Calculate the [Zn2 ] in the solution if the cell potential, Ecell, is -1.061 V.

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

The equation is:  E_cell = E^o _cell - (0.0592/n) log Q

 

You know all values except Q (n = 2).  Solve for Q.  This is the ratio of the concentration of the products over the reactants, raised to the power of their coefficients.  Since, Ag and Zn are solids, their concentrations are constant and do not enter into the calculation.  

The only conundrum is whether the Zn2+ is a product or reactant.  Based on the given info, it says that a Zn wire is attached to the positive terminal: the anode.  Oxidation occurs at the anode, so the oxidation reaction is Zn  -->  Zn²⁺  + 2 e^-.  So, the Q ratio is [Zn²⁺]/ unity.  Hence, 10^Q = [Zn2^]