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Table 4 shows some electrode half-equations and their standard electrode potentials - AQA - A-Level Chemistry - Question 10 - 2017 - Paper 1

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Table 4 shows some electrode half-equations and their standard electrode potentials. Electrode half-equation E° / V Cl₂(g) + 2e⁻ → 2Cl⁻(aq) ... show full transcript

Worked Solution & Example Answer:Table 4 shows some electrode half-equations and their standard electrode potentials - AQA - A-Level Chemistry - Question 10 - 2017 - Paper 1

Step 1

Deduce the oxidation state of nitrogen in NO₃⁻ and in NO.

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Answer

Nitrogen in NO₃⁻: The oxidation state of nitrogen can be calculated considering the charge of the ion. The overall charge is -1, and knowing that each oxygen has an oxidation state of -2:

Let the oxidation state of nitrogen be x. The equation is:

x+3(−2)=−1x + 3(-2) = -1

Solving for x gives: x−6=−1x=+5x - 6 = -1 \\ x = +5

Nitrogen in NO: Here, the equation is simpler:

x+(−2)=0x + (-2) = 0

Solving gives: x=+2x = +2

Step 2

State the weakest reducing agent in Table 4.

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Answer

The weakest reducing agent in Table 4 is Fe²⁺, as it has the lowest standard electrode potential (+0.77 V).

Step 3

Write the conventional representation of the cell that has an EMF of +0.43 V.

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Answer

The cell representation would be:

Anode∣Cu2+(aq)/Cu(s)∣∣NO3−(aq)/NO(g)∣Cathode\text{Anode} | \text{Cu}^{2+}(aq) / \text{Cu}(s) || \text{NO}_3^{-}(aq) / \text{NO}(g) | \text{Cathode}

Step 4

Use data from Table 4 to identify an acid that will oxidise copper. Explain your choice of acid. Use these data to suggest a possible equation for the reaction. Calculate the EMF of the cell that has the same overall reaction.

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Answer

The suitable acid that can oxidise copper is nitric acid (HNO₃). This is due to the higher oxidizing potential of the nitrate ion (NO₃⁻).

Possible reaction equation:

3Cu(s)+8H+(aq)+2NO3−(aq)→3Cu2+(aq)+2NO(g)+4H2O(l)\text{3Cu}(s) + 8H^+(aq) + 2NO_3^{-}(aq) → 3Cu^{2+}(aq) + 2NO(g) + 4H_2O(l)

To calculate the EMF for the reaction:

Using the half-reactions:

  • Reduction:
    NO3−(aq)+4H+(aq)+3e−→NO(g)+2H2O(l)\text{NO}_3^{-}(aq) + 4H^+(aq) + 3e^- → \text{NO}(g) + 2H2O(l)
    E° = +0.96 V
  • Oxidation: Cu(s)→Cu2+(aq)+2e−\text{Cu}(s) → \text{Cu}^{2+}(aq) + 2e^-
    E° = -0.34 V

Overall EMF:

  • The overall EMF can be calculated using: EMF=Ecathode−Eanode=0.96−0.34=0.62VEMF = E_{cathode} - E_{anode} = 0.96 - 0.34 = 0.62 V

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