JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
Consider the following two half-cell reactions along with the standard reduction potential given:
\(CO_2 + 6H^+ + 6e^- \rightarrow CH_3 OH + H_2 O \quad E°_{red} = 0.02\) V
\(\dfrac{1}{2} O_2 + 2H^+ + 2e^- \rightarrow H_2 O \quad E°_{red} = 1.23\) V
A fuel cell was set up using the above two reactions such that the cell operates under the standard condition of \(1\) bar pressure and \(298\) K temperature. The fuel cell works with \(80\%\) efficiency. If the work derived from the cell using \(1\) mol of \(CH_3 OH\) is used to compress an ideal gas isothermally against a constant pressure of \(1\) kPa, then the change in the volume of the gas, \(\Delta V = \) _____ m\(^3\). (nearest integer)
Given: \(F = 96500\) C mol\(^{-1}\)
- A -560
- B -565
- C -570
- D -575
Answer & Solution
Correct Answer
(A) -560
Step-by-step Solution
Detailed explanation
The half-cell reactions for the fuel cell are: Anode (Oxidation): \(CH_3OH + H_2O \rightarrow CO_2 + 6H^+ + 6e^- \quad E^{\circ}_{ox} = -0.02\) V Cathode (Reduction): \(\dfrac{3}{2}O_2 + 6H^+ + 6e^- \rightarrow 3H_2O \quad E^{\circ}_{red} = 1.23\) V The overall cell reaction is:…
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