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 0.52 V ?

Zn(s)∣∣Zn2+(1M)∣∣∣∣H+(?M)∣∣H2(1atm)|Pt(s)

Cell Potential at 25 C and 0.52 V

Answer 1

Here's the reaction for the voltaic cell described,

`Zn(s) + 2H^(+)(aq) rightleftharpoons H_2(g) + Zn^(2+)(aq)`

Using the (simplified) Nernst equation, and a table of standard reduction potentials, we may solve this,

`E = E^° - (0.0592/n)*logQ`,

`Q_(eq) = (P_(H_2)[Zn^(2+)])/([H^+]^2)`, and

`E^° = 0.763"V"` per standard tables.

The concentration of hydronium ions is,

`0.52"V" = 0.763"V" - (0.0592/2)* (log(1"atm" * 1"M")-log[H^+]^2)`
`therefore [H^+] approx 7.9*10^-5"M"`

Hence,

`"pH" = -log[H^+] approx 4.10`