circ }_l = - 1.66V$. So ${E^ \circ }_{cell} = - 0.14 - ( - 1.66) = 1.52V$.
And if we try to write down net cell reaction then it will be as:
$2Al + 3S{n^{2 + }} \to 2A{l^{3 + }} + 3Sn$.
Here in the reaction we have total $6$ transfer electrons.
Nernst equation: It is defined as the relation of ${E^ \circ }_{cell}$ and ${E_{cell}}$ which is as:
${E_{cell}} = {E^ \circ }_{cell} - \dfrac{{0.059}}{n}\log \dfrac{{{{[{A^{q + }}]}^p}}}{{{{[{B^{p + }}]}^q}}}$ if the equation is as $pA + q{B^{p + }} \to p{A^{q + }} + qB$ where $[{A^{q + }}]$ is the concentration of the ion which is oxidised and $[{B^{p + }}]$ is the concentration of the ion which is reduced in the reaction. And $p$ is the coefficient of the compound which is oxidised and $q$ is the coefficient of the compound which is reduced in the reaction and $n$ is the total number of electrons transferred.
Here in the question we are given with the reaction as $2Al + 3S{n^{2 + }} \to 2A{l^{3 + }} + 3Sn$. Here we have $A = Al,B = Sn,p = 2,q = 3,[A{l^{3 + }}] = 0.01M,[S{n^{2 + }}] = 0.015M$ and the value of $n$ is $6$
Substituting these values in the Nernst equation we get:
${E_{cell}} = 0.152 - \dfrac{{0.059}}{n}\log \dfrac{{{{[0.01]}^2}}}{{{{[0.015]}^3}}}$
Then if we solve further after taking log of this then log of the required term is $1.47$.

So the ${E_{cell}} = 1.52 - 0.01447 = 1.50553V$.

Note:
${E_{cell}}$ of a reaction is defined as electrode potential of the cell and ${E^ \circ }_{cell}$ of a reaction is defined as electrode potential measured at $1$ atmosphere pressure, $1$ molar solution at ${25^ \circ }C$ also known as standard electrode potential.