In a galvanic cell, the silver compartment contains a silver electrode and excess AgCl(s) (Ksp = 1.6x10^-10) The copper compartment contains a copper electrode and [Cu2+]=2M
Ag+(aq) + e- -> Ag(s) E= 0.80V
Cu2+(aq) + 2e- -> Cu(s) E=0.34V
Assuming 1.0 L of 2.0 Cu2+ in the copper compartment, calculate how many moles of ammonia would have to be added to establish the cell potential of 0.52V at 25 Celsius (assume no volume change on addition of Ammonia)
Cu2+(aq) + 4NH3(aq) -> Cu(NH3)4 2+ (aq) Kf=1.0 x 10^13
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Here's what i did.
I found K for the cell rxn by 0.52=0.46-(0.0591/2)Log K. K was 9.3x10^-3
Then to find [Cu2+] to solve for concentration of Ammonia in Cu2+(aq) + 4NH3(aq) -> Cu(NH3)4 2+ (aq) Kf=1.0 x 10^13 equation, I first found concentration of Ag+ from Ksp equation which was found to be 1.26 x 10^-5. Then i put that value in cell reaction equation to find concentration of Cu3+ and used this value to find concentration of NH3 in Kf equation.
Please help me!
thanks in advance