Electrochemistry: Concentrations in a Voltaic Cell

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loreben

Full Member
10+ Year Member
Hi Sdners,

Question

What is the pH of the solution in the cathode compartment of a cell for the reaction shown below, when P(H2)=1atm , [Zn2+] in the anode = 0.10M, and the cell EMF is 0.542V.

Zn(s) + 2H(+) (aq) ----> Zn(2+) aq + H2 (g)

E red(oxidation process) = -0.763V [Zinc is the oxidized in this reaction]

E red (reduction process) = 0V [ hydrogen is reducedin this reaction]

(The answer is pH = 4.19)

Chowdder

Full Member
10+ Year Member
15+ Year Member
Hi Sdners,

Question

What is the pH of the solution in the cathode compartment of a cell for the reaction shown below, when P(H2)=1atm , [Zn2+] in the anode = 0.10M, and the cell EMF is 0.542V.

Zn(s) + 2H(+) (aq) ----> Zn(2+) aq + H2 (g)

E red(oxidation process) = -0.763V [Zinc is the oxidized in this reaction]

E red (reduction process) = 0V [ hydrogen is reducedin this reaction]

(The answer is pH = 4.19)

Kind of a complicated problem and a little bit too much calculation to be actual MCAT, but in any case I think this requires a bit more information that was provided, which should be given on the real MCAT.

I tried solving it but I can't seem to get the answer but mine came out pretty close.

I tried solving this using the Nernest equation E = E0 - 0.05916/z*log10Q
First of all E = 0.542 as given.
E0 = 0.763 according to the cell potential as given by the half reactions.
z = 2 since 2 moles of electrons are being transferred per 1 mole of Zn oxidation or hydrogen reduction.
Now you need to find Q which is the reaction quotient
according to the equation Q = [Product]/[Reactant]
......[Zn2+]*Pressure of Hydrogen gas
Q = --------------------------------
...........[H+]^2*1
Since Zn solid drops out because of it's activity as a solid is 1

We know [Zn2+] and pressure of hydrogen gas and you can directly find the [H+]

So just plug in all the values and I got [H+] = 5.8*10^-5
pH = -log10[H+] so I got a pH of 4.23

Not exactly your answer, but this is how I'd approach to solve it. Maybe I got a values mixed up there, but this is close enough to be picked as a correct answer on the actual MCAT.

loreben

Full Member
10+ Year Member
Hi,
I got the same answer as well. Thanks for your help. I just wanted to make sure i was not missing any minute details.