The answer for a) 1E-7, b)-0.415 plz explain a,b,c The reduction potential of pu

Question

The answer for a) 1E-7, b)-0.415 plz explain a,b,c

The reduction potential of pure water is - 0.414 V under 1.0 atm H2 pressure. If the reduction for pure water is considered to be 2 H + (aq) + 2 e - rightarrow H2 (g), calculate the hydrogen ion concentration in pure water. The reduction potential for pure water could also be considered to be: 2 H2O (1) + 2 e - rightarrow H2 (g) + 2 OH - (aq). Knowing the hydrogen ion concentration calculated in part a, calculate the reduction potential for pure water using this alternate half reaction. Should the half reaction potential be the same for the reduction of water regardless of the version of the half reaction used? Explain.

Explanation / Answer

formulae :

Ecell = Eocell - (RT/nF) ln Q

Ecell = cell potential at non-standard state conditions
            Eocell = standard state cell potential = 0 for pure water.
            R = constant (8.31 J/mole K)
            T = absolute temperature (Kelvin scale) = 298K room tempreature.
            F = Faraday's constant (96,485 C/mole e-)
            n = number of moles of electrons transferred in the balanced equation for the reaction occurring
                  in the cell = 2 in this reaction.

Q = [H2] / [H+]^2 [H2] = 1 atm.

-0.414 = - RT/nFln([1/[H+]^2 )

[H+] = 1 x10^-7 M. pH = 7

part 2: Ecell = Eocell - (RT/nF) ln Q we have to find E given

  pure water pH = 7.0 and pOH =7 [OH-] = 10^-7.

E = -0.415 v

part c : yes regardless of the half rection reduction potential of pue water should remain constant

  

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