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MCQEasyJEE 2025Batteries & Fuel Cells

JEE Chemistry 2025 Question with Solution

The standard cell potential (Ecell∘E^\circ_{cell}) of a fuel cell based on the oxidation of methanol in air that has been used to power a television relay station is measured as 1.21 V1.21 \, \text{V}. The standard half cell reduction potential for O2/H2OO_2/H_2O (EO2/H2O∘E^\circ_{O_2/H_2O}) is 1.229 V1.229 \, \text{V}. Choose the correct statement:

  • A

    The standard half cell reduction potential for the reduction of CO2CO_2 (ECO2/CH3OH∘E^\circ_{CO_2/CH_3OH}) is 19 mV19 \, \text{mV}

  • B

    Oxygen is formed at the anode.

  • C

    Reactants are fed at one go to each electrode.

  • D

    Reduction of methanol takes place at the cathode.

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Ecell∘=1.21 VE^\circ_{cell} = 1.21 \, \text{V} and the standard reduction potential for O2/H2OO_2/H_2O is EO2/H2O∘=1.229 VE^\circ_{O_2/H_2O} = 1.229 \, \text{V}.

Find: The correct statement about the methanol fuel cell.

In a fuel cell, oxidation occurs at the anode and reduction occurs at the cathode. Here, oxygen is reduced at the cathode and methanol is oxidized at the anode.

Use the relation

Ecell∘=Ecathode∘−Eanode∘E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}

Substituting the given values,

1.21=1.229−Eanode∘1.21 = 1.229 - E^\circ_{anode}

So,

Eanode∘=1.229−1.21=0.019 VE^\circ_{anode} = 1.229 - 1.21 = 0.019 \, \text{V}

Therefore,

ECO2/CH3OH∘=0.019 V=19 mVE^\circ_{CO_2/CH_3OH} = 0.019 \, \text{V} = 19 \, \text{mV}

Now check the statements:

  • Option A matches the calculated value, so it is correct.
  • Oxygen is not formed at the anode; it is reduced at the cathode.
  • In a fuel cell, reactants are supplied continuously, not at one go.
  • Methanol undergoes oxidation at the anode, not reduction at the cathode.

Therefore, the correct option is A.

Reaction-Based Explanation

Given: A methanol fuel cell has Ecell∘=1.21 VE^\circ_{cell} = 1.21 \, \text{V} and EO2/H2O∘=1.229 VE^\circ_{O_2/H_2O} = 1.229 \, \text{V}.

Find: The correct statement among the given options.

The overall reaction of the fuel cell is

CH3OH(l)+32O2(g)→CO2(g)+2H2O(l)CH_3OH(l) + \frac{3}{2} O_2(g) \rightarrow CO_2(g) + 2H_2O(l)

Since oxygen is reduced, the oxygen electrode is the cathode. Methanol is oxidized, so the methanol electrode is the anode.

Using

Ecell∘=Ecathode∘−Eanode∘E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}

we get

ECO2/CH3OH∘=EO2/H2O∘−Ecell∘E^\circ_{CO_2/CH_3OH} = E^\circ_{O_2/H_2O} - E^\circ_{cell}

Thus,

ECO2/CH3OH∘=1.229−1.21=0.019 VE^\circ_{CO_2/CH_3OH} = 1.229 - 1.21 = 0.019 \, \text{V}

Converting to millivolts,

0.019 V=19 mV0.019 \, \text{V} = 19 \, \text{mV}

Therefore, the standard half cell reduction potential for the reduction of CO2CO_2 is 19 mV19 \, \text{mV}, so the correct option is A.

Common mistakes

  • Using Ecell∘=Eanode∘−Ecathode∘E^\circ_{cell} = E^\circ_{anode} - E^\circ_{cathode}. This reverses the galvanic cell convention and gives the wrong sign. Always use Ecell∘=Ecathode∘−Eanode∘E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} when reduction potentials are given.

  • Assuming methanol is reduced at the cathode. In a fuel cell, methanol acts as the fuel and is oxidized at the anode. Identify oxidation and reduction before assigning electrodes.

  • Thinking oxygen is produced at the anode. Here oxygen is a reactant and is consumed at the cathode. Check the role of each species in the overall reaction before judging the statement.

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