JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Redox Reactions and Electrochemistry

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Consider a 70% efficient hydrogen-oxygen fuel cell working under standard conditions at 1 bar and 298 K. Its cell reaction is

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (ℓ) .

The work derived from the cell on the consumption of

1.0 \times 1 0^{- 3} mol

H_{2} (g)

is used to compress 1.00 mol of a monoatomic ideal gas in a thermally insulated container. What is the change in the temperature (in K) of the ideal gas?

The standard reduction potentials for the two half-cells are given below.

O_{2} (g) + 4 H^{+} (aq.) + 4 e^{-} \to 2 H_{2} O (ℓ), E^{\circ} = 1.23 V 2 H^{+} (aq.) + 2 e^{-} \to H_{2} (g), E^{o} = 0.00 V

Use

F = 96500 C mol^{- 1}, R = 8.314 J mol^{- 1} K^{- 1}

Answer: 13.3

Solution

Cell potential:

E_{ce l l}^{\circ} = 1.23 - 0.00 = 1.23

V

Gibbs energy change:

Δ G^{\circ} = - n F E^{\circ} = - 2 \times 96500 \times 1.23

J

Work from fuel cell (70% efficient) for

1.0 \times 1 0^{- 3}

mol

H X_{2}

w = 0.70 \times 2 \times 96500 \times 1.23 \times 1 0^{- 3} = 166.11 J

For the insulated container (adiabatic process),

q = 0

, so

w = Δ U

.

For monoatomic ideal gas:

C_{V, m} = \frac{3 R}{2}

166.11 = 1.00 \times \frac{3 \times 8.314}{2} \times Δ T

Δ T = \frac{166.11}{12.471} = 13.32 K

JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Numerical answer

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JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Redox Reactions and Electrochemistry

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