Answer:
0.193 M
Explanation:
We need to calculate the Zn²⁺ concentration at the cathode where reduction occurs which is the right side in the expression:
Zn2+(aq,0.100 M) ‖ Zn2+(aq,? M) | Zn(s)
Zn²⁺ (aq,?) + 2 e⁻ ⇒ Zn (s)
and oxidation will occur in the anode
Zn (s) ⇒ Zn²⁺ (aq, 0.100 M ) + 2 e⁻
and the overall reaction is
Zn²⁺ (aq,?) ⇒ Zn²⁺ (0.100 M )
The driving force is the difference in concentration and E the electromotive force will be given by
E = Eº - 0.0592/2 log [Zn²⁺ (0.100 M) / Zn²⁺ (M) ]
Plugging the value for E and knowing Eº is cero because we have the same electrodes, we have
17.0 x 10⁻³ = 0 - 0.0592 log 0.100/ X =
- 17.0 x 10⁻³ / 0.0592 = log 0.100 / X
- 0.287 = log (0.100 / X)
Taking inverse log to both sides of the equation
0.516 = 0.100 / X ⇒ X = 0.100 / 0.516 = 0.193 M
