Question

What is the cell potential of the following voltaic cell if the Ksp of Ag₂SO₄ is 1.4x10−5?

Cu(s) | Cu+(1.0 M) || Ag+(Saturated Ag₂SO₄) | Ag(s)

The answer is 0.19 V , please explain how and step by step.

Answer 1

First, with the Ksp value, let's calculate concentration of the Ag⁺ and write an ICE chart:
r: Ag₂SO_4(s) --------> 2Ag⁺ + SO₄^2-
i: s 0 0
e: 2s s

1.4x10^-5 = (2s)² * s
1.4x10^-5 = 4s³
s = (1.4x10^-5/4)^1/3
s = 0.0151 M
[Ag⁺] = 0.0152*2 = 0.0304 M

Now, let's write the half reactions:
Cu ----------> Cu⁺ + 1e^-    E° = -0.52 V
Ag⁺ + 1e^- ---------> Ag E° = 0.80 V
E° = 0.80-0.52 = 0.28 V

Finally, calculate the cell potential with the nerst equation:
E = E° - 0.059/n logK
E = 0.28 - 0.059log(1/0.0304)
E = 0.28 - 0.09
E = 0.19 V

Hope this helps