need help to solve

Question

For the cell shown, the measured cell potential, 𝐸 cell is 0.3629 V at 25 °c

Pt(s) | H2(g,0.707 atm) | H+(aq,? M) || Cd2+(aq,1.00 M) | Cd(s)

The balanced reduction half-reactions for the cell, and their respective standard reduction potential values, , are

2H+(aq)+2e−⟶H2(g)𝐸o=0.00 V

Cd2+(aq)+2e−⟶Cd(s)𝐸o=−0.403 V

Calculate the H+ concentration.

Answer 1

Oxidation (anode): H₂(g) ==> 2H⁺ + 2e-

Reduction (cathode): Cd²⁺(aq) + 2e- ==> Cd(s)

Overall redox: H₂(g) + Cd²⁺(aq) ==> 2H⁺(aq) + Cd(s)

Eºcell = cathode - anode = -0.403 - 0.00 = -0.403 V

Using the Nernst equation: Ecell = Eºcell - 0.0592/n log [H+]²/(PH₂)[Cd²⁺]

Now, I'm going to use the value you gave for Ecell of 0.3629 V, but I believe it should be -0.3629. If my assumption is correct, you will have to go back and re-do the calculations plugging in the negative value where I have the positive 0.3629.

0.3629 = -0.403 - 0.0592/2 log [H+]² / (0.707)(1)

0.0592/2 log [H⁺]² / 0.707 = -0.403 - 0.3629

0.0296 log [H⁺]² /0.707 = -0.7659

0.0296 log [H⁺]² = -0.5415

2 log [H⁺] = -18.3

log [H⁺] = -9.15

[H⁺] = 7.1x10^-10 M (which doesn't make sense, so my assumption that the original Ecell should be negative was probably correct, and you can now re-do the calculations using the negative value).