ElectrochemistryHard
Question
The standard potential of the reaction H2O + e- → 
H2 + OH- at 298 K by using Kw (H2O) = 10-14, is :
H2 + OH- at 298 K by using Kw (H2O) = 10-14, is : Options
A.- 0.828 V
B.0.828 V
C.0 V
D.- 0.5 V
Solution
H+ + e- → 
H2 , Eo = 0 ,ᐃGo = 0
H2O ⇋ H+ + OH-, ᐃGo = - 8.314 × 298 ln 10-14
H2O + e- →
H2 + OH- , -1 × Eo × 96500 = - 8.314 × 298 ln10-14
Eo = - 0.828 Volt.
H2 , Eo = 0 ,ᐃGo = 0H2O ⇋ H+ + OH-, ᐃGo = - 8.314 × 298 ln 10-14
H2O + e- →
H2 + OH- , -1 × Eo × 96500 = - 8.314 × 298 ln10-14Eo = - 0.828 Volt.
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