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Electrochemistry: Question 10

Syllabus 24.1

Structured A2 9 marks

Dilute sulfuric acid is electrolysed using inert platinum electrodes. A constant current of 0.500 A0.500\ \text{A} is passed for 40.040.0 minutes. (F=96500 C mol−1F = 96500\ \text{C mol}^{-1}; 1 mole of gas occupies 24000 cm324000\ \text{cm}^3 at room temperature and pressure, rtp.)

(a) Write half-equations for the reactions occurring at the cathode and at the anode. [2]

(b) Calculate the quantity of electric charge, QQ, passed during electrolysis. [1]

(c) Calculate the amount, in mol, of electrons transferred, and hence the volume of gas produced at the cathode, measured at rtp. [3]

(d) Calculate the volume of gas produced at the anode over the same time period, measured at rtp, and state the simplest whole-number ratio of the volume of gas at the cathode to the volume of gas at the anode. [3]

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Worked solution

Part (a): Half-equations at each electrode

In dilute sulfuric acid with inert platinum electrodes, H+(aq)\text{H}^+(aq) is discharged at the cathode, while at the anode it is water (not the sulfate ion, which is very difficult to oxidise) that is discharged:

Cathode (reduction): 2H+(aq)+2e−→H2(g)2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g)

Anode (oxidation): 2H2O(l)→O2(g)+4H+(aq)+4e−2\text{H}_2\text{O}(l) \rightarrow \text{O}_2(g) + 4\text{H}^+(aq) + 4e^-

Part (b): Charge passed

First convert the time to seconds: 40.0 min=40.0×60=2400 s40.0\ \text{min} = 40.0\times60 = 2400\ \text{s}.

Q=It=0.500×2400=1200 CQ = It = 0.500\times2400 = 1200\ \text{C}

Part (c): Moles of electrons and volume of gas at the cathode

n(e−)=QF=120096500=1.2435×10−2 mol≈1.24×10−2 moln(e^-) = \frac{Q}{F} = \frac{1200}{96500} = 1.2435\times10^{-2}\ \text{mol} \approx 1.24\times10^{-2}\ \text{mol}

From the cathode half-equation, 22 mol of electrons produce 11 mol of H2\text{H}_2, so:

n(H2)=n(e−)2=1.2435×10−22=6.2176×10−3 moln(\text{H}_2) = \frac{n(e^-)}{2} = \frac{1.2435\times10^{-2}}{2} = 6.2176\times10^{-3}\ \text{mol}

V(H2)=n(H2)×24000=6.2176×10−3×24000=149.2 cm3≈149 cm3V(\text{H}_2) = n(\text{H}_2)\times24000 = 6.2176\times10^{-3}\times24000 = 149.2\ \text{cm}^3 \approx 149\ \text{cm}^3

Check (independent recomputation): 1200×2400096500×2=28 800 000193 000=149.2 cm3\dfrac{1200\times24000}{96500\times2} = \dfrac{28\,800\,000}{193\,000} = 149.2\ \text{cm}^3, consistent.

Part (d): Volume of gas at the anode and the ratio

The same charge passes through the anode as through the cathode (it is a single circuit), so n(e−)=1.2435×10−2 moln(e^-) = 1.2435\times10^{-2}\ \text{mol} still applies. From the anode half-equation, 44 mol of electrons produce 11 mol of O2\text{O}_2, so:

n(O2)=n(e−)4=1.2435×10−24=3.1088×10−3 moln(\text{O}_2) = \frac{n(e^-)}{4} = \frac{1.2435\times10^{-2}}{4} = 3.1088\times10^{-3}\ \text{mol}

V(O2)=n(O2)×24000=3.1088×10−3×24000=74.61 cm3≈74.6 cm3V(\text{O}_2) = n(\text{O}_2)\times24000 = 3.1088\times10^{-3}\times24000 = 74.61\ \text{cm}^3 \approx 74.6\ \text{cm}^3

Check (independent recomputation): 1200×2400096500×4=28 800 000386 000=74.61 cm3\dfrac{1200\times24000}{96500\times4} = \dfrac{28\,800\,000}{386\,000} = 74.61\ \text{cm}^3, consistent.

Comparing the two volumes:

V(H2)V(O2)=149.274.61=2.00\frac{V(\text{H}_2)}{V(\text{O}_2)} = \frac{149.2}{74.61} = 2.00

So V(H2):V(O2)=2:1V(\text{H}_2):V(\text{O}_2) = \boxed{2:1}, which matches the overall stoichiometry of water electrolysis (2H2O→2H2+O22\text{H}_2\text{O} \rightarrow 2\text{H}_2 + \text{O}_2), since each electron does twice as much “work” at the cathode (2e−2e^- per gas molecule) as it needs to at the anode relative to that ratio (4e−4e^- per gas molecule, i.e. half the moles of gas per electron).

Final answers

  • (a) Cathode: 2H+(aq)+2e−→H2(g)2\text{H}^+(aq) + 2e^- \rightarrow \text{H}_2(g). Anode: 2H2O(l)→O2(g)+4H+(aq)+4e−2\text{H}_2\text{O}(l) \rightarrow \text{O}_2(g) + 4\text{H}^+(aq) + 4e^-
  • (b) Q=1200 CQ = \boxed{1200\ \text{C}}
  • (c) n(e−)=1.24×10−2 moln(e^-) = 1.24\times10^{-2}\ \text{mol}; V(H2)=149 cm3V(\text{H}_2) = \boxed{149\ \text{cm}^3}
  • (d) V(O2)=74.6 cm3V(\text{O}_2) = \boxed{74.6\ \text{cm}^3}; ratio V(H2):V(O2)=2:1V(\text{H}_2):V(\text{O}_2) = \boxed{2:1}