A voltaic cell utilizes the following reaction:

Question

2Fe3+(aq)+H2(g)→2Fe2+(aq)+2H+(aq).

What is the emf for this cell when [Fe3+]= 3.70M , PH2= 0.95atm , [Fe2+]= 7.0×10−4M , and the pH in both compartments is 4.05?

Express your answer using two significant figures.

(emf of this cell under standard conditions is 0.771 V)

E=_____V

Answer 1

2Fe3+(aq)+H2(g)→2Fe2+(aq)+2H+(aq).

Ecell = Eocell - (0.5916/2*log Q
and I know that confuses you but don't be. Plug in the numbers Eo and n and for Q plug in concns/pressures(in atm) as follows:
Q = [(H^)^2*(Fe^2+)^2]/[(Fe^3+)^2*pH2]

I write it this way because I'm limited with spacing so I shorten that by using Q, then defining what Q is. Just plug in for Q, evaluate log Q, the multiply by -(0.05916/n) and add algebraically Eocell.