- Cu+2+2e−⟶Cu
(1 faraday = charge on 1 mole electron )t=01.51 molet=t1−0.5 mole
[Cu+2]=1M after electrolysis
- Ag⊕+e−⟶Ag
t=0t=t0.20.10.1 mole −−
[Ag+]=0.1M after electrolysis
Cell Cu(s)+2Ag(aq)+→Cu(aq)+2+2Ag(s)
reaction
E=E∘−n0.06log[Ag+]2[Cu+2]E=(0.8−0.34)−20.06log(0.1)21=0.4 V
Correct answer =400mV