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Electrochemistry

Question
CBSEENCH12006018

The emf of a cell corresponding to the reaction
Zn(s) + 2H+ (aq) → Zn2+ (0.1 M) + H2(g) (1 atm) is 0.28 V at 15° C.Write the half cell reactions and calculate the pH of the solution at the hydrogen electrode.

Solution
We have given the cell reaction
Zn(s) + 2H+ (aq) → Zn2+ (0.1 M) + H2(g) (1 atm) 
thus 
E0Zn2+/Zn = -0.76 V,   E0H+/H2 = 0
Half call reaction will be
            Zn2++2e-  Zn                    ...(i)
     H++e- = 12 H2
 or                2H+ + 2e-  H2         ...(ii)
        EZn/Zn2+ = E°Zn/Zn2+-RTnFlnZn2+Zn
Here          
          R = 8.314 J mol-1 log-1,T = 298 K,F = 96500 coulomb n = 2,E0Zn/Zn2+ = 0.76.
Therefore,
EZn/Zn2+ = 0.76 - 8.314 × 2982 × 96500 ln 0.11                   = 0.79 V
Similarly,
          EH+/H2 = E0H+/H2 - RTnFlnH2H+2             =0-0.314×2982×96500 ln IH+2             = 0.05915 log10H+              = 0.05915 pH                                       [ - log10H+ = pH]
Now since
            E = EZn/Zn2+ + EH+/H2    = 0.28 = 0.79 - 0.05915 pHpH = 0.150.05915 = 8.62.