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Electrochemistry question

2022 · 29 Jun · Shift 1 · Q13
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  5. /2022 · 29 Jun · Shift 1 · Q13

Electrochemistry question

2022 · 29 Jun · Shift 1 · Q13

JEE MainChemistryElectrochemistryNumerical+4 / −1
A dilute solution of sulphuric acid is electrolysed using a current of 0.10 A for 2 hours to produce hydrogen and oxygen gas. The total volume of gases produced a STP is ‾\underline{\hspace{2cm}}​ cm3. (Nearest integer) [Given : Faraday constant F = 96500 C mol −-− 1 at STP, molar volume of an ideal gas is 22.7 L mol −-− 1]
Numerical answer
View written solutionFree

Correct answer: 127

  1. Electrolysis reactions in dilute H2SO4\mathrm{H_2SO_4}H2​SO4​

At the electrodes:

  • Cathode: 2H++2e−→H22H^+ + 2e^- \rightarrow H_22H++2e−→H2​
  • Anode: 2H2O→O2+4H++4e−2H_2O \rightarrow O_2 + 4H^+ + 4e^-2H2​O→O2​+4H++4e−

Overall, hydrogen and oxygen are produced in the mole ratio: H2:O2=2:1H_2:O_2 = 2:1H2​:O2​=2:1

Also, for the overall decomposition:

  • 444 mol electrons produce 222 mol H2+1H_2 + 1H2​+1 mol O2=3O_2 = 3O2​=3 mol gas

So, 4F charge →3 mol gas4F \text{ charge } \rightarrow 3 \text{ mol gas}4F charge →3 mol gas

  1. Calculate total charge passed

Given:

  • Current, I=0.10 AI = 0.10\,\mathrm{A}I=0.10A
  • Time, t=2 h=2×3600=7200 st = 2\,\mathrm{h} = 2 \times 3600 = 7200\,\mathrm{s}t=2h=2×3600=7200s

Hence, Q=It=0.10×7200=720 CQ = It = 0.10 \times 7200 = 720\,\mathrm{C}Q=It=0.10×7200=720C

  1. Calculate moles of electrons

Using Faraday constant F=96500 C mol−1F = 96500\,\mathrm{C\,mol^{-1}}F=96500Cmol−1, n(e−)=QF=72096500n(e^-) = \frac{Q}{F} = \frac{720}{96500}n(e−)=FQ​=96500720​

  1. Calculate total moles of gases produced

Since 444 mol electrons give 333 mol gas, n(total gas)=34⋅72096500n(\text{total gas}) = \frac{3}{4}\cdot \frac{720}{96500}n(total gas)=43​⋅96500720​

n(total gas)=2160386000≈0.005596 moln(\text{total gas}) = \frac{2160}{386000} \approx 0.005596\,\mathrm{mol}n(total gas)=3860002160​≈0.005596mol

  1. Convert moles to volume at STP

At STP, molar volume =22.7 L mol−1= 22.7\,\mathrm{L\,mol^{-1}}=22.7Lmol−1

V=0.005596×22.7 LV = 0.005596 \times 22.7\,\mathrm{L}V=0.005596×22.7L

V≈0.1270 LV \approx 0.1270\,\mathrm{L}V≈0.1270L

Converting to cm3\mathrm{cm^3}cm3: 0.1270 L=127.0 cm30.1270\,\mathrm{L} = 127.0\,\mathrm{cm^3}0.1270L=127.0cm3

Nearest integer: 127\boxed{127}127​

  1. Comparison with stored answer

Stored correct answer = 127127127

Our derived answer = 127127127

So, they agree.

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