1 litre aqueous solution of NaCl was electrolysed bètweern Pt electrodes passing a direct current of 12.87 A for 100s with a current efficiency of 75%. Calculate pH of the solution after electrolysis assuming no change in volume of solution.

Explanation

`NaCl_((aq)) to Na_((aq))^(+) +Cl_((aq))^(-)`
`H_(2)O Leftrightarrow H_((aq))^(+)+OH_((aq))^(+)`
Half reactions at electrodes:
`H^(+)+e Leftrightarrow 1/2 H_(2)("at cathode")`
`Cl^(-) to 1/2 Cl_(2)+e ("at anode")`
Due to discharge of `H^(+)` ion, the dissociation equilibrium of water will be disturbed and in order to maintain the constancy of `K_(w), OH^(-)` ions concentration will increase rendering the solution alkaline.
`H_(2)O Leftright H^(+)+OH^(-)`
`H^(+) +e Leftrightarrow 1/2 H_(2)("at cathode")`
`Cl^(-) to 1/2 Cl_(2)+e(" at anode")`
Due to discharge of `H^(+)` ion, the dissociation equilibrium of water will be disturbed and in order to maintain the constancy of `K_(w), OH^(-)` ion concentration will increase rendering the solution alkaline.
`H_(2)O Leftrightarrow H^(+) +OH^(-)`
`H^(+) +e Leftrightarrow 1/2 H_(2)`
`=H_(2)O+underset("1 mole")(e) Leftrightarrow 1/2 H_(2)+underset("1 mole")(OH^(-))`
For every mole of electrons (i.e, 1F of charge), 1 mole of `OH^(-)` ion will be produced
Charged passed `=(12.87 xx 100 xx 0.75)/(96500) =0.01F`
Amount of `OH^(-)` ion produced =0.01 mole
`[OH^(-)]=("No. of mole")/("Volume of solution in litre")=(0.01)/(1)=0.01M`
pOH=2, pH=12