a 100 mL of a2.00 M NaCl was electrolyzed for 25 min with 1.45 A what is the concentration of hypochlorite produced?

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  • 6 months ago

    Chlorination of water by electrolysis

    The electrolysis of brine produces H2 gas and Cl2 gas in a basic solution. Cl2 gas reacts with water to produce HOCl(aq) and HCl(aq). But in basic solution the products are OCl- and Cl- (which is one of the reactants). Without going through all the steps, the overall reaction is:

    .... NaCl(aq) + H2O(l) --electricity --> NaOCl(aq) + H2(g)

    .............. Cl- + H2O(l) --electricity --> OCl- + H2(g)

    ..............100 mL ............ ....................?M

    Cl- + 2OH- --> OCl- + H2O + 2e- .......... oxidation half-reaction

    2 mol e- per mol of Cl- and OCl-

    25 min x (60s / 1 min) x (1.45 C / 1s) x (1 mol e- / 96485C) x (1 mol OCl- / 2 mol e-) / 0.100L = 0.113M OCl- ................. 0.0113 mol OCl- is produced

    ========= =========

    Reduction half-reaction

    2H2O + 2e- --> H2 + 2OH-

    25 min x (60s / 1 min) x (1.45 C / 1s) x (1 mol e- / 96485C) x (1 mol H2O / 1 mol e-) = 0.0225 mol H2O ............ 0.0225 mol H2O is reduced, 0.0113 mol H2 is produced

    2H2O + 2e- --> H2 + 2OH- ................... reduction half-reaction

    Cl- + 2OH- --> OCl- + H2O + 2e- .......... oxidation half-reaction

    ----------------- ----------------- -------------

    Cl- + 2OH- + 2H2O --> OCl- + H2O + H2 + 2OH-

    simplify

    Cl- + H2O --> OCl- + H2

    Cl- .......+...... H2O -------> OCl- ...+... H2

    0.0113 mol .. 0.0113 mol ....0...............0 .................. before reaction

    .0........... .......0............ 0.0113 mol .. 0.0113 mol...... after reaction

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