Question

Calculate the reduction potential of a half cell consisting of a platinum electrode immersed in $2.0M$ $F{e^{ + 2}}$ and $0.02M$ $F{e^{ + 3}}$solution. Given, $E_{F{e^{ + 2}}/F{e^{ + 3}}}^ \circ = 0.771V$

Answer 1

The Nernst equation is employed to calculate the voltage of an electrochemical cell or to seek out the concentration of 1 of the components of the cell. The Nernst equation relates the equilibrium cell potential (also known as the Nernst potential) to its concentration gradient across a membrane.

Complete step by step answer:
The equation may be written as follows:
${E_{cell}}\; = {\text{ }}{E^0}_{cell}\; - {\text{ }}\left( {RT/nF} \right)lnQ$
${E_{cell}}\; =$cell potential under nonstandard conditions (V)
${E^0}_{cell}\; =$cell potential under standard conditions
$R{\text{ }} =$gas constant, which is $8.31$(volt-coulomb)/(mol-K)
$T{\text{ }} =$Temperature (Kelvin)
$n{\text{ }} =$number of moles of electrons exchanged in an electrochemical reaction (unit-mol)
$F{\text{ }} =$expressed as Faraday's constant, $96500$ coulombs/mol
$Q{\text{ }} =$the reaction quotient, which is the equilibrium expression with the initial concentrations rather than the equilibrium concentrations
The Nernst equation can also be represented differently:
${E_{cell}}\; = {\text{ }}{E^0}_{cell}\; - {\text{ }}\left( {2.303*RT/nF} \right)logQ$
at $298$K, ${E_{cell}}\; = {\text{ }}{E^0}_{cell}\; - {\text{ }}\left( {0.0591{\text{ }}V/n} \right)log{\text{ }}Q$
The half reaction of iron at the electrode is as follows:
$F{e^{ + 3}} + {e^ - }\; \to F{e^{ + 2}}$
Given $E_{F{e^{ + 2}}/F{e^{ + 3}}}^ \circ = 0.771V$
The molar concentrations of the ferrous and ferric ions are given as follows:
$\left[ {F{e^{ + 2}}} \right] = 2.0M$
$\left[ {F{e^{ + 3}}} \right] = 0.02M$
Substituting the values in the Nernst equation as follows:
$E = {E^ \circ } - 0.0591\;log\dfrac{{\left[ {F{e^{ + 2}}} \right]}}{{[F{e^{ + 3}}]}}$
$= 0.771 - 0.0591log(\dfrac{2}{{0.02}})$
∴On solving, we get the electrode potential of a half cell of platinum electrode as :
$E = 0.6528volt$

Note: A half-reaction is defined as the incomplete transfer of electrons. In the oxidation half-reaction, a substance loses some free electrons. In the reduction half-reaction, a substance gains some free electrons. Although, in both the cases, none of the electrons completely transfer from one chemical to another.
Each half-reaction contains a standard reduction potential. The word "potential" comes from the fact that this value measures the potential a half-reaction has to create electricity. The half-reaction has a standard reduction potential which is measured for the reduction form of a half-reaction.