I need help solving this chemistry problem.

Question

A voltaic cell using Pb²⁺/Pb and Ni²⁺/Ni half-cells is set up at standard conditions, and each compartment has a volume of 355 mL. What is the cell potential, in V, after it has delivered 0.150 A for 31.0 hours at 25°C? (E° for Pb²⁺/Pb = -0.130 V and E° for Ni²⁺/Ni = -0.250 V.)

Answer 1

0.150 A = 0.150 C/sec

31.0 hours x 60 min/hour x 60 sec/min = 111,600 sec

0.150 C/sec x 111,600 sec = 16,740 C

16,740 C x 1 mol electrons / 96,485 C = 0.173 mols of electrons were transferred

Looking at the reaction of the galvanic cell ...

Pb²⁺ + Ni ==> Ni²⁺ + Pb Eº = 0.13 + 0.25 = +0.12 V (using standard reduction potentials)

0.173 mols e- transferred x 1 mol Pb / 2 mols e- = 0.0865 mols Pb²⁺ and Ni reacting

Convert this to molar concentration of Pb²⁺ and Ni²⁺ (moles / 0.355 L)

Use the Nernst equation to find E_cell (Eºcell = Eº - 0.0592/n log Q where Q = [ Ni²⁺]/[Pb²⁺]

(If you have a problem finishing this, drop a comment and hopefully I'll see it and respond)