Homework Answers
4) Zn + 2Ag+ -> 2Ag + Zn2+
E°cell = E°cathode - E°anode = E°(Ag+/Ag) - E°(Zn2+/Zn)
= 0.80-(-0.76) = 1.56 V
Ecell = E°cell - 0.059/n log([Zn 2+]/[Ag+]2)
1.48 = 1.56 - 0.059/2 log(1/[Ag+]2)
log[Ag+]2 = (1.48-1.56)×2/0.059 = -2.71
[Ag+]2 = 10-2.71 = 0.00194
[Ag+] = √(0.00194)
= 0.044 M
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