Question

The standard cell potential (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 V. The standard half cell reduction potential for O\( _2 \)/H\( _2 \)O (E\(^\circ_{O_2/H_2O}\)) is 1.229 V. Choose the correct statement:

• A. The standard half cell reduction potential for the reduction of CO\( _2 \) (E\(^\circ_{CO_2/CH_3OH}\)) is 19 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.

Hint:

In a fuel cell, oxidation occurs at the anode and reduction occurs at the cathode. The standard cell potential is the difference between the standard reduction potentials of the cathode and the anode. Pay close attention to the signs and conventions for oxidation and reduction potentials.

Answer 1

The Correct Option is A

To solve this problem, we need to determine the correct statement about the fuel cell based on the given standard cell potential and the standard reduction potentials provided in the options. Let's analyze the information step-by-step:

1. The standard cell potential (\( E^\circ_{\text{cell}} \)) of the fuel cell is given as 1.21 V. This potential is for the overall reaction in the fuel cell.
2. The standard reduction potential for the \( \text{O}_2/\text{H}_2\text{O} \) half cell (\( E^\circ_{O_2/H_2O} \)) is 1.229 V. This indicates that the reduction of oxygen to water happens at the cathode.
3. In any galvanic cell (such as this fuel cell), the overall cell potential is the difference between the cathode and anode potentials:
   • Using the formula: \(E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}\)
   • Given, \( E^\circ_{\text{cell}} = 1.21 \text{ V} \), and \( E^\circ_{\text{cathode}} = 1.229 \text{ V} \) for \( \text{O}_2/\text{H}_2\text{O} \).
   • Substituting the values: \(1.21 = 1.229 - E^\circ_{\text{anode}}\)
   • Solving for \( E^\circ_{\text{anode}} \), we get: \(E^\circ_{\text{anode}} = 1.229 - 1.21 = 0.019 \text{ V} = 19 \text{ mV}\)
4. This calculated value, 19 mV, corresponds to the standard half-cell potential for the oxidation of methanol to carbon dioxide and water at the anode, signifying that the calculated potential matches with the statement regarding carbon dioxide reduction.

Now, let's verify each statement:

• The standard half cell reduction potential for the reduction of CO\(_2\) (E\(^\circ_{CO_2/CH_3OH}\)) is 19 mV: As calculated above, this statement is correct.
• Oxygen is formed at the anode: In a fuel cell using methanol, oxygen is consumed at the cathode, not formed. Wrong statement.
• Reactants are fed at one go to each electrode: Typically, reactants are continuously fed to maintain the reaction consistently in a fuel cell, not at one go. Assuming this refers to traditional continuous feeding, this option doesn't fit well within the standard scenarios. Likely incorrect in this context.
• Reduction of methanol takes place at the cathode: Methanol is oxidized at the anode, not reduced at the cathode. This statement is incorrect.

Therefore, the correct answer is: The standard half cell reduction potential for the reduction of CO\(_2\) (E\(^\circ_{CO_2/CH_3OH}\)) is 19 mV.

Answer 2

Let's analyze the fuel cell based on the oxidation of methanol in air. 

The overall reaction is:

CH₃OH(l) + 3/2 O₂(g) → CO₂(g) + 2H₂O(l)

We are given that the standard cell potential E°cell = 1.21 V, and the standard reduction potential for O₂/H₂O is E°(O₂/H₂O) = 1.229 V. 
We want to find the standard reduction potential for CO₂/CH₃OH, which we'll denote as E°(CO₂/CH₃OH).

The overall cell potential is given by:

E°cell = E°(cathode) - E°(anode)

In this fuel cell, oxygen is reduced at the cathode, and methanol is oxidized at the anode. Therefore:

E°cell = E°(O₂/H₂O) - E°(CO₂/CH₃OH)

We want to find E°(CO₂/CH₃OH), so we can rearrange the equation:

E°(CO₂/CH₃OH) = E°(O₂/H₂O) - E°cell

Plugging in the given values:

E°(CO₂/CH₃OH) = 1.229 V - 1.21 V

E°(CO₂/CH₃OH) = 0.019 V

Converting this to mV:

E°(CO₂/CH₃OH) = 0.019 V * 1000 mV/V = 19 mV

Therefore, the standard half cell reduction potential for the reduction of CO₂ (E°(CO₂/CH₃OH)) is 19 mV.

Final Answer: The final answer is The standard half cell reduction potential for the reduction of CO\( _2 \) (E\(^\circ_{CO_2/CH_3OH})\) is 19 mV