1 Faraday of electricity was passed through Cu2+(1.5 M, 1 L)/Cu and 0.1 Faraday was passed through Ag+(0.2 M, 1 L)/Ag electrolytic cells. After this the two cells were connected as shown below to make an electrochemical cell. The emf of the cell thus formed at 298 K is mV (nearest integer) [figure] Given : E Cu2+/Cu = 0.34 V E Ag+/Ag = 0.8 V 2.303RTF = 0.06 V
Answer
400
Solution
Electrolysis (cathode deposits metal from the solution):
Cu2+ + 2e- Cu: 1 F deposits 0.5 mol Cu
[Cu2+] = 1.5 − 0.5 = 1.0 M
Ag+ + e- Ag: 0.1 F deposits 0.1 mol Ag
[Ag+] = 0.2 − 0.1 = 0.1 M
Galvanic cell: Cu | Cu2+(1.0 M) || Ag+(0.1 M) | Ag
Cell reaction: Cu + 2Ag+ Cu2+ + 2Ag, n = 2
E cell = 0.80 − 0.34 = 0.46 V
Ecell = E cell - 0.062 [Cu2+][Ag+]2
= 0.46 - 0.03 1.0(0.1)2
= 0.46 - 0.03 2 = 0.40 V
emf = 400 mV
Answer: 400