Electrochemestry and battery

[nassas91]

Hi (sorry if my english isn't good)

I have a task with which goes like this: 

A battery is based on oxidation of Magnesium and reduction of Copper-ions. The starting consentration of Mg²⁺-ions and Cu²⁺-ions are 1,0 * 10^-4 M and 1,5 M. The volume of each halfcell is 1 liter. 

a ) Calculate the battery-voltage at the start. 

b) Calculate the battery-voltage after the battery has delivered 5,0 A for 8,0 hours.   

 

I think i got a) right. I just calculated the cellpotential and put it in Nernst forumla like this: 

Oxidation: Mg²⁺ + 2e^-  --> Mg(s)      : -2.36V

Reduction: 2Cu²⁺ + 2e^- --> 2Cu       : 0.16 V

Cell potential: E = 0.16 - (-2.36) = 2,52 V

E_cell  = 2,52 V - (0,0592 V / 2 ) * log([1,5 M] / 2*[1,0 * 10^-4 M]) = 2,4053 V = 2,4 V        

 

My problem is b). I tried different ways to solve it but i can't figure it out. I suspect i have to use Faraday's forumal: n_p = (I * t) / (F * n), but i can't figure out how to use it to find the battery-potensial. 

Can anyone help me?

 

Sorry for but terrible translation xD

Edited April 21, 2020 by nassas91
typo

[joigus]

I'm not sure whether this is related to the problem you're having, but your oxidation looks to me like an reduction. The magnesium is capturing electrons.