Question

Calculate the standard oxidation potentials of zinc electrode from the following data. The EMF of the cell:
${\text{Li|L}}{{\text{i}}^{\text{ + }}}{\text{(0}}{\text{.05 M)||Zn|Z}}{{\text{n}}^{{\text{2 + }}}}{\text{(0}}{\text{.5M),}}{{\text{E}}^{\text{o}}}_{{\text{(L}}{{\text{i}}^{\text{ + }}}{\text{|Li)}}}{\text{ = - 3}}{\text{.045V}}$ is 2.35 V at ${\text{2}}{{\text{5}}^{\text{o}}}{\text{C}}$.

Answer 1

The standard reduction potential can be calculated from the standard cell potential using the formula, ${{\text{E}}_{{\text{cell}}}}{\text{ = }}{{\text{E}}^{\text{o}}}_{{\text{cell}}}{\text{ - }}\dfrac{{{\text{0}}{\text{.0591}}}}{{\text{n}}}{\text{log}}\dfrac{{{\text{[product]}}}}{{{\text{[reactant]}}}}$.
Formula used:
${{\text{E}}_{{\text{cell}}}}{\text{ = }}{{\text{E}}^{\text{o}}}_{{\text{cell}}}{\text{ - }}\dfrac{{{\text{0}}{\text{.0591}}}}{{\text{n}}}{\text{log}}\dfrac{{{\text{[product]}}}}{{{\text{[reactant]}}}}$
Here, ${{\text{E}}_{{\text{cell}}}}$ = EMF of the cell
${{\text{E}}^ \circ }_{{\text{cell}}}$ = standard cell potential
n = number of electrons transferred

Complete step by step answer:
We can modify cell representation from the question as follows:
${\text{Li|L}}{{\text{i}}^{\text{ + }}}{\text{(0}}{\text{.05 M)||Z}}{{\text{n}}^{{\text{2 + }}}}{\text{(0}}{\text{.5M)|Zn}}$
From the above cell representation, we know that oxidation is taking place on the left side at the anode (Lithium) and reduction is taking place on the right side at the cathode (Zinc).
Now, we can write the half-cell reactions as follows:
Oxidation half-cell
${\text{2Li}} \to 2{\text{L}}{{\text{i}}^{\text{ + }}} + 2{{\text{e}}^{\text{ - }}}$
${{\text{E}}^{\text{o}}}_{{\text{(Li|L}}{{\text{i}}^{\text{ + }}}{\text{)}}}{\text{ = 3}}{\text{.045V}}$
In the question, we were given the standard reduction potential of lithium. So, we converted it into standard oxidation potential.The standard oxidation potential is equal to the negative of standard reduction potential.
${{\text{E}}^{\text{o}}}_{{\text{(L}}{{\text{i}}^{\text{ + }}}{\text{|Li)}}} = - {{\text{E}}^{\text{o}}}_{{\text{(L}}{{\text{i}}^{\text{ + }}}{\text{|Li)}}}$

Reduction half-cell:
${\text{Z}}{{\text{n}}^{{\text{2 + }}}}{\text{ + 2}}{{\text{e}}^{\text{ - }}} \to {\text{Zn}}$
Here, ${{\text{E}}^{\text{o}}}_{{\text{(Z}}{{\text{n}}^{{\text{2 + }}}}{\text{/Zn)}}}$= Standard reduction potential
${{\text{E}}^{\text{o}}}_{{\text{(Zn/Z}}{{\text{n}}^{{\text{2 + }}}}{\text{)}}}$= Standard oxidation potential

