Electrochemical cells (2) (AHL) answers

Answers to Electrochemical cells (2) (AHL) questions (b) 2H+(aq) + 2e– → H2(g) and Pb2+(l) + 2e– → Pb(l) so the same amount of lead (in mol) will be formed as hydrogen produced.Amount of hydrogen = 500/22700 = 0.0220 mol; Ar for Pb = 207.20Mass of lead = 0.0220 x 207.20 = 4.56 g(c) It would have no effect. The cells were connected in series so the same quantity of electricity passed through both cells.


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