Electrolysis
in solutions exercises solution 1
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.

1) An aqueous solution of copper sulfate is electrolysed. Use the reactivity series on the right.

a)What are the products formed at the anode and at the cathode?
At the anode (+) oxygen and hydrogen ions are formed according to the equation below.

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

At the cathode(-) copper metal is deposited according to the equation below.
Cu2+(aq)+ 2e => Cu(s)

b) What happens to the pH of the solution?
As hydrogen ions are formed the pH will slowly decrease.

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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
Each oxygen atom has an oxidation number of -2 Reductants in order of increasing strength as we go down the column Oxidants in order of increasing strength as we go up the column. Anode Cathode Solution turns light blue as the copper ions are slowly reduced to copper metal.