(1) $$2C{l^ - }(aq) + 2{H_2}O \to 2O{H^ - }(aq) + {H_2}(g) + C{l_2}(g)$$

Cathode reaction : $$2{H_2}O + 2e \to 2O{H^ - } + {H_2}$$

Anode reaction : $$2C{l^ - } \to C{l_2} + 2e$$

(2) According to Faraday's first law, $$W = Z \times It$$ ...... (1)

Current efficiency = 62% = 0.62

Effective electric current (I) = 25 $$\times$$ 0.62 A

Mass of $$C{l_2} = {10^3}g$$ $$\therefore$$ Z for $$C{l_2} = {{eq.\,weight} \over {96500}} = {{35.5} \over {96500}}$$

Substituting in equation (1), $${10^3} = {{35.5} \over {96500}} \times 25 \times 0.62 \times t$$

or, $$t = {{{{10}^3} \times 96500} \over {35.5 \times 25 \times 0.62}} = 175374.83$$ sec = 48.71 hrs

(3) No. of equivalents of OH^$$-$$ = no. of equivalents of Cl₂ = $${{{{10}^3}} \over {35.5}} = 28.17$$ (equivalent weight of Cl₂ = 35.5)

Equivalent weight and formula mass of OH^$$-$$ are equal.

$$\therefore$$ Number of moles of OH^$$-$$ = 28.17

$$\therefore$$ $$[O{H^ - }] = {{number\,of\,moles} \over {volume\,(in\,L)}} = {{28.17} \over {20}} = 1.408\,M$$