A concentration cell is set up using two zinc wires and two solution, one containing 0.450 M ZnCl2 solution and the other containing 1.30 M Zn(NO3)2 solution. Compute the potential of this cell.

Answer 1

Zn | Zn²⁺(0.450 M) || Zn²⁺(1.30 M) | Zn

Zn(s) ==> Zn²⁺(0.450 M) oxidation on the left side

Zn²⁺(1.30M) + 2e- ==> Zn(s) reduction on the right side

-------------------------------------------

Zn²⁺(1.30M) ==> Zn²⁺(0.450M)

E_cell = Eº_cell - RT ln Q and assuming this situation is at 25ºC (298K), we have...

E_cell = Eº_cell - 0.0592 / 2 log Q

Q = 0.450 / 1.30 = 0.346

E_cell = 0 - 0.0296 log 0.346

E_cell = -0.0296 * -0.461

E_cell = +0.0136 V (note the potential is positive, meaning the reaction is spontaneous)

Also note that this is the INSTANTANEOUS cell potential. As time goes on, the cell potential will become ZERO