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Write the half-reaction for copper deposition.

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13

48 g\text{48 g}

Step-by-step solution

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Part 13· 4 steps

Calculate the mass of copper metal that was obtained.

  1. Write the half-reaction for copper deposition.

    During electrolysis, copper(II) ions gain electrons to form solid copper.

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

    This reaction shows that 2 moles of electrons2 \text{ moles of electrons} are required to deposit 1 mole of Cu1 \text{ mole of Cu}.

  2. Calculate the moles of electrons transferred.

    Given charge Q=144,750 CQ = 144,750 \text{ C} and Faraday's constant F=96,500 C/mol e−F = 96,500 \text{ C/mol e}^-.

    Moles of electrons=QF\text{Moles of electrons} = \frac{Q}{F} Moles of electrons=144750 C96500 C/mol e−\text{Moles of electrons} = \frac{144750 \text{ C}}{96500 \text{ C/mol e}^-} Moles of electrons=1.5 mol e−\text{Moles of electrons} = 1.5 \text{ mol e}^-

  3. Calculate the moles of copper deposited.

    From the half-reaction, 1 mole of Cu1 \text{ mole of Cu} requires 2 moles of electrons2 \text{ moles of electrons}.

    Moles of Cu=Moles of electrons2\text{Moles of Cu} = \frac{\text{Moles of electrons}}{2} Moles of Cu=1.5 mol2\text{Moles of Cu} = \frac{1.5 \text{ mol}}{2} Moles of Cu=0.75 mol\text{Moles of Cu} = 0.75 \text{ mol}

  4. Calculate the mass of copper deposited.

    Given molar mass of Cu = 64 g/mol64 \text{ g/mol}.

    Mass of Cu=Moles of Cu×Molar mass of Cu\text{Mass of Cu} = \text{Moles of Cu} \times \text{Molar mass of Cu} Mass of Cu=0.75 mol×64 g/mol\text{Mass of Cu} = 0.75 \text{ mol} \times 64 \text{ g/mol} Mass of Cu=48 g\text{Mass of Cu} = 48 \text{ g}

    The mass of copper metal obtained is 48 g\boxed{\text{48 g}}.

Part 14

State one physical and one chemical difference between Na₂O₂ and Na₂O.

Physical difference:
Sodium peroxide (Na2O2Na_2O_2) is a pale yellow solid, while sodium oxide (Na2ONa_2O) is a white solid.

Chemical difference:
Sodium peroxide (Na2O2Na_2O_2) reacts with water to produce hydrogen peroxide (H2O2H_2O_2) and sodium hydroxide, showing its peroxidic nature.

Na2O2(s)+2H2O(l)→2NaOH(aq)+H2O2(aq)Na_2O_2(s) + 2H_2O(l) \to 2NaOH(aq) + H_2O_2(aq)

Sodium oxide (Na2ONa_2O) reacts with water to produce only sodium hydroxide, as it is a basic oxide.

Na2O(s)+H2O(l)→2NaOH(aq)Na_2O(s) + H_2O(l) \to 2NaOH(aq)

Part 15

The question states "The diagram below shows an electrochemical cell:". However, the diagram itself is not visible in the image provided. Therefore, this question cannot be answered without the diagram.

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