Calculate the operating potential of the battery produced

Question

While working in the lab, a student was designing a battery using the reaction below:

Ni(s) + 2 Ag⁺(aq) → Ni²⁺(aq) + 2 Ag(s), E_cell^o=1.03 V, n = 2 e-

 

They wanted to produce a cell with an operating potential of 1.05 V and calculated that Q for the reaction should be 0.211. Unfortunately, they diluted the wrong half cell and rather than having 0.5 M Ag⁺ ion and 0.053M Ni²⁺, they reversed the concentrations and used 0.50 M Ni²⁺ and 0.053M Ag⁺. 

Use the Nernst Equation to calculate the operating potential of the battery they produced.  

Give your answer to the nearest 0.01 V.

Answer 1

Nernst equation:

Ecell = Eºcell - RT/nF ln Q

The problem DOES NOT provide the temperature so we'll assume it to be 298K and then the Nernst is ....

Ecell = Eºcell -0.0592/n log Q

Ecell = 1.03 - 0.0296 log [Ni2+]/[Ag+]²

Ecell = 1.03 - 0.0296 log (0.50/0.0028) = 1.03 -0.0296 *2.25

Ecell = 1.03 - 0.067

Ecell = 0.96 V