Electrolysis
in solutions exercises solution 3

Special note. Marks will be deducted in the exam if equilibrium arrows appear on half equatins at each electrod. The reduction half-equations provided in the electrochemical series include equilibrium arrows because the direction of each half-equation is determined by the combination of oxidant and reductant used in the reaction. Equilibria between an oxidant, its conjugate reductant and electrons are not established at the electrodes.
Students should be aware that oxidation (the loss of electrons) occurs at the anode and reduction (the gain of electrons) occurs at the cathode.

 

3) An aqueous solution of lead nitrate is electrolysed.
a) Using the table on the right write the half equations that occur at the anode and the cathode.
Since lead(II) ions are the strongest oxidant present they will react at the cathode.
At the cathode(-) reduction of lead ions takes place according to the reaction below.

Pb2+(aq)+ 2e => Pb(s)

At the anode(+) oxidation takes place. Water is the only reductant present and this will react according to the equation below.

2H2O(l) => O2(g) + 4H+(aq) + 4e


b) If the current through the electrodes is maintained for a long time predict any changes in product formation.

Once the lead ions are fully deposited at the cathode water then becomes the strongest oxidant present and will react at the cathode according to the equation below.

2H2O(l) + 2e =>H2(g) + 2OH-(aq)

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Balance for charge by adding electrons tot he most positive side. Balance for hydrogen by adding Hydrogen ions to the hydrogen deficient side. Balance for oxygen by adding water to the oxygen deficient side. Balance for Ag atoms Balance for oxygen by adding water to the oxygen deficient side. Balance for hydrogen by adding Hydrogen ions to the hydrogen deficient side. Balance for charge by adding electrons tot he most positive side. Home
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