The voltaic cell:

Pt, H2(g, 1 atm), H+(xM),,H+ (1M), H2(g, 1 atm), Pt

registers an Ecell=+0.108 V. What is the pH of the soln?

I know that E*cell=0 for this. I was not sure if i place the [xM] on the top of the equation or the bottom of
Ecell=E*cell-0.0592/2*log[0.1M]/[xM]. If I do it with this equation, I get a pH of 2.64 and book says pH=2.82??? Thanks.

To find the pH of the solution based on the given voltaic cell, you can use the Nernst equation:

Ecell = E*cell - (0.0592/n) * log([H+]/[H2])

Where:
- Ecell is the observed cell potential (+0.108 V in this case)
- E*cell is the standard cell potential (0 V in this case because you mentioned E*cell=0)
- n is the number of moles of electrons transferred in the balanced cell reaction (in this case, it is 2)
- [H+] is the concentration of H+ ions in the solution (x M in this case)
- [H2] is the concentration of H2 gas (1 atm in this case)

Now, let's plug in the values and solve for [H+] (x M):

0.108 V = 0 V - (0.0592/2) * log([H+]/1)
0.108 = -0.0296 * log([H+])

Solving for log([H+]):

log([H+]) = 0.108 / (-0.0296)
log([H+]) = -3.65

Now, to find the actual concentration [H+], we need to take the antilog of -3.65:

[H+] = 10^(-3.65)

[H+] ≈ 2.85 × 10^(-4) M

So, the concentration of H+ ions in the solution is approximately 2.85 × 10^(-4) M.

Now, to find the pH of the solution, we take the negative logarithm of [H+]:

pH = -log([H+])
pH = -log(2.85 × 10^(-4))
pH ≈ 2.45

So, according to the calculations, the pH of the solution should be around 2.45. The value of 2.82 in the book seems to be different, which might be due to rounding differences or a calculation error.