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Define and write the units of the following terminologies: (i) Specific conductivity (ii) equivalent conductance (iii) molar conductance.

Chemistry

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Step-by-step solution

5 steps

To determine the current strength required, we first need to calculate the total charge needed to liberate 10g of iodine.

  1. Write the half-reaction for the liberation of iodine.

    Iodide ions (I−I^-) are oxidized to iodine (I2I_2) at the anode.

    2I−→I2+2e−2I^{-} \to I_2 + 2e^{-}
  2. Calculate the molar mass of I2I_2 and the moles of I2I_2 to be liberated.

    Given atomic weight of iodine = 127 g/mol127 \text{ g/mol}.
    Molar mass of I2=2×127 g/mol=254 g/molI_2 = 2 \times 127 \text{ g/mol} = 254 \text{ g/mol}.
    Mass of I2I_2 to be liberated = 10 g10 \text{ g}.

    Moles of I2=massmolar mass=10 g254 g/mol≈0.03937 mol\text{Moles of } I_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{10 \text{ g}}{254 \text{ g/mol}} \approx 0.03937 \text{ mol}
  3. Determine the moles of electrons required.

    From the half-reaction, 1 mole of I2I_2 requires 2 moles of electrons.

    Moles of electrons=2×moles of I2=2×0.03937 mol≈0.07874 mol\text{Moles of electrons} = 2 \times \text{moles of } I_2 = 2 \times 0.03937 \text{ mol} \approx 0.07874 \text{ mol}
  4. Calculate the total charge (Q) required.

    Given Faraday's constant = 96,500 C/mol96,500 \text{ C/mol}.

    Q=moles of electrons×Faraday’s constantQ = \text{moles of electrons} \times \text{Faraday's constant} Q=0.07874 mol×96,500 C/mol≈7600.8 CQ = 0.07874 \text{ mol} \times 96,500 \text{ C/mol} \approx 7600.8 \text{ C}

    The total charge required to liberate 10 g10 \text{ g} of iodine is 7600.8 C\boxed{7600.8 \text{ C}}.

  5. Determine the current strength in amperes.

    Current (II) is defined as charge (QQ) per unit time (tt): I=QtI = \frac{Q}{t}.
    Since the time (tt) over which the liberation occurs is not provided in the question, the current strength in amperes cannot be uniquely determined. To calculate the current, a specific time duration must be given.

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