JEE Main 2025 · 3 April · evening · Single correct · +4 / −1
The standard cell potential (E_(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 O₂(E_(O₂ / H₂ O)°) is 1.229 V .
Choose the correct statement :
- (a) Reactants are fed at one go to each electrode.
- (b) Oxygen is formed at the anode.
- (c) The standard half cell reduction potential for the reduction of CO₂(E_(CO₂ / CH₃ OH)°) is 19 mV
- (d) Reduction of methanol takes place at the cathode.
Answer and solution
Answer (c) The standard half cell reduction potential for the reduction of CO₂(E_(CO₂ /…
Solution
Chemistry ma’am · 4 steps
Methanol is the fuel, so it sits at the anode: subtract its potential from oxygen's and you get 19 mV.
- Who does what. In a fuel cell the fuel is oxidised at the anode and oxygen is reduced at the cathode. So CH₃OH goes to CO₂ at the anode, and O₂ goes to H₂O at the cathode.
- Half reactions. Anode: CH₃OH + H₂O → CO₂ + 6H⁺ + 6e⁻. Cathode: 3/2 O₂ + 6H⁺ + 6e⁻ → 3H₂O. The anode couple, written as a reduction, is CO₂/CH₃OH.
E⊖_(cell) = E⊖_(O₂/H₂O) - E⊖_(CO₂/CH₃OH)
- Put in numbers. 1.21 = 1.229 - E⊖_(CO₂/CH₃OH), so E⊖_(CO₂/CH₃OH) = 0.019 V = 19 mV.
- So. Statement (c) is correct. The other three get the electrode roles or the fuel supply wrong.
The idea to keep. Fuel always goes to the anode and is oxidised, oxygen to the cathode and is reduced. Then E_(cell) = E_(cathode) - E_(anode), both as reduction potentials.
Answer from the NTA final answer key. Solution by PrepWiser.