Question

Calculate the emf of the following cell at 298 K:
$F{e_{(s)}}|F{e^{2 + }}(0.001M)||{H^ + }(1M)|{H_{2(g)}}(1\,\,bar),\,P{t_{(s)}}$
(Given: $E_{cell}^0$ = +0.44V)

Answer 1

To solve the given question, follow these steps: First, start with writing the cell reactions and the net cell reaction. After that, find the value of ‘n’ and substitute it in Nernst equation. Nernst's equation give us the final required answer.

Complete Step-by-Step Answer:
The cell given to us is $F{e_{(s)}}|F{e^{2 + }}(0.001M)||{H^ + }(1M)|{H_{2(g)}}(1\,\,bar),\,P{t_{(s)}}$ , and the standard potential of this cell has been given to +0.44 V

$Fe \to F{e^{2 + }} + 2{e^ - }$
$2{H^ + } + 2{e^ - } \to {H_2}$
Since the total electrons transferred in these reactions are 2, we can say that n =2
Now the net chemical reaction for the entire process can be given as:
$2{H^ + } + Fe \to F{e^{2 + }} + {H_2}$
Now, we know that Nernst equation can be calculated as:
${E_{cell}}$ = $E_{cell}^0$ - $\dfrac{{0.0591}}{n}\log {k_c}$ = $E_{cell}^0$ - $\dfrac{{0.0591}}{n}\log \dfrac{{[F{e^{2 + }}]}}{{{{[{H^ + }]}^2}}}$
Hence,
${E_{cell}}$ = $E_{cell}^0$ - $\dfrac{{0.0591}}{n}\log \dfrac{{[F{e^{2 + }}]}}{{{{[{H^ + }]}^2}}}$
= 0.44 - $\dfrac{{0.0591}}{n}\log \dfrac{{[0.001]}}{{{{[1.0]}^2}}}$
= 0.44 - $\dfrac{{(0.0591)\,\,( - 3)}}{2}$
\begin{array}{*{20}{l}} { = {\text{ }}0.44{\text{ }} + {\text{ }}0.089} \\\ { = {\text{ }}0.529{\text{ }}V} \end{array}

Hence, the value of the EMF for the given cell has been calculated to be 0.529 V.

Note: Nernst Equations are used to find the potential of the cell when the conditions are non - standard. It basically establishes a relation between the measured cell potential and the reaction quotient.