Question

What will be the E.M.F of the cell containing \[Z{n^{2 + }}/Zn\] and hydrogen gas electrode half cells at \[{25^ \circ }C\] if \[\left[ {Z{n^{2 + }}} \right]{\text{ = 0}}{\text{.24 M}}\], \[\left[ {{H^ + }} \right]{\text{ = 1}}{\text{.6 M}}\] and \[{p_{{H_2}}}{\text{ = 1}}{\text{.8 atm}}\]. It is also given that \[{E^ \circ }_{Zn}{\text{ = 0}}{\text{.763 V}}\] , also find \[\Delta G\].

Answer 1

Hint: We will find the E.M.F for cathode and anode separately with the help of Nernst equation of cell. After finding E.M.F of each cell we will find the E.M.F of the whole cell by subtracting E.M.F of anode from E.M.F of anode. Also the value of \[{E^ \circ }\] for hydrogen gas is \[0.00{\text{ V}}\].
Formula Used:
\[{E_{{\text{cathode/anode}}}}{\text{ = }}{{\text{E}}^ \circ }_{{\text{cathode/anode}}}{\text{ - }}\dfrac{{0.0591}}{n}{\text{ log}}\left[ Q \right]\]

Complete answer:
The cell consists of a zinc electrode and a hydrogen electrode. Zinc electrodes will undergo oxidation and hereby hydrogen will act as a reduction electrode. Therefore we can say that zinc electrodes act as anode and hydrogen electrodes act as cathode. We will write half-cell reactions for both electrodes and then find their respective E.M.F as,
At anode:
At anode oxidation of zinc takes place as,
\[Zn(s){\text{ }} \to {\text{ Z}}{{\text{n}}^{2 + }}{\text{(aq}}{\text{.) + 2e}}\]
Also we are given with \[{E^ \circ }_{Zn}{\text{ = 0}}{\text{.763 V}}\] , from Nernst equation we know that,
\[{E_{{\text{cathode/anode}}}}{\text{ = }}{{\text{E}}^ \circ }_{{\text{cathode/anode}}}{\text{ - }}\dfrac{{0.0591}}{n}{\text{ log}}\left[ Q \right]\]
Therefore we can write as,
\[{E_{Zn}}{\text{ = }}{{\text{E}}^ \circ }_{Zn}{\text{ - }}\dfrac{{0.0591}}{n}{\text{ log}}\left[ Q \right]\]
Here n is the number of electrons which are involved in the reaction and Q is the reaction quotient of the reaction. We observe that two electrons take place in the reaction. Thus the Nernst equation can be deduced as,
\[{E_{Zn}}{\text{ = }}{{\text{E}}^ \circ }_{Zn}{\text{ - }}\dfrac{{0.0591}}{2}{\text{ log}}\left[ {Z{n^{2 + }}} \right]\]
\[{E_{Zn}}{\text{ = 0}}{\text{.763 - }}\dfrac{{0.0591}}{2}{\text{ log}}\left[ {0.24} \right]\] , using values \[\left[ {Z{n^{2 + }}} \right]{\text{ = 0}}{\text{.24 M}}\] and \[{E^ \circ }_{Zn}{\text{ = 0}}{\text{.763 V}}\]
\[{E_{Zn}}{\text{ = 0}}{\text{.781 V}}\]
At cathode:
Similarly we can find for cathode electrode as,
\[{\text{2}}{{\text{H}}^ + }{\text{(aq}}{\text{.) + 2e }} \to {\text{ }}{{\text{H}}_2}{\text{(g) }}\]
We can write Nernst equation for above cell reaction as,
\[{E_{{H_2}}}{\text{ = }}{{\text{E}}^ \circ }_{{H_2}}{\text{ - }}\dfrac{{0.0591}}{n}{\text{ log}}\left[ Q \right]\]
The reaction quotient for reaction can be written as,
\[Q{\text{ = }}\dfrac{{{p_{{H_2}}}}}{{{{\left[ {{H^ + }} \right]}^2}}}\]
According to question, \[{p_{{H_2}}}{\text{ = 1}}{\text{.8 atm}}\] and \[\left[ {{H^ + }} \right]{\text{ = 1}}{\text{.6 M}}\] we get the reaction quotient as,
\[Q{\text{ = }}\dfrac{{1.8}}{{1.6{\text{ }} \times {\text{ 1}}{\text{.6}}}}\]
\[Q{\text{ = 0}}{\text{.703}}\]
Hence the Nernst equation can be written as,
\[{E_{{H_2}}}{\text{ = }}{{\text{E}}^ \circ }_{{H_2}}{\text{ - }}\dfrac{{0.0591}}{n}{\text{ log}}\left[ Q \right]\]
Here the number of electrons is two and also \[{E_{{H_2}}}{\text{ = 0}}{\text{.00 V}}\].
\[{E_{{H_2}}}{\text{ = 0 - }}\dfrac{{0.0591}}{2}{\text{ log}}\left[ {0.703} \right]\]
\[{E_{{H_2}}}{\text{ = 0}}{\text{.004 V}}\]
Thus we get E.M.F for cathode and anode reaction. Now we will calculate E.M.F for whole cell as,
\[{E_{cell}}{\text{ = }}{{\text{E}}_{cathode}}{\text{ - }}{{\text{E}}_{anode}}\]
\[{E_{cell}}{\text{ = }}\left( {{\text{0}}{\text{.004 - 0}}{\text{.703}}} \right){\text{ V}}\]
\[{E_{cell}}{\text{ = 0}}{\text{.699 V}}\]
We can also calculate the value of \[\Delta G\] for the cell as,
\[\Delta G{\text{ = - nF}}{{\text{E}}_{cell}}\]
\[\Delta G{\text{ = - 2 }} \times {\text{ 96500 }} \times {\text{ 0}}{\text{.699 J}}\]
\[\Delta G{\text{ = - 135 kJ}}\]
Hence the EMF of the cell is 0.699V.

Note:
The value of one faraday is equal to \[96500{\text{ C}}\]. We can also find E.M.F of the cell by first finding the \[{E^ \circ }_{cell}\] and then by applying the Nernst equation. The standard reduction potential of a hydrogen cell is always zero, this is why it can be used as a reference electrode.