Chemistry
posted by Alanna on .
The following voltaic cell registers an E cell = 0.100V.
PtH2(g 1 atm)H^+(x M)H^+(1.0 M)H2(g 1 atm)
What is the pH of the unknown solution?
I am completely baffled as I was not aware until I read this question that pH could be found from voltaic cells. Any help would be greatly appreciated.

The anode is H2 ==> 2H^+ + 2e
The cathode is 2H^+ + 2e ==> H2

The cell is H2+2H^+==>2H^+ + H2
Ecell = Eocell (0.0592/2)logQ
log Q = (H^+)^2*pH2/(H^+)^2*pH2
H^+ in numerator is the unknown; H^+ in denominator is 1M. pH2 is 1atm in both cases. Solve for (H^+) and convert to pH. Ecell is 0.1. 
How am I able to solve for [H^+] when I don't have Q and can't rearrange?

Ecell=E*cell0.0592/2*log[xM]/[1.0]
3.378 = log[xM]/[1.0M]
x=0.0000410 M
pH=3.38
What did I do wrong because this is not the right answer 
3.378 is right but x is not.
First, 3.378 = log X^2
4.19E4 = x^2 and not E5 as you have.
Second, you didn't take the square root.
Third, if 4.10E5 were the right answer, log of that for pH is not right.
So 3.378 = log x^2 and solve for x, then convert to pH.