The apparatus shown in the diagram can be used to electrolyse a cadmium sulfate
ID: 474870 • Letter: T
Question
The apparatus shown in the diagram can be used to electrolyse a cadmium sulfate solution. The overall reaction is: 2CdSO_4 + 2H_2O rightarrow 2Cd + 2H_2SO_4 + O_2 Briefly explain why cadmium ml fate conducts a current when in aqueous solution but does not conduct when solid. At which graphite elect rode (right hand side or left hand vide on the diagram) would cadmium metal form? Briefly justify your choice. Briefly explain why the pH decreases during the electrolysis of cadmium sulfate solution. If the electrolysis was 100% efficient, calculate the mass of cadmium metal which would form after 0.540 moles of electrons were delivered by the battery. Calculate the volume of oxygen gas (measured at R.T.P.) produced at the same time as the cadmium in (d).Explanation / Answer
(a) CdSO4 conducts electricity in aqueous solution. In the solid state, cadmium sulfate exists as a lattice where the dipositive cadmium ions, Cd2+ and the dinegative sulfate anions, SO42- are held to each other by strong forces of attraction. The ions are not free to move. Since ions are the charge carriers, hence CdSO4 doesn’t conduct electric current in the solid state.
In aqueous solution, the lattice breaks down into Cd2+ and SO42- ions. The ions are free to move in the solution and hence cadmium sulfate conducts electricity in aqueous solution.
(b) The right hand electrode is connected to the negative terminal of the battery and hence is the cathode (an electrode which loves cations). Naturally, Cd2+ ions will migrate to the cathode (right electrode) and be deposited as Cd metal by the reaction:
Cd2+ (aq) + 2 e- -------> Cd (s)
(c) As can be seen from the balanced equation, sulfuric acid, H2SO4 is formed during the course of the electrolysis. Sulfuric acid is a strong acid; hence as the electrolysis progresses, more H2SO4 is formed and the pH of the solution reduces. Note that a drop in pH of the solution means that the solution becomes more acidic.
(d) We have 0.540 mole of electrons delivered by the battery.
As per the stoichiometric reaction in part (b) above,
1 mole Cd = 2 mole electrons.
Therefore, 0.540 mole electrons = (0.540 mole e-)*(1 mole Cd/2 mole e-) = 0.270 mole Cd.
Molar mass of Cd = 112.411 g/mol.
Therefore, mass of Cd produced = (0.270 mole)*(112.411 g/mol) = 30.35097 g 30.351 g (ans).
(e) Note the balanced stoichiometric reaction given:
2 mole Cd = 1 mole O2.
Therefore, 0.270 mole Cd = (0.270 mole Cd)*(1 mole O2/2 mole Cd) = 0.135 mole O2.
We know that 1 mole of O2 = 22400 mL O2 at RTP (or STP).
Therefore, 0.135 mole O2 = (0.135 mole)*(22400 mL/1 mole) = 3024 mL O2 = (3024 mL)*(1 L/1000 mL) = 3.024 L O2 at STP (ans).
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