Question

X) The answers: a) 0.315 V c) 0.273 V d) 1.80 M

10. For the voltaic cell: Cd(s) Cd2 (0.100 M, 25 mL) || Pb2 (2.000 M, 25 mL) | Pb(s) a. What is Eel initially? (E 0.277 v) b. If the cell is allowed to operate spontaneously, will Eel increase, decrease or remain constant with time? Explain. c. What will be Ee when [Pb2"] has fallen to 0.900 M? [Hint: Both concentrations change. Why?] d. What will be [Cd2"] at the point when El 0.254 V? cell cell

Answer 1

10)

a)

Cd(s) + Pb2+(aq) ------->   Cd2+(aq)   +   Pb(s)

E0cell = E0(Pb2+/Pb) - E0(Cd2+/Cd)

E0cell =    0.277 V

Ecell = E0cell + ( 0.059 / n ) log ( Pb2+ / Cd2+)

Ecell = 0.277 + ( 0.059 / 2 ) * log ( 2 / 0.1)

Ecell = 0.277 + ( 0.059 / 2 ) * log ( 20 )

Ecell = 0.277 +   0.038

Ecell = 0.315 V

so Ecell = 0.315 V

b)

during the operation the Cd will oxidize to Cd2+ and Pb2+ will reduce to Pb

so the concentration of Pb2+ will decrease and Cd2+ will increase

log ( Pb2+ / Cd2+)  will decrease and hence the Ecell will decrease during the operation of cell

c)

when [ Pb2+] = 0.9 M

so change in [Pb2+] = 2 - 0.9 = 1.1 M

since both oxidation of Cd to Cd2+ and reduction Pb2+ to Pb will happen

so the concentration of Cd2+ will change corresponding to concentration change in Pb2+

so the concentration of Cd2+ increase by 1.1 M

[ Cd2+ ] = 0.1 + 1.1 = 1.2 M

Ecell = E0cell + ( 0.059 / n ) log ( Pb2+ / Cd2+)

Ecell = 0.277 + ( 0.059 /  2 ) log ( 0.9 / 1.2 )

Ecell = 0.277 + ( 0.059 /  2 ) log ( 0.75)

Ecell = 0.277 - 0.004

Ecell = 0.273 V

Ecell = 0.273 V

d)

Ecell = E0cell + ( 0.059 / n ) log ( Pb2+ / Cd2+)

0.254   = 0.277 + ( 0.059 /  2 ) log ( Pb2+ / Cd2+)

0.254 - 0.277 = 0.0295 * log ( Pb2+ / Cd2+)

-0.023 / 0.0295 = log ( Pb2+ / Cd2+)

- 0.7797   = log ( Pb2+ / Cd2+)

Pb2+ / Cd2+ = 10 ^ ( - 0.7797 )

Pb2+ / Cd2+   = 0.166

Pb2+ = 0.166 * Cd2+

at any time

[ Pb2+] + [Cd2+] = 2.1

0.166 * Cd2+    +   Cd2+   = 2.1

1.166 Cd2+ = 2.1

[ Cd2+] = 2.1 / 1.166

[Cd2+] = 1.80 M

so the [ Cd2+ ] = 1.80 M     At Ecell = 0.254 V