Net cell reaction:
${\text{2Li + Z}}{{\text{n}}^{{\text{2 + }}}} \to 2{\text{L}}{{\text{i}}^{\text{ + }}} + {\text{Zn}}$
Here, n=2 as two electrons participate in this redox reaction.
Now, we know that
${{\text{E}}^{\text{o}}}_{{\text{cell}}}{\text{ = }}{{\text{E}}^{\text{o}}}_{{\text{SOP}}}\left( {{\text{anode}}} \right){\text{ - }}{{\text{E}}^{\text{o}}}_{{\text{SOP}}}\left( {{\text{cathode}}} \right)$
Here, SOP=standard oxidation potential
To find the standard oxidation potential of the cathode (Zinc), we have to calculate the standard cell potential first (${{\text{E}}^{\text{o}}}_{{\text{cell}}}$).
Standard cell potential can be calculated by the Nernst equation which is given as follows:
${{\text{E}}_{{\text{cell}}}}{\text{ = }}{{\text{E}}^{\text{o}}}_{{\text{cell}}}{\text{ - }}\dfrac{{{\text{0}}{\text{.0591}}}}{{\text{n}}}{\text{log}}\dfrac{{\left[ {{\text{product}}} \right]}}{{\left[ {{\text{reactant}}} \right]}}$
By substituting the value we get:
${\text{2}}{\text{.35V = }}{{\text{E}}^{\text{o}}}_{{\text{cell}}}{\text{ - }}\dfrac{{{\text{0}}{\text{.0591}}}}{2}{\text{log}}\dfrac{{{{\left[ {{\text{L}}{{\text{i}}^{\text{ + }}}} \right]}^2}}}{{\left[ {{\text{Z}}{{\text{n}}^{{\text{2 + }}}}} \right]}}$
The concentration of product and reactant is given in the question. Substitute the concentrations in the formula.
${\text{2}}{\text{.35V = }}{{\text{E}}^{\text{o}}}_{{\text{cell}}}{\text{ - }}\dfrac{{{\text{0}}{\text{.0591}}}}{2}{\text{log}}\dfrac{{{{\left[ {{\text{0}}{\text{.05}}} \right]}^2}}}{{\left[ {{\text{0}}{\text{.5}}} \right]}}$
Now, by solving the right side of the equation, we get:
${\text{2}}{\text{.35V = }}{{\text{E}}^{\text{o}}}_{{\text{cell}}}{\text{ - 0}}{\text{.0295log}}\dfrac{{2.5 \times {{10}^{ - 3}}}}{{\left[ {{\text{0}}{\text{.5}}} \right]}}$
By dividing the numerical value associated with the log term we get:
${\text{2}}{\text{.35V = }}{{\text{E}}^{\text{o}}}_{{\text{cell}}}{\text{ - 0}}{\text{.0295 log}}\left( {{\text{5}} \times {{10}^{ - 3}}} \right)$
By takin the log of $0.005$ we get:
${\text{2}}{\text{.35V = }}{{\text{E}}^{\text{o}}}_{{\text{cell}}}{\text{ - 0}}{\text{.0295 }}\left( {{\text{ - 2}}{\text{.3}}} \right)$
By multiplying the numerical value on the right side of the equation:
${\text{2}}{\text{.35V = }}{{\text{E}}^{\text{o}}}_{{\text{cell}}} + {\text{ 0}}{\text{.068}}$
By taking the numerical value on left side we will get the standard cell potential:
${\text{2}}{\text{.35 - 0}}{\text{.068 = }}{{\text{E}}^{\text{o}}}_{{\text{cell}}}$
By subtracting the value we get:
${{\text{E}}^{\text{o}}}_{{\text{cell}}}{\text{ = 2}}{\text{.282 V}}$
We know that the standard cell potential is given as :
${{\text{E}}^{\text{o}}}_{{\text{cell}}}{\text{ = }}{{\text{E}}^{\text{o}}}_{{\text{SOP}}}{\text{(anode) - }}{{\text{E}}^{\text{o}}}_{{\text{SOP}}}{\text{(cathode)}}$
By substituting the value we get:
${\text{2}}{\text{.282V = 3}}{\text{.045V - }}{{\text{E}}^{\text{o}}}_{{\text{SOP}}}{\text{(cathode)}}$
By taking the numerical value on one side:
${{\text{E}}^{\text{o}}}_{{\text{SOP}}}{\text{(cathode) = 3}}{\text{.045V - 2}}{\text{.282V}}$
By subtracting we get:
${{\text{E}}^{\text{o}}}_{{\text{SOP}}}{\text{(cathode) = 0}}{\text{.763V}}$
Therefore, the standard oxidation potential of the Zinc electrode for the following data is $0.763$ V.

Note:
Standard reduction potential is always in negative and standard oxidation potential is always positive.
In the Nernst equation, the product and reactant are finalized from the net cell reaction and not the cell representation.