posted by Alanna on .
The following voltaic cell registers an E cell = 0.100V.
Pt|H2(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?
-3.378 = log[xM]/[1.0M]
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
4.19E-4 = x^2 and not E-5 as you have.
Second, you didn't take the square root.
Third, if 4.10E-5 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.