A tin electrode in .619M Sn(NO3)2 is connected to a hydrogen electrode in which the pressure of H2 is 1 bar

Question

Sn I Sn^2 II H^ I H2 I Pt

If the cell potential is .061 V at 25 deg C, what is the pH of the electrolyte at the hydrogen electrode?

Answer 1

Sn --> Sn+2 ==> + 0.136 volts

nernst:

E = Eo - (0.0592 / n) (logQ)

0.061  = 0.136 - (0.0592 / 2) (log[ H2] [Sn+2] / [H+])

0.075  =   (0.0296) (log[ 0.987 atm] [0.619] / [H+])

0.075 / (0.0296) =    (log[ 0.987 atm] [0.619] / [H+])

2.533 =    (log[ 0.987 atm] [0.619] / [H+])

10^x both sides

341.8 = [ 0.987 atm] [0.619] / [H+])

H+ = [0.987 atm] [0.619] / 341.8

H+ = 1.79 e-3

pH = 2.75