Question

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

Answer 1

In electrochemistry, ${E_{cell}}$ is the measure of the potential difference between the two half cells. This potential difference is the reason that electrons flow from one half cell to another. It can be calculated using the Nernst equation.
Formula Used: ${E_{cell}} = {E^ \circ }_{cell} - \dfrac{{0.059}}{n}\log \dfrac{{{{\left[ {A{l^{3 + }}} \right]}^2}}}{{{{\left[ {S{n^{2 + }}} \right]}^3}}}$
Where n is the number of electrons transferred and ${E^ \circ }_{cell} = {E^o}_{oxidation} + {E^o}_{reduction}$

Complete step by step answer:
The reaction that takes place is as follows,
$2Al + 3S{n^{2 + }} \to 2A{l^{3 + }} + 3Sn$
In the above reaction, aluminium is undergoing oxidation and tin that is Sn is undergoing reduction. Aluminium is undergoing oxidation because the oxidation number of aluminium is increasing and tin is undergoing reduction because the oxidation number of tin is decreasing. The net transfer taking place is of 6 moles of electrons. Therefore, the value of n factor for this reaction is 6.
It is given that, ${E^ \circ }_{A{l^{3 + }}/Al} = - 1.66{\text{ V}}$
${E^ \circ }_{Al/A{l^{3 + }}} = 1.66{\text{ V}}$
${E^ \circ }_{S{n^{2 + }}/Sn} = - 0.14{\text{ V}}$
${E^ \circ }_{cell} = {E^o}_{oxidation} + {E^o}_{reduction}$
${E^o}_{cell} = 1.66 + \left( { - 0.14} \right) = 1.52{\text{ V}}$
The concentration of aluminium ion is $0.01{\text{ M}}$. The concentration of $S{n^{2 + }}$ ion is $0.015{\text{ M}}$.
The value of ${E^ \circ }_{cell}$ found is $1.52{\text{ V}}$.
Now to find ${E_{cell}}$, we will use Nernst equation. According to the Nernst equation at 298 K,
${E_{cell}} = {E^ \circ }_{cell} - \dfrac{{0.059}}{n}\log \dfrac{{{{\left[ {A{l^{3 + }}} \right]}^2}}}{{{{\left[ {S{n^{2 + }}} \right]}^3}}}$
${E_{cell}} = 1.52{\text{ }} - \;\dfrac{{0.059}}{6}\log \dfrac{{{{\left[ {0.01} \right]}^2}}}{{{{\left[ {0.015} \right]}^3}}}$
$\Rightarrow \;1.52 - 0.01447$
${E_{cell}} = 1.5055{\text{ V}}$

The answer of the given question is $1.5055{\text{ V}}$.
Note:
Nernst equation is the equation that relates half or full cell reaction to temperature concentration of the species and the standard electrode potential. These species are the ones who undergo oxidation and reduction. Nernst's equation was named after Walther Nernst who was a German physical chemist.