Electrolysis
in solutions exercises solution 2
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.

 

 

2) A solution of lead(II)iodide is electrolysed.
a) What are the products formed at the cathode and at the anode?

b) Give the reactions that occur at the anode and at the cathode.
Use the reactivity series on the right.


Looking at the electrochemical series on the right we see that lead(II) ion is the strongest oxidant present. So lead will be deposited at the cathode according to the equation below.

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

The strongest reductant present is I- .

2I -(aq)=> I2 (aq) + 2e

 
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