Question

Consider the following cell reaction:
$2Fe(s)+{{O}_{2}}(g)+4{{H}^{+}}(aq)\to 2F{{e}^{2+}}(aq)+2{{H}_{2}}O(l);\,{{E}^{\circ }}=1.67V$
At $[F{{e}^{2+}}]={{10}^{-3}}M,\text{ }{{P}_{{{o}_{2}}}}=0.1atm$ and $pH = 3$, the cell potential at ${{25}^{\circ }}C$ is:
(A) 1.47 V
(B) 1.77 V
(C) 1.87 V
(D) 1.57 V

Answer 1

Hint: This question can be solved by formula $E_{cell}^{\circ }+\dfrac{0.059}{n}\log [conc.]$, n is the number of electrons involved in the reaction. This equation has to be applied to the oxidation half-reaction and reduction half-reaction.

Complete step by step answer:
So the reaction given in the question is:
$2Fe(s)+{{O}_{2}}(g)+4{{H}^{+}}(aq)\to 2F{{e}^{2+}}(aq)+2{{H}_{2}}O(l);\,{{E}^{\circ }}=1.67V$
So, in this reaction iron is oxidized because the $Fe$ loses electrons to form $F{{e}^{2+}}$ ion and the oxygen is reduced.
So, the above reaction can be split into:
Oxidation half-reaction: $2Fe\to 2F{{e}^{2+}}+4{{e}^{-}}$
And reduction half-reaction: $4{{e}^{-}}+{{O}_{2}}+4{{H}^{+}}\to 2{{H}_{2}}O$
Since it has oxidation and reduction going in the process, the cell potential of the reaction is calculated by adding the cell potential of oxidation and reduction.
The formula for calculating the cell potential is: $E_{cell}^{\circ }-\dfrac{0.059}{n}\log [conc.]$
So in the question $E_{cell}^{\circ }$ is given $1.67 V$.
So for oxidation half-reaction, the formula will be:
$E_{cell}^{\circ }-\dfrac{0.059}{4}\log {{[F{{e}^{2+}}]}^{2}}$
Because 4 electrons are involved in the reaction. The concentration of $[F{{e}^{2+}}]={{10}^{-3}}M$
So for reduction half-reaction, the formula will be:
$E_{cell}^{\circ }-\dfrac{0.059}{4}\log {{P}_{{{o}_{2}}}}\text{ x }{{[{{H}^{+}}]}^{4}}$
Since the pH is 3, the concentration of $[{{H}^{+}}]$ ions will be ${{10}^{-3}}$.
For cell potential of overall reaction the formula will be:
${{E}_{cell}}=E_{cell}^{\circ }-\dfrac{0.059}{4}\log {{[F{{e}^{2+}}]}^{2}}+E_{cell}^{\circ }-\dfrac{0.059}{4}\log {{P}_{{{o}_{2}}}}\text{ x }{{[{{H}^{+}}]}^{4}}$
On simplifying the above equation we get,
$\Rightarrow {{E}_{cell}}=E_{cell}^{\circ }-\dfrac{0.059}{4}\log \dfrac{{{P}_{{{o}_{2}}}}\text{ x }{{[{{H}^{+}}]}^{4}}}{{{[F{{e}^{2+}}]}^{2}}}$
So, putting all the values, we get:
$\Rightarrow {{E}_{cell}}=E_{cell}^{\circ }-\dfrac{0.059}{4}\log \dfrac{\text{0}\text{.1 x }{{[{{10}^{-3}}]}^{4}}}{{{[{{10}^{-3}}]}^{2}}}$
$\Rightarrow {{E}_{cell}}=1.67-\dfrac{0.059}{4}\log {{10}^{-7}}$
Putting the value of ${{\log }^{-7}}$, we get
$\Rightarrow {{E}_{cell}}=1.67-\dfrac{0.059\text{ x (-7)}}{4}$
$\Rightarrow {{E}_{cell}}=1.67-0.103=1.57\text{ V}$
Therefore the cell potential is $1.57 V$.

So, the correct answer is an option (D) $1.57 V$.

Note: The pH of the solution was given 3. And we know that the pH is the negative logarithm of the concentration of hydrogen ions in the solution. So, the concentration of the hydrogen ions is taken ${{10}^{-3}}$.