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Question: Consider the following cell reaction: \(2Fe(s)+{{O}_{2}}(g)+4{{H}^{+}}(aq)\to 2F{{e}^{2+}}(aq)+2{{...

Consider the following cell reaction:
2Fe(s)+O2(g)+4H+(aq)→2Fe2+(aq)+2H2O(l); E∘=1.67V2Fe(s)+{{O}_{2}}(g)+4{{H}^{+}}(aq)\to 2F{{e}^{2+}}(aq)+2{{H}_{2}}O(l);\,{{E}^{\circ }}=1.67V
At [Fe2+]=10−3M, Po2=0.1atm[F{{e}^{2+}}]={{10}^{-3}}M,\text{ }{{P}_{{{o}_{2}}}}=0.1atm and pH=3pH = 3, the cell potential at 25∘C{{25}^{\circ }}C is:
(A) 1.47 V
(B) 1.77 V
(C) 1.87 V
(D) 1.57 V

Explanation

Solution

This question can be solved by formula Ecell∘+0.059nlog⁡[conc.]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)+O2(g)+4H+(aq)→2Fe2+(aq)+2H2O(l); E∘=1.67V2Fe(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 FeFe loses electrons to form Fe2+F{{e}^{2+}} ion and the oxygen is reduced.
So, the above reaction can be split into:
Oxidation half-reaction: 2Fe→2Fe2++4e−2Fe\to 2F{{e}^{2+}}+4{{e}^{-}}
And reduction half-reaction: 4e−+O2+4H+→2H2O4{{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: Ecell∘−0.059nlog⁡[conc.]E_{cell}^{\circ }-\dfrac{0.059}{n}\log [conc.]
So in the question Ecell∘E_{cell}^{\circ } is given 1.67V1.67 V.
So for oxidation half-reaction, the formula will be:
Ecell∘−0.0594log⁡[Fe2+]2E_{cell}^{\circ }-\dfrac{0.059}{4}\log {{[F{{e}^{2+}}]}^{2}}
Because 4 electrons are involved in the reaction. The concentration of [Fe2+]=10−3M[F{{e}^{2+}}]={{10}^{-3}}M
So for reduction half-reaction, the formula will be:
Ecell∘−0.0594log⁡Po2 x [H+]4E_{cell}^{\circ }-\dfrac{0.059}{4}\log {{P}_{{{o}_{2}}}}\text{ x }{{[{{H}^{+}}]}^{4}}
Since the pH is 3, the concentration of [H+][{{H}^{+}}] ions will be 10−3{{10}^{-3}}.
For cell potential of overall reaction the formula will be:
Ecell=Ecell∘−0.0594log⁡[Fe2+]2+Ecell∘−0.0594log⁡Po2 x [H+]4{{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,
⇒Ecell=Ecell∘−0.0594log⁡Po2 x [H+]4[Fe2+]2\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:
⇒Ecell=Ecell∘−0.0594log⁡0.1 x [10−3]4[10−3]2\Rightarrow {{E}_{cell}}=E_{cell}^{\circ }-\dfrac{0.059}{4}\log \dfrac{\text{0}\text{.1 x }{{[{{10}^{-3}}]}^{4}}}{{{[{{10}^{-3}}]}^{2}}}
⇒Ecell=1.67−0.0594log⁡10−7\Rightarrow {{E}_{cell}}=1.67-\dfrac{0.059}{4}\log {{10}^{-7}}
Putting the value of log⁡−7{{\log }^{-7}}, we get
⇒Ecell=1.67−0.059 x (-7)4\Rightarrow {{E}_{cell}}=1.67-\dfrac{0.059\text{ x (-7)}}{4}
⇒Ecell=1.67−0.103=1.57 V\Rightarrow {{E}_{cell}}=1.67-0.103=1.57\text{ V}
Therefore the cell potential is 1.57V1.57 V.

So, the correct answer is an option (D) 1.57V1.57 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{{10}^{-3}}.