For the electrode Pt, H2 (1 atm) | H+ (aq) (XM), the reduction electrode potential at 25 degrees C os - Chemistry - Electrochemistry

Question

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

Geetha · Accepted Answer

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

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