Correct Answer - (a) `V(O_(2)) =6.2L`, (b) `V(O_(2)) =0.824L`
`I : 2H_(2)O rarr 4H^(+) +O_(2) +4e^(-)`
`Ag^(+) +e^(-) rarr Ag`
`n_(E) = n_(Ag) = 1`
`n_(O_(2)) = (1)/(4)`
From `PV = nRT`
`V_(O_(2)) = 6.2L`
II: `n_(Ag^(+)) = n_(E) = 0.133`
`n_(O_(2)) = (1)/(4) xx 0.133`
From `PV = nRT`
`V_(O_(2)) = 0.824L`