Chem help please?

Question

Calculate the pH of a cathod compartment solution if the cell emf at 298k is measured to be 0.680 V when [Zn^2+]=0.33M and PH2=0.92atm

Answer 1

2H⁺(aq) + 2e- ===> H2(g) ... cathode reduction reaction Eº = 0

Zn(s) ===> Zn²⁺(aq) ... anode oxidation reaction Eº = -0.76 V

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Zn(s) + 2H⁺(aq) ==> Zn²⁺ + H2(g) ... overall reaction Eº = +0.76 V

Using a version of the Nernst equation for 298K, we can solve for [H+] and then pH:

E = Eº - 0.0592/n log Q where n = moles of electrons = 2, and Q = reaction quotient

Q = [Zn²⁺](P_H2) / [H⁺]² = (0.33)(0.92) / [H⁺]² = 0.3036/[H⁺]²

0.680 = -0.76 - (0.0296)log 0.3036/[H⁺]²

1.44 = - (0.0296)log 0.3036 / [H⁺]²

At this point, it's more algebra than chemistry to solve for [H⁺]. Then take the negative log of that value to obtain the pH in the cathode compartment. My math skills are not what they should be, but when I solved this I obtained pH = 6.93. You should most definitely check the math, however.