Question

For the following reaction: NiO2(s) + 4 H+(aq) + 2 Ag(s) → Ni2+(aq) + 2 H2O(l) + 2 Ag+(aq) script E° = 2.48 V Calculate the pH of the solution if script E = 2.32 V and [Ag+] = [Ni2+] = 0.012 M. Use the Standard Reduction Table.

Answer 1

NiO2(s) + 4 H+(aq) + 2 Ag(s) → Ni2+(aq) + 2 H2O(l) + 2 Ag+(aq)

so Ka = [Ni+2 ]*[2H2O ]^2*[Ag+ ]^2

           -----------                                            = (0.012)(1)^2 (0.012)^2/(1)* (H+)^4 (1)^2    = (0.012)^3/(H+)^4

              [NiO) ]*[H+ ]^4*[Ag ]^2

                so Ka =1.726*10^-6 (H+)^4

so ln Ka = ln (1.726*10^-6) +ln(H+)^4    ------------(1)

Again we know E° = RTlnK/nF

                  so 2.48 = 8.314*298/2*96500 lnK = 0.0128 lnK

ln k = 2.48/.0128 = 193.75          --------------(2)

From(1) and(2) , we get ln 1.726*10^-6 + ln(H+)^4 = 193.75

              From here calculate H+ ion concl

Then pH = - log (H+ion conc)