Question

Electrode potential of cadmium is -0.4V and electrode potential of cadmium is -0.74V. \[[C{{d}^{2+}}]\,=\,0.1M\] and $[C{{r}^{3+}}]\,\,=\,0.01M$. Calculate the ${{E}_{cell}}$ and ${{E}^{\circ }}_{cell}$.

Answer 1

Hint: In the given question we need to calculate ${{E}_{cell}}$ and ${{E}^{\circ }}_{cell}$. For this we will use Nernst equation.
Nernst equation: ${{E}_{cell}}\,=\,{{E}^{\circ }}_{cell}\,-\,\dfrac{2.303RT}{nF}\,\log \dfrac{[oxidised\,state]}{[reduced\,state]}$
Where n = number of electrons transferred
              F = Faraday

Complete step by step solution:
It is given that electrode potential of cadmium is -0.4V and electrode potential of cadmium is – 0.74V. The required concentration is also given. We need to calculate ${{E}_{cell}}$ and ${{E}^{\circ }}_{cell}$.
First we will calculate the value of ${{E}^{\circ }}_{cell}$.
${{E}^{\circ }}_{cell}$ = Electrode potential of reduction – Electrode potential of oxidation

${{E}^{\circ }}_{cell}$ = Electrode potential of cadmium – Electrode potential of chromium
${{E}^{\circ }}_{cell}$ = -0.4 V– (-0.74 V)
${{E}^{\circ }}_{cell}$ = 0.34V
So from the above equation we get one of the required answers.
Next we will calculate${{E}_{cell}}$. For this we will use Nernst equation.
The cell reaction is:
$3C{{d}^{2+}}\,+\,2Cr\,\to \,3Cd\,+\,2C{{r}^{3+}}$
In this reaction the number of electrons transferred is n = 6.
\[[C{{d}^{2+}}]\,=\,0.1M\] and $[C{{r}^{3+}}]\,\,=\,0.01M$
We will now put the values in the Nernst equation.

$\Rightarrow$ ${E_{cell}}\, = \,{E^ \circ }_{cell}\, + \,\dfrac{{0.059}}{n}\log \dfrac{{{{[C{d^{2 + }}]}^3}}}{{{{[C{r^{3 + }}]}^2}}}$
$\Rightarrow$ ${E_{cell}}$= $0.34 + \dfrac{{0.059}}{6}\log \dfrac{{{{(0.1)}^3}}}{{{{(0.01)}^2}}}$
$\Rightarrow$ ${E_{cell}} = 0.35\;V$

So, the answer is $0.35\;V$

Additional Information:
In electrochemistry, the Nernst equation is an equation that relates the reduction potential of an electrochemical reaction to the standard electrode potential, temperature, and activities (often approximated by concentrations) of the chemical species undergoing reduction and oxidation. It may be a half-cell or full cell reaction. It was named after Walther Nernst, a German physical chemist who formulated the equation.

Note: By applying Nernst equation we can solve this equation quickly. It should be noted that for calculating standard electrode potential oxidation potential should be subtracted from the reduction potential.