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Electrochemistry question

2020 · Shift 1 · Q15
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Electrochemistry question

2020 · Shift 1 · Q15

JEE AdvancedChemistryElectrochemistryNumerical+4 / −1
Consider a 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){H_2}(g) + {1 \over 2}{O_2}(g)\xrightarrow{} {H_2}O(l)H2​(g)+21​O2​(g)​H2​O(l) The work derived from the cell on the consumption of 1.0 ×\times× 10 −-− 3 mole of H2H_2H2​(g) is used to compress 1.00 mole 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),{O_2}(g) + 4{H^ + }(aq) + 4{e^ - }\xrightarrow{} 2{H_2}O(l),O2​(g)+4H+(aq)+4e−​2H2​O(l), Eo=1.23V2H+(aq)+2e−→H2(g),Eo=0.00 V{E^o} = 1.23V2{H^ + }(aq) + 2{e^ - }\xrightarrow{} {H_2}(g),{E^o} = 0.00\,VEo=1.23V2H+(aq)+2e−​H2​(g),Eo=0.00V Use, F=96500 C mol−1F = 96500\,C\,mo{l^{ - 1}}F=96500Cmol−1, R=8.314 J mol−1 K−1R = 8.314\,J\,mo{l^{ - 1}}\,{K^{ - 1}}R=8.314Jmol−1K−1.
Numerical answer
View written solutionFree

Correct answer: 13.32

  1. Find the standard cell potential

The cell reaction is

H2(g)+12O2(g)→H2O(l)H_2(g)+\frac{1}{2}O_2(g)\to H_2O(l)H2​(g)+21​O2​(g)→H2​O(l)

Given standard reduction potentials:

O2(g)+4H+(aq)+4e−→2H2O(l),E∘=1.23 VO_2(g)+4H^+(aq)+4e^-\to 2H_2O(l),\quad E^\circ=1.23\,\text{V}O2​(g)+4H+(aq)+4e−→2H2​O(l),E∘=1.23V 2H+(aq)+2e−→H2(g),E∘=0.00 V2H^+(aq)+2e^-\to H_2(g),\quad E^\circ=0.00\,\text{V}2H+(aq)+2e−→H2​(g),E∘=0.00V

So,

Ecell∘=Ecathode∘−Eanode∘=1.23−0.00=1.23 VE^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}=1.23-0.00=1.23\,\text{V}Ecell∘​=Ecathode∘​−Eanode∘​=1.23−0.00=1.23V
  1. Calculate maximum electrical work from consumption of 1.0×10−31.0\times 10^{-3}1.0×10−3 mol H2H_2H2​

For the reaction,

H2→2H++2e−H_2\to 2H^+ +2e^-H2​→2H++2e−

So 1 mol of H2H_2H2​ transfers 2 mol of electrons.

For 1.0×10−31.0\times 10^{-3}1.0×10−3 mol H2H_2H2​,

ne−=2×10−3 moln_{e^-}=2\times 10^{-3}\,\text{mol}ne−​=2×10−3mol

Total charge passed:

q=nF=(2×10−3)(96500)=193 Cq=nF=(2\times 10^{-3})(96500)=193\,\text{C}q=nF=(2×10−3)(96500)=193C

Maximum electrical work under standard conditions:

Wmax⁡=qE∘=193×1.23=237.39 JW_{\max}=qE^\circ=193\times 1.23=237.39\,\text{J}Wmax​=qE∘=193×1.23=237.39J

Since the fuel cell is 70% efficient, actual work obtained is

W=0.70×237.39=166.173 JW=0.70\times 237.39=166.173\,\text{J}W=0.70×237.39=166.173J
  1. Use this work to compress 1 mole of monoatomic ideal gas adiabatically

The gas is in a thermally insulated container, so

q=0q=0q=0

From the first law,

ΔU=q+w=w\Delta U=q+w=wΔU=q+w=w

The work done on the gas is 166.173 J166.173\,\text{J}166.173J, hence

ΔU=166.173 J\Delta U=166.173\,\text{J}ΔU=166.173J

For 1 mole of monoatomic ideal gas,

ΔU=nCVΔT\Delta U=nC_V\Delta TΔU=nCV​ΔT

with

CV=32RC_V=\frac{3}{2}RCV​=23​R

Thus,

166.173=1×32(8.314)ΔT166.173=1\times \frac{3}{2}(8.314)\Delta T166.173=1×23​(8.314)ΔT ΔT=166.17312.471=13.324 K\Delta T=\frac{166.173}{12.471}=13.324\,\text{K}ΔT=12.471166.173​=13.324K
  1. Final answer
13.32 K\boxed{13.32\,\text{K}}13.32K​
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