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

Options

A. dil . `H_(2)SO_(4)` with pt electrode
B. aqueous `AgNo_(3)` with pt electrode
C. dil. `H_(2)SO_(4)` with Cu electrode
D. fused `NaOH` with Fe cathod and Ni anode

Answer

A, B

Explanation

Clearly `O_(2)` Will be evolved in (A) and (B)
In (C ) ,Cu Will get oxidised. `[Cu(s)rightarrowCu^(2+) (aq)+2e^(-) E_(Cu//Cu^(2+))^(@) =- 0.34V]`
`[xx2H_(2)O_(1)rightarrow O_(2) (g) + 4H^(+) (aq) + 4e^(-) e_(H_(2)O //O_(2))= -1.23V]`
In (D) `[ Ni(s) rightarrow Ni^(2+) (aq) + 2e^(-) E_(Ni//Ni^(2+))= -0.25V]`
`[ xx4OH^(-) (aq) rightarrow O_(2) (g) + 2H_(2)O (1) + 4e^(-) E_(OH^(-)// O_(2))=-0.4V]`