Chemistry · Electrochemistry

JEE Advanced 2020 — Paper 1 — Question 26

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

reaction is

H2( g)+12O2( g)⟶H2O(ℓ)\mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\ell)

The work derived from the cell on the consumption of 10×10−3 mol10 \times 10^{-3} \mathrm{~mol} of H2( g)\mathrm{H}_{2}(\mathrm{~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.

O2( g)+4H+(aq)+4e−⟶2H2O(ℓ),E0=1.23 V\mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{H}^{+}(\mathrm{aq})+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(\ell), \mathrm{E}^{0}=1.23 \mathrm{~V},

2H+(aq)+2e−⟶H2( g),E0=0.00 V2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}(\mathrm{~g}), \mathrm{E}^{0}=0.00 \mathrm{~V}.

Use F=96500Cmol−1,R=8.314 J mol−1 K−1F=96500 \mathrm{C} \mathrm{mol}^{-1}, \mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}.

Answer: 13.32

Numerical answer — enter this value.

Step-by-step solution

I. 12O2( g)+2H++2e−⟶H2O(ℓ),E∘=1.23 V\quad \frac{1}{2} \mathrm{O}_{2}(\mathrm{~g})+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2} \mathrm{O}(\ell), \quad \mathrm{E}^{\circ}=1.23 \mathrm{~V}

II. H2( g)⟶2H++2e−,E0=0.00 V\quad \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{H}^{+}+2 \mathrm{e}^{-}, \quad \mathrm{E}^{0}=0.00 \mathrm{~V}

H2( g)+12O2( g)⟶H2O,E0=1.23 V\mathrm{H}_{2}(\mathrm{~g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}, \quad \mathrm{E}^{0}=1.23 \mathrm{~V}

ΔG0=−2×96500×1.23×1×10−3×0.7=−166.173 J\Delta G^{0}=-2 \times 96500 \times 1.23 \times 1 \times 10^{-3} \times 0.7=-166.173 \mathrm{~J} W=166.173 J\mathrm{W}=166.173 \mathrm{~J}

Wadiabatic =nR(γ−1)(T2−T1)W_{\text {adiabatic }}=\frac{n R}{(\gamma-1)}\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right)

166.173=1×8.314(53−1)(T2−T1)ΔT=166.1738.314×23=13.32\begin{aligned} & 166.173=\frac{1 \times 8.314}{\left(\frac{5}{3}-1\right)}\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right) \\& \Delta \mathrm{T}=\frac{166.173}{8.314} \times \frac{2}{3}=13.32 \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2020
Paper
Paper 1
Subject
Chemistry
Chapter
Electrochemistry
Topic
Basics of Galvanic Cell
Consider a 70 \% efficient hydrogen-oxygen fuel cell working under… | JEE Advanced 2020 PYQ with Solution · DhiX AI