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800 mg of copper at I=180mA should be removed from a copper sulphate solution at the cathode. Calculate (a) the time duration of electrolysis and (b) determines the volume of oxygen that is created simultaneously under normal conditions (298 K, 101300 N/m2).
The general gas equation is p·V=n·R·T, whereasR = 8.314 J/molK .
Note : identical exercise also at the same time, subjects salts.
Solution: - · Involved reactions:
- Cu2+ + 2·e– → Cu, z = 2
- 6·H2O → 4·H3O+ + O2 + 4·e– ; z = 4
- · m/M = I·t/(z·F)
- t = F·m·z/(I·M)
- t = 96485·0.8·2/(0.18·63.5)
- t = 13506 s = about 4 h
- · m/M = n(O2) = I·t/(z·F)
- n(O2) = 0.18·13506/(4·96485) = 0.00629 mol
- · p·V = n·R·T
- V(O2)= n·R·T/p
- V(O2) = 0.00629·8.314·298/101300
- V(O2) = 0.000154 m3 = approx. 154 ml
- · Units: [mol·J/(mol·K) ·K / N/m2]
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