With PH2 = 1.0 atm and Zn2+ concentration of 1.0 M, the cell potential is 0.43 V. The concentration of H+ in the cathode compartment is ________ M.

Question

The standard cell potential (E°) of a voltaic cell constructed using the cell reaction below is 0.76 V:

Zn (s) + 2H⁺ (aq) → Zn²⁺ (aq) + H₂ (g)

Answer 1

The cell can be written as Zn(2) | Zn⁺(1.0M ) || H¹⁺ | H2(g) 1.0 atm | Pt(s).

According to the Nernst equation, the cell potential is given by

E_cell = E_{0 cell} - RT ln(Q) where Q is the ratio of product concentration to reactant concentration.

Q =[Zn²⁺] / [H¹⁺] / Substitute into Nernst equation.

0.43 = 0.76 - 8.314 x 298 x ln [1.0] / [H⁺] Solve for [H¹⁺]. As the absolute temperature wasn't given, I assumed 25°Celsius.

0.43 V - 0.76 V / -8.314 x 298 = ln [1.0 M] - ln [H¹⁺]

ln [H^1+] = 0.000133 and [H¹⁺] = 1.00 M after taking antilog.

It has been a long time since I did a problem of this type; please let me know if this is correct. Thanks.