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Answer the following questions regarding electrolysis and iodine sublimation: a) Identify the electrode that decreased in mass during electrolysis and explain why. b) Calculate the quantity of charge

Chemistry

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Answer

b

193000 C\text{193000 C}

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Here are the answers for question 39:

Part a

The electrode that decreased in mass during electrolysis is the anode (positive electrode).
Explanation: At the anode, copper metal is oxidized to copper(II) ions (Cu(s)→Cu2+(aq)+2e−Cu(s) \rightarrow Cu^{2+}(aq) + 2e^-), causing the copper electrode to dissolve into the solution and its mass to decrease.

Part b

193000 C\text{193000 C}

  1. Write the half-reaction for the deposition of copper.

    The deposition of copper at the cathode is given by the half-reaction:

    Cu2+(aq)+2e−→Cu(s)Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)

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

    From the balanced half-reaction, 2 moles of electrons are required to deposit 1 mole of copper.

  3. Calculate the total charge.

    The charge of 1 mole of electrons is 1 Faraday, which is 96500 Coulombs.
    Therefore, the quantity of charge needed to deposit 1 mole of copper is:

    Charge=2 moles of electrons×96500Cmol\text{Charge} = 2 \text{ moles of electrons} \times 96500 \frac{\text{C}}{\text{mol}} Charge=193000 C\text{Charge} = 193000 \text{ C}

    The quantity of charge needed to deposit 1 mole of copper is 193000 C\boxed{\text{193000 C}}.

Part c

Observation: When crystals of iodine are heated gently, purple/violet vapor is observed.
Explanation: Iodine undergoes sublimation, which is the direct change from a solid to a gaseous state without passing through a liquid phase. The iodine vapor has a characteristic purple color.

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