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A) Identify the Cathode and Anode. b) If a current of 0.5 A was passed through the solution for 20 minutes, determine the volume of Oxygen liberated at STP. (Molar gas volume = 24 dm³)

Chemistry

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Answer

b

0.0373 dm30.0373 \text{ dm}^3

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Here's the solution to your questions:

Part a

On the diagram, identify the Cathode and Anode.

In an electrolytic cell, the electrodes are connected to a power source.

  • The cathode is the negative electrode, connected to the negative terminal of the power supply. Reduction occurs here.
  • The anode is the positive electrode, connected to the positive terminal of the power supply. Oxidation occurs here.

Based on the diagram:

  • The electrode on the left is connected to the negative terminal (shorter line) of the power source, so it is the cathode.
  • The electrode on the right is connected to the positive terminal (longer line) of the power source, so it is the anode.

Part b

0.0373 dm30.0373 \text{ dm}^3

If a current of 0.5 A0.5 \text{ A} was passed through the solution for 20 minutes20 \text{ minutes}, determine the volume of Oxygen liberated at STP. (Molar gas volume =24 dm3= 24 \text{ dm}^3).

The electrolysis of dilute copper(II) sulfate involves the following reaction at the anode (positive electrode), where water is oxidized to produce oxygen gas:

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

From the reaction, 4 moles of electrons are required to produce 1 mole of oxygen gas.
We will use Faraday's constant F=96500 C/molF = 96500 \text{ C/mol} as indicated in your notes.

  1. Calculate the total charge (QQ) passed.

    The time must be in seconds.

    t=20 minutes×60 s/minute=1200 st = 20 \text{ minutes} \times 60 \text{ s/minute} = 1200 \text{ s} Q=I×tQ = I \times t Q=0.5 A×1200 sQ = 0.5 \text{ A} \times 1200 \text{ s} Q=600 CQ = 600 \text{ C}

  2. Calculate the number of moles of electrons (nen_e).

    ne=QFn_e = \frac{Q}{F} ne=600 C96500 C/moln_e = \frac{600 \text{ C}}{96500 \text{ C/mol}} ne≈0.0062176 moln_e \approx 0.0062176 \text{ mol}

  3. Calculate the number of moles of Oxygen (nO2n_{\text{O}_2}) produced.

    From the balanced equation, 4 moles of electrons produce 1 mole of O2_2.

    nO2=ne4n_{\text{O}_2} = \frac{n_e}{4} nO2=0.0062176 mol4n_{\text{O}_2} = \frac{0.0062176 \text{ mol}}{4} nO2≈0.0015544 moln_{\text{O}_2} \approx 0.0015544 \text{ mol}

  4. Calculate the volume of Oxygen liberated.

    Given Molar gas volume =24 dm3/mol= 24 \text{ dm}^3/\text{mol}.

    Volume of O2=nO2×Molar gas volume\text{Volume of O}_2 = n_{\text{O}_2} \times \text{Molar gas volume} Volume of O2=0.0015544 mol×24 dm3/mol\text{Volume of O}_2 = 0.0015544 \text{ mol} \times 24 \text{ dm}^3/\text{mol} Volume of O2≈0.0373056 dm3\text{Volume of O}_2 \approx 0.0373056 \text{ dm}^3

    Rounding to three significant figures:

    Volume of O2≈0.0373 dm3\text{Volume of O}_2 \approx \boxed{0.0373 \text{ dm}^3}
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