Electrolysis |
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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. |
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3) An aqueous solution of
lead nitrate is electrolysed. 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
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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