What is the pH of the solution in the anode compartment of the following cell if

Question

What is the pH of the solution in the anode compartment of the following cell if the measured cell potential at 25 degree C is 0.28 V: Pt(s)|H_2(1 atm)| H^+(xM|| Pb^2+(1 M) | Pb(s)? Some Standard Reduction (Electrode) Potentials at 25 degree C 2 H^+(aq) + 2 e^- rightarrow H_2(g) E degree = 0 V; Al^3 + (aq) + 3e^- rightarrow Al(s) E degree = -1.66 V Cr^3^+(aq) + 3 e^- rightarrow Cr(s) E degree = -0.74 V Fe^2+ (aq) + 2e^- rightarrow Fe(s) E degree = -0.44 V O_2(g) + 4 H^+ (aq) + 4 e^- rightarrow 2 H_2O(l) E degree = 1.23 V Cu^2+(aq) + 2e^- rightarrow Cu(s) E degree = 0.34 V; Sn^4+ (aq) + 2e^- rightarrow Sn^2+ (aq) E degree = 0.15 v Sn^2+ (aq) + 2e^- rightarrow Sn(s) E degree = -0.14 V Ni^2+ (aq) + 2e^- rightarrow Ni(s) E degree (s) E degree = -0.26 V

Explanation / Answer

Given

Ecell = 0.28V

We will apply here Nernst equation , which is

Ecell= E0cell - RT/nF ln Q

R = gas constant = 8.314 J / mol K

T = temperature = 298 K

F = 96485 C

n = number of electrons = 2(here)

Q = product of concentration of products / product of concentration of reactants

on solving

Ecell = E0cell - 0.0592 / n log Q

E0cell = E0cathode -E0anode

E0cathode = -0.113 V

E0anode = 0

E0cell = -0.113 - 0 = -0.113 V

Reaction will be [as per given cell]

H2 + Pb+2 --> Pb(s) + 2H+

Q = [H+]2 / [Pb+2]

Putting all values in Nernst equaiton

0.28 = -0.113 - 0.0592 / 2 log [[H+]2 / [Pb+2] ]

-0.393 = 0.0296 log [[H+]2 / [Pb+2] ]

-13.277 = log [[H+]2 / [Pb+2] ]

taking antilog

5.28 X 10^-14 = [[H+]2 / [Pb+2] ] = [H+]2

[H+] = 2.297 X 10^-7

pH = 6.64

Q =

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