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IBDP Chemistry - Reactivity 3.2 Electron transfer reactions- IB Style Questions For SL Paper 1A - FA 2025

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Question

A reactivity series of selected elements, arranged from highest activity to lowest, is shown. Which reaction is spontaneous?
Zn > Fe > Cd > Co > Ni > Sn > Pb > Cu
(A) \(Cd^{2+}(aq)+Pb(s)\rightarrow Pb^{2+}(aq)+Cd(s)\)
(B) \(Fe^{2+}(aq)+Sn(s)\rightarrow Sn^{2+}(aq)+Fe(s)\)
(C) \(Zn^{2+}(aq)+Cu(s)\rightarrow Cu^{2+}(aq)+Zn(s)\)
(D) \(Ni^{2+}(aq)+Co(s)\rightarrow Co^{2+}(aq)+Ni(s)\)
▶️ Answer/Explanation
Detailed solution

1. Principle of Displacement:
A more reactive metal (higher on the list) will displace a less reactive metal ion from solution.
Spontaneous: \(\text{Metal}_{more reactive} + \text{Ion}_{less reactive} \rightarrow \text{Ion}_{more reactive} + \text{Metal}_{less reactive}\)

2. Evaluate Options:

  • (A) \(Pb\) + \(Cd^{2+}\): \(Pb\) is lower than \(Cd\). Non-spontaneous.
  • (B) \(Sn\) + \(Fe^{2+}\): \(Sn\) is lower than \(Fe\). Non-spontaneous.
  • (C) \(Cu\) + \(Zn^{2+}\): \(Cu\) is lower than \(Zn\). Non-spontaneous.
  • (D) \(Co\) + \(Ni^{2+}\): \(Co\) is higher than \(Ni\) in the list. Spontaneous.

✅ Answer: (D)

Question

Which change involves oxidation of N?
A. \(\text{NH}_{3}\) to \(\text{N}_{2}\)
B. \(\text{NO}_{2}\) to \(\text{NO}\)
C. \(\text{N}_{2}\) to \(\text{AlN}\)
D. \(\text{NO}_{2}\) to \(\text{N}_{2}\text{O}_{4}\)
▶️ Answer/Explanation
Detailed solution

Oxidation of nitrogen means its oxidation state becomes more positive.

Option A: \(\text{NH}_{3} \rightarrow \text{N}_{2}\)
In \(\text{NH}_{3}\): N has oxidation state \(-3\).
In \(\text{N}_{2}\): N has oxidation state \(0\).
\(-3 \rightarrow 0\) is an increase in oxidation state ⇒ oxidation.

Option B: \(\text{NO}_{2} \rightarrow \text{NO}\)
N in \(\text{NO}_{2}\): \(+4\); N in \(\text{NO}\): \(+2\).
\(+4 \rightarrow +2\): decrease ⇒ reduction.

Option C: \(\text{N}_{2} \rightarrow \text{AlN}\)
N in \(\text{N}_{2}\): \(0\); N in \(\text{AlN}\): \(-3\).
\(0 \rightarrow -3\): decrease ⇒ reduction.

Option D: \(\text{NO}_{2} \rightarrow \text{N}_{2}\text{O}_{4}\)
N is \(+4\) in both \(\text{NO}_{2}\) and \(\text{N}_{2}\text{O}_{4}\) ⇒ no change.

Only option A shows oxidation of N.
✅ Answer: (A)

Question

What is the formula of copper(I) sulfide?
A. \(\text{CuS}\)
B. \(\text{Cu}_{2}\text{S}\)
C. \(\text{CuSO}_{3}\)
D. \(\text{Cu}_{2}\text{SO}_{3}\)
▶️ Answer/Explanation
Detailed solution

Copper(I) means the oxidation state of copper is \(+1\), so the ion is \(\text{Cu}^{+}\).
Sulfide is the \(\text{S}^{2-}\) ion.

To form a neutral compound, the total positive and negative charges must balance:
\(2 \times (+1) + (-2) = 0\).

Therefore the empirical formula is \(\text{Cu}_{2}\text{S}\).

✅ Answer: (B)

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