A Zn wire and Ag/AgCl reference electrode (E° = 0.197 V) are placed into a solution of ZnSO4.

Question

A Zn wire and Ag/AgCl reference electrode (E° = 0.197 V) are placed into a solution of ZnSO4. The Zn wire is attached to the positive terminal and the Ag/AgCl electrode is attached to the negative terminal of the potentiometer. Calculate the [Zn2+] in the solution if the cell potential, Ecell, is -1.061 V. The standard reduction potential of the Zn2+/Zn half-reaction is –0.762 V.

[Zn2+]= ? M

Answer 1

E_cell = E₊ - E_- where E₊ is potential at positive terminal and E_- is potential at negative terminal

From this, find potential of Zn half cell at positive terminal and use the Nernst equation to find [Zn²⁺]

E_{(positive terminal)} = Eº_{(positive terminal)} - (0.0592/n) log Q where n = # electrons transferred and Q is the reaction quotient (Zn2+/Zn) for the reduction reaction. This assumes the reaction is at 298K which isn't stated in the problem.  Also, I think other information is omitted from the problem such as the concentration of the ZnSO4 solution.  Is it meant to be 1 M?