Correct Answer - A
Process at Anode `H_(2)O rarr (1)/(2)O_(2)+2H^(+)+2e`
Process at Cathode `2H_(2)O +2e rarr H_(2)uarr +2OH^(-)`
Total change passed `= 9.65 xx 10^(4)`
`= 96500` Calorie `= 1F`
`rArr` mole of `H^(+)` produced in Anode `= 1` mole
mole of `OH^(-)` produced in Cathode `= 1` mole
`{:("Anode":,HPO_(4)^(2-)+,H^(+)rarr,H_(2)PO_(4)^(-)),(,(1)/(3),1,(1)/(3)),(,-,(2)/(3),(2)/(3)):}`
`{:(H_(2)PO_(4)^(-)+,H^(+)rarr,H_(3)PO_(4)),((2)/(3),(2)/(3),0),(0,0,(2)/(3)):}`
`{:("Cathode":,H_(2)PO_(4)^(-)+,OH^(-)rarr,HPO_(4)^(2-)),(,(1)/(3),1,(1)/(3)),(,-(1)/(3),-(1)/(3),+(1)/(3)),(,=(2)/(3),=(2)/(3),):}`
`{:(HPO_(4)^(2-)+,OH^(-)rarr,PO_(4)^(3-)),((2)/(3),(2)/(3),(2)/(3)):}`
`K_(a_(1)) = 10^(-4)`
`K_(a_(2)) = 10^(-7)`
`K_(a_(3)) = 10^(-10)`
`pH` in Anodic compartment:
`pH = (1)/(2)pK_(a_(1)) -(1)/(2)logC`
`= (1)/(2)xx4 -(1)/(2) (log2 -log3)`
`= 2 +0.09 = 2.09`