Table 4 shows some electrode half-equations and their standard electrode potentials. Table 4 Electrode half-equation E° / V Cl2(g) + 2e⁻ → 2Cl⁻(aq) +1.36 NO3⁻(aq) + 4H⁺(aq) + 3e⁻ → NO(aq) + 2H2O(l) +0.96 Fe3+(aq) + 6e⁻ → Fe(s) +0.77 Cu2+(aq) + 2e⁻ → Cu(s) +0.34 SO4²⁻(aq) + 4H⁺(aq) + 2e⁻ → SO2(g) + 2H2O(l) +0.17 2H⁺(aq) + 2e⁻ → H2(g) 0.00 Fe²⁺(aq) + 2e⁻ → Fe(s) -0.44 1.0.1 Deduce the oxidation state of nitrogen in NO3⁻ and in NO. Nitrogen in NO3⁻: Nitrogen in NO: 1.0.2 State the weakest reducing agent in Table 4. 1.0.3 Write the conventional representation of the cell that has an EMF of +0.43 V. 1.0.4 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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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. Table 4 Electrode half-equation E° / V Cl2(g) + 2e⁻ → 2Cl⁻(aq) +1.36 NO3⁻(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

For the oxidation state of nitrogen in NO3⁻:

The total charge of the ion is -1. Each oxygen has an oxidation state of -2.
Therefore, for three oxygen atoms, we have: $3(-2) + x = -1$ Solving for x gives us: $x = +5$ Thus, nitrogen in NO3⁻ has an oxidation state of +5.

For nitrogen in NO:

Oxygen has an oxidation state of -2. This leads to: $x - 2 = 0$ Which means: $x = +2$ Thus, nitrogen in NO has an oxidation state of +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⁺(aq) because it has the lowest electrode potential (-0.44 V), indicating that it is less favorable to reduce compared to the others.

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 is:
$ext{Cu(s)} | ext{Cu}^{2+}(aq) || ext{Fe}^{3+}(aq), ext{Fe}(s)$
This shows copper as the anode and iron as the cathode, where the half-cell reactions correspond to their respective standard electrode potentials.

Step 4

Use data from Table 4 to identify an acid that will oxidise copper. Explain your choice of acid.

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Answer

The appropriate acid that can oxidise copper is nitric acid (HNO3). Nitric acid contains the nitrate ion (NO3⁻), which has a higher electrode potential (+0.96 V) than copper (+0.34 V). This potential indicates that nitrate ion can act as an oxidizing agent, thus being capable of oxidizing copper.

The possible equation for the reaction is:
$3 ext{Cu}(s) + 8 ext{H}^{+}(aq) + 2 ext{NO}_3^−(aq) \rightarrow 3 ext{Cu}^{2+}(aq) + 2 ext{NO}(g) + 4 ext{H}_2O(l)$

To calculate the EMF of the cell for the reaction:

The EMF for the reaction can be computed by taking the difference between the electrode potentials of the cathode and anode:
$ext{EMF} = E_{cathode} - E_{anode}$
Where:
$E_{cathode}$ (for NO3⁻) = +0.96 V
$E_{anode}$ (for Cu) = +0.34 V Thus: $ext{EMF} = 0.96 ext{V} - 0.34 ext{V} = 0.62 ext{V}$

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

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

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

State the weakest reducing agent in Table 4.

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

Use data from Table 4 to identify an acid that will oxidise copper. Explain your choice of acid.

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