ElectrochemistryHard
Question
How much the potential of a hydrogen electrode will change when its solution initially at pH = 0 is neutralised to pH = 7 ?
Options
A.Increase by 0.059 V
B.Decrease by 0.059 V
C.Increase by 0.41 V
D.Decrease by 0.41 V
Solution
H+ + e- → 
H2. E = 0 - 
log10 
= + 0.591 log10[H+].
E1 = 0 {pH = 0}.
E2 = + 0.0591log10[10-7] = - .0591 × 7 {at pH = 7} = - 0.41 V.
H2. E = 0 - log10 = + 0.591 log10[H+].E1 = 0 {pH = 0}.
E2 = + 0.0591log10[10-7] = - .0591 × 7 {at pH = 7} = - 0.41 V.
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