Question

What is the pH of the solution at the cathode if the standard electrode potential equals –0.362 V for the following electrochemical cell at 25 °C?

                                                 Pt | H₂(g, 1.0 atm) | H⁺(aq, 1.00 M) || H⁺(aq) | H₂(g, 1.0 atm) | Pt

Answer 1

For concentration cell, cathode and anode are same electrode. So, this is a concentration cell.

So, Eo = 0

Number of electron being transferred in balanced reaction is 2 since H+ is changing to H2

So, n = 2

Use:

E = Eo - (2.303*RT/nF) log {[H+] at anode/[H+]at cathode}

Here:

2.303*R*T/n

= 2.303*8.314*298.0/F

= 0.0591

So, above expression becomes:

E = Eo - (0.0591/n) log {[H+] at anode/[H+]at cathode}

-0.362 = 0 - (0.0591/2) log (1.00/[H+])

0.362 = (0.0591/2) log (1.00/[H+])

log (1.00/[H+]) = 12.25

1.00/[H+] = 1.78*10^12

[H+] = 5.62*10^-13 M

Answer: 5.62*10^-13 M