What is the pH of the solution in the cathode compartment of the following cell

Question

What is the pH of the solution in the cathode compartment of the following cell if the measured cell potential at 25 C is 0.60 V?

I have no idea where to even start with this problem. Please list step by step with correct answers so I can try to understand.

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What is the pH of the solution in the cathode compartment of the following cell if the measured cell potential at 25 C is 0.60 V? I have no idea where to even start with this problem. Please list step by step with correct answers so I can try to understand.

Explanation / Answer

oxidation: Zn(s) ---> Zn2+ & 2 e- Eo = + 0.763 volts
reduction: 2 H+(1M) @ 2 e- ---> H2(1atm) Eo = 0 volts
the standard Eo for your reaction would have been + o.763 volts

the measured cell potential got cut off
email me with it
or try using this

nernst equation:
E = Eo - (0.0592 / n )(log Q)

E = +0.763 - (0.0592 / 2 moles of e-) (log [prod] / [reactants]

E = +0.763 - (0.0296) (log [Zn+2] [H2] / [H+]^2 )

E = +0.763 - (0.0296) (log [1M] [1 atm] / [H+]^2 )

E =.5
.5= +0.763 - (0.0296) (log [1M] [1 atm] / [H+]^2 )

0.5 - 0.763 = - (0.0296) (log [1 ] / [H+]^2 )

- 0.263= - (0.0296) (log [1 ] / [H+]^2 )

- 0.263 / - (0.0296) = (log [1 ] / [H+]^2 )

0.8.885= log ([1 ] / [H+]^2 )

do a 10^x on both sides:
767361489.4 = ([1 ] / [H+]^2 )

[H+]^2 = 1.303*10^-9

[H+]^2=3.6055*10^-5

pH = 4.443

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