Question

What is the single electrode potential of a half-cell for a zinc electrode dipped in $0.01M\text{ }ZnS{{O}_{4}}$ solution at $25{}^\circ C$? The standard electrode potential of a ${Zn}/{Z{{n}^{2+}}}\;$ system is 0.763$V$ at $25{}^\circ C$.

Answer 1

Hint: Analyze what is happening in the problem carefully, we have to find the single electrode potential of the oxidation half of the reaction. We can use a derivative of the Nernst equations and arrive at the answer.

Complete step by step solution:
Zinc is considered to be a good reducing agent due to the fact that it is a metal. Metals tend to lose electrons to another species and reduce it. In this process, metals themselves get oxidized. Hence, as zinc here is going from $Zn$ to $Z{{n}^{2+}}$, it is getting oxidized. Therefore, we will use the ${{E}_{oxi}}$ form of the Nernst equation.
The Nernst equation is used to find out the cell potential under non-standard conditions if the potential under standard conditions is given.
The Nernst equation is as follows:
\[{{E}_{oxi}}={{E}^{0}}_{oxi}-\dfrac{0.0592}{n}{{\log }_{10}}[ion]\]
Where,
${{E}_{oxi}}$= electrode potential
${{E}^{0}}_{oxi}$= standard electrode potential
$n$= number of moles of electrons exchanged
[ion] = concentration of ions
Here, the ${{E}^{0}}_{oxi}$ is the standard electrode potential, it is obtained when the concentration of all the species involved in the half-cell is 1M, at a certain temperature. If the concentration is anything other than 1M, we have to find out the electrode potential (${{E}_{oxi}}$) using the standard electrode potential.
Thus, from the given information, we know that
${{E}^{0}}_{oxi}$= 0.763 $V$
$n$= 2 (from ${Zn}/{Z{{n}^{2+}}}\;$)
[ion] = 0.01$M$
Therefore, plugging in all the values in the equation, we get:
\[{{E}_{oxi}}=0.763-\dfrac{0.0592}{2}{{\log }_{10}}(0.01)\]
${{E}_{oxi}}=0.763-[(\dfrac{0.0592}{2})\times (-2)]$
${{E}_{oxi}}=0.763+0.0592$
${{E}_{oxi}}=0.8222$

Hence, the answer for this equation is ${{E}_{oxi}}=0.8222$

Additional information:
The Nernst equation has many forms and any one of them can be used as per the requirements of the sums. If the reaction is a reduction half reaction and the standard electrode(reduction) potential is given, then it can be converted to the standard oxidation potential and used in the same formula.

Note: Always check whether the reaction is an oxidation half reaction or a reduction half reaction before applying any formulae. This may cause confusion regarding which sign should be applied. An easy way to remember this is OIL RIG whose full form is ‘Oxidation Is Losing (electrons), Reduction Is Gaining (electrons)’. Here the electrode potential is given as ${Zn}/{Z{{n}^{2+}}}\;$ where zinc is losing electrons.

Hence, it is an oxidation half reaction.