For the electrode Pt, H2 (1 atm) | H+ (aq) (XM), the reduction electrode potential at 25 degrees C os -0.34V. Write the electrode reaction equation and calculate the value of X. How will you deduce the pH of the solution from this result

The given cell isPt, H2 (1 atm) // H+(aq) (X M)At anode : 12H2  H+ + e-At Cathode :  H+ + e- 12H2Net cell reaction is12H2(1 atm)   H+(X M) + e-The Emf of the cell at std condition = Eox0 + ERed0 = 0According to Nernst equation,E = E0 - 0.0592n log [anode][cathode]-0.34 = 0 - 0.05921 log 1[H+]-0.34 = -0.0592 (log 1 -log [H+]-0.34 = -0.0592 ( -log [H+])-0.34 = -0.0592 pHpH = 0.340.0592 = 5.74

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