The cell potential of the following electrochemical cell depends on the pH of the solution in the anode half-cell:Pt(s)|H2(g, 1atm)|H+(aq, ?M)||Cu2+(aq,1.0M)|Cu(s)What is the pH of the solution if Ecell = 355 mV?

0.51

Explanation:

Given the Nernst equation;

E= E° - 0.0592/n logQ

E= 355 mV or 0.355 V

E° = 0.34 - 0= 0.34 V

n= 2(two electrons were transferred in the process)

Equation of the reaction;

H2(g) + Cu^2+(aq) > 2H^+(aq) + Cu(s)

Substituting values;

0.355 = 0.34 - 0.0592/2 log([H^+]/1)

0.355 - 0.34 = - 0.0296 log [H^+]

0.015/-0.0296 = log [H^+]

Antilog (-0.5068) = [H^+]

[H^+] = 0.311 M

pH = -log[H^+]

pH= - log(0.311 M)

pH = 0.51

e

explanation: