During electrolysis, `O_(2)(g)` is evolved at anode in

Options

A. Dilute `H_(2)SO_(4)` with `Pt` electrode
B. Aqueous `AgNO_(3)` with `Pt` electrode
C. Dilute `H_(2)SO_(4)` with `Cu` electrode
D. Fused `NaOH` with an `Fe` cathode and `Ni` anode

Answer

a,b

Explanation

Clearly, `O_(2)` will be evolved in `(a)` and `(b)`
In `(c), Cu` will get oxidized. `[(sqrt())Cu(s)rarr Cu^(2+)(aq)+2e^(-)
E^(c-)._(Cu|Cu^(2+))=-0.34V
`(X)2H_(20O(l)rarr O_(2)(g)+4H^(o+)(aq)+4e^(-)`
`E^9c-)._(H_(2)|O_(2))=-1.23V`
In `(d)[((sqrt())Ni(s) rarr Ni^(2+)(aq)+2e^(-)`
`E^(c-).Ni|Ni^(2+)=-0.25V`
`(X)4overset(c-)(O)H(aq)rarrO_(2)(g)+2H_(2)O(l)+4e^(-)`
`E^(c-)._(overset(c-)(O)H|O_(2))=-0.4V)`