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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⁻: +5

Nitrogen in NO: +2

Step 2

State the weakest reducing agent in Table 4.

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Answer

The weakest reducing agent in Table 4 is Fe³⁺, as it has the lowest standard electrode potential.

Step 3

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

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Answer

The conventional representation of the cell is:

Cu | Cu²⁺ || 2H⁺ | H2

Step 4

Use data from Table 4 to identify an acid that will oxidise copper.

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Answer

An acid that will oxidise copper is nitric acid (HNO3).

Explanation: Nitric acid is a stronger oxidizing agent as it contains NO3⁻ which can readily react with copper.

Balanced Equation: 3Cu + 8H⁺ + 2NO3⁻ → 3Cu²⁺ + 2NO + 4H2O

EMF Calculation: The overall EMF for the reaction is calculated as follows:

For the half-reactions:

  • Cu²⁺ + 2e⁻ → Cu (E° = +0.34 V)
  • 2NO3⁻ + 4H⁺ + 2e⁻ → NO(g) + 2H2O (E° = +0.96 V)

Combining these, we get:

EMF = (E° for reduction) - (E° for oxidation) = 0.96 V - 0.34 V = 0.62 V.

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