JEE Main · 2025 · Shift-IImediumEC-130

The standard cell potential (Ecell) of a fuel cell based on the oxidation of methanol in air that has been used to…

Electrochemistry · Class 12 · JEE Main Previous Year Question

Question

The standard cell potential (EcellE^\circ_{\text{cell}}) of a fuel cell based on the oxidation of methanol in air that has been used to power television relay station is measured as 1.21 V. The standard half cell reduction potential for \ceO2\ce{O2} (E\ceO2/H2OE^\circ_{\ce{O2/H2O}}) is 1.229 V. Choose the correct statement:

Options
  1. a

    The standard half cell reduction potential for the reduction of \ceCO2\ce{CO2} (E\ceCO2/CH3OHE^\circ_{\ce{CO2/CH3OH}}) is 19 mV

  2. b

    Oxygen is formed at the anode

  3. c

    Reactants are fed at one go to each electrode

  4. d

    Reduction of methanol takes pla\ce at the cathode

Correct Answera

The standard half cell reduction potential for the reduction of \ceCO2\ce{CO2} (E\ceCO2/CH3OHE^\circ_{\ce{CO2/CH3OH}}) is 19 mV

Detailed Solution

Strategy: In a fuel cell, the standard cell potential is the differen\ce between the reduction potentials of the cathode and anode (Ecell=EcatEanE^\circ_{ \text{cell}} = E^\circ_{ \text{cat}} - E^\circ_{ \text{an}}).

Calculation: Methanol fuel cell: \ceCH3OH\ce{CH3OH} is oxidized (Anode) and \ceO2\ce{O2} is redu\ced (Cathode). Ecell=E\ceO2/H2OE\ceCO2/CH3OHE^\circ_{ \text{cell}} = E^\circ_{\ce{O2/H2O}} - E^\circ_{\ce{CO2/CH3OH}} 1.21=1.229E\ceCO2/CH3OH1.21 = 1.229 - E^\circ_{\ce{CO2/CH3OH}} E\ceCO2/CH3OH=1.2291.21=0.019 V=19 mVE^\circ_{\ce{CO2/CH3OH}} = 1.229 - 1.21 = 0.019 \text{ V} = 19 \text{ mV}

Answer: (a) \boxed{\text{Answer: (a) }}

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