Question

Calculate the e.m.f. of the following cell reaction at 298 K.
\[2C{r_{(s)}} + \mathop {3F{e^{2 + }}_{(aq)}}\limits_{0.05M} \to \mathop {2C{r^{3 + }}_{(aq)}}\limits_{0.005M} + 3F{e_{(s)}}\]
Given : $E_{cell}^ \circ $ = 0.30 V

Answer 1

Hint: we can find the potential of the cell using the Nernst equation.
\[{E_{cell}} = E_{cell}^ \circ - \dfrac{{RT}}{{nF}}\ln \dfrac{{[{\text{Products]}}}}{{[{\text{Reactants]}}}}\]

Complete step by step solution:
The reaction is given here. We will try to write the oxidation and reduction half reactions in order to find the number of electrons involved in the reaction.
Oxidation half reaction : $2Cr \to 2C{r^{3 + }} + 6{e^ - }$
Reduction half reaction : $3F{e^{2 + }} + 6{e^ - } \to 3Fe$
Thus, we can say that a total 6 electrons are involved in the given reaction.
- We are provided with the standard potential of the given cell which is given as 0.30 V.
- We will now find the potential of the cell at given concentrations. We will use the Nernst equation to find the potential of the cell. The Nernst equation can be given as
\[{E_{cell}} = E_{cell}^ \circ - \dfrac{{RT}}{{nF}}\ln \dfrac{{[{\text{Products]}}}}{{[{\text{Reactants]}}}}\]
Here, we know that universal gas constant,R = 8.314 $J/molK$
Temperature T = 298 K
Faraday constant F = 96500 C
number of electrons involved in the reaction n = 6
And we can take a log at the base of 10 by multiplying the natural log by 2.303.
So, we can rewrite the above equation as
\[{E_{cell}} = E_{cell}^ \circ - \dfrac{{(8.314)(298)(2.303)}}{{(6)(96500)}}\log \dfrac{{{{[C{r^{3 + }}]}^2}}}{{{{[F{e^{2 + }}]}^3}}}\]
Here we can neglect the concentration of solids. Thus, we just take into consideration the concentration of species in aqueous medium. So, we can write that
\[{E_{cell}} = 0.30 - 0.009854\log \dfrac{{{{[0.005]}^2}}}{{{{[0.05]}^3}}} = 0.30 - 0.009854\log (0.2)\]
Thus, we obtain that
\[{E_{cell}} = 0.30 - 0.009854( - 0.6989) = 0.30 - 0.006887 = 0.2931V\]

Thus, we obtained that the potential of the cell will be 0.29 V.

Note: Note that we do not need to write the concentrations of the solids present in the reactant or product side in the reaction while writing the Nernst equation. The standard potential of the cell can be found out using the standard potentials of the given half cells.