5) Calculate Gnn for the reaction at 25 °C, (2 pts) 2Fe3+(aq) + 3Sn(s) 2Fe(s) +

Question

5) Calculate Gnn for the reaction at 25 °C, (2 pts) 2Fe3+(aq) + 3Sn(s) 2Fe(s) + 3Sn2+(aq) 6) A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 °C. The initial concentrations of Pb2+ and Cu2+ are 0.0500 M and 1.50 M respectively. (3 pts) a) What is the initial cell potential? b) What is the cell potential when the concentration of Cu2" has fallen to 0.200 M? c) What are the concentrations of Pb2 and Cu* when the cell potential falls to 0.35 V? 7) Silver can be electroplated at the cathode of an electrolysis cell by the half-reaction: Ag+(aq) + e-(aq) Ag(s)

Explanation / Answer

6) electrochemical cell

cell reaction,

Cu2+ + Pb <==> Cu + Pb2+

a) initial cell potential

Q = [Pb2+]/[Cu2+]

n = 2

Using Nernst equation,

Ecell = Eo - 0.0502/nlogQ

        = (0.337 - (-0.126) - 0.0592/2 log(0.05/1.5)

        = 0.507 V

b) change in [Cu2+] = 1.5 - 0.2 = 1.3 M

equilibrium [Pb2+] = 0.05 + 1.3 = 1.35 M

So,

Ecell = (0.337 - (-0.126) - 0.0592/2 log(1.35/0.2)

        = 0.438 V

c) When Ecell = 0.35 V

let x be the change in concentration

Ecell = Eo - 0.0502/nlogQ

0.35 = (0.337 - (-0.126) - 0.0592/2 log(0.05+x/1.5-x)

9855 - 6570x = 0.05 + x

x = 9854.95/6571 = 1.45 M

Equilibrium concentration of [Cu2+] = 1.5 - 1.45 = 0.05 M

Equilibrium concentration of [Pb2+] = 0.05 + 1.45 = 1.50 M

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