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What is the density of dry air at 101 KPa? What is the density of oxygen? Calculate during the electrolysis of dilute Sodium Chloride. If 96500 C of charge is passed?

Chemistry

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Answer

48250 C\text{48250 C}

Step-by-step solution

4 steps

  1. Write the half-reaction for the liberation of oxygen gas.

    During the electrolysis of dilute sodium chloride, oxygen gas is produced at the anode from the oxidation of water.

    2H2O(l)→O2(g)+4H+(aq)+4e−2\text{H}_2\text{O}(l) \to \text{O}_2(g) + 4\text{H}^+(aq) + 4\text{e}^-

  2. Determine the moles of electrons required per mole of oxygen.

    From the balanced half-reaction, 4 moles of electrons are required to produce 1 mole of oxygen gas.

  3. Calculate the total moles of electrons needed for 0.125 mole of oxygen.

    Moles of electrons=0.125 mol O2×4 mol e−1 mol O2=0.500 mol e−\text{Moles of electrons} = 0.125 \text{ mol O}_2 \times \frac{4 \text{ mol e}^-}{1 \text{ mol O}_2} = 0.500 \text{ mol e}^-

  4. Calculate the quantity of electricity in coulombs.

    Using Faraday's constant (F=96500 C/mol e−F = 96500 \text{ C/mol e}^-):

    Quantity of electricity=Moles of electrons×F\text{Quantity of electricity} = \text{Moles of electrons} \times F Quantity of electricity=0.500 mol e−×96500 C/mol e−\text{Quantity of electricity} = 0.500 \text{ mol e}^- \times 96500 \text{ C/mol e}^- Quantity of electricity=48250 C\text{Quantity of electricity} = 48250 \text{ C}

    The quantity of electricity required is 48250 C\text{48250 C}.

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