To tackle this problem, we need to understand the underlying chemistry of the reactions involved, particularly focusing on the reduction of ferric sulfate (Fe2(SO4)3) by zinc (Zn) and another metal (M), and how these reactions relate to the stoichiometry of potassium permanganate (KMnO4) used for reoxidation.
Understanding the Reactions
When ferric sulfate is reduced by a metal, it converts ferric ions (Fe³⁺) into ferrous ions (Fe²⁺). The reduction reaction can be represented as follows:
In this case, zinc acts as a reducing agent. The balanced reaction for zinc reducing ferric ions can be expressed as:
- Zn + 2Fe³⁺ → Zn²⁺ + 2Fe²⁺
Here, one mole of zinc reduces two moles of ferric ions, meaning the valence of zinc in this reaction is +2.
Analyzing the Role of KMnO4
Potassium permanganate is a strong oxidizing agent. In acidic medium, it can oxidize ferrous ions back to ferric ions. The half-reaction for KMnO4 in acidic conditions is:
- MnO4⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H2O
From this, we see that 1 mole of KMnO4 can accept 5 moles of electrons. Therefore, if V1 ml of 0.1 N KMnO4 is required to reoxidize the ferrous ions back to ferric ions, we can calculate the moles of electrons involved:
- Moles of KMnO4 = (0.1 N) × (V1 ml / 1000) = 0.0001 × V1
- Moles of electrons = 5 × Moles of KMnO4 = 5 × 0.0001 × V1 = 0.0005 × V1
Comparing the Two Metals
Now, when the same volume of ferric sulfate solution is reduced by another metal M, we denote the volume of KMnO4 required for reoxidation as V2. Given that V1 : V2 = 1 : 1.5, we can express V2 as:
Following the same logic as before, the moles of electrons required for reoxidation when using metal M can be calculated:
- Moles of KMnO4 = (0.1 N) × (V2 ml / 1000) = 0.0001 × V2 = 0.0001 × (1.5 × V1) = 0.00015 × V1
- Moles of electrons = 5 × Moles of KMnO4 = 5 × 0.00015 × V1 = 0.00075 × V1
Determining the Valence of Metal M
Now, we can compare the moles of electrons transferred in both reactions. For zinc, we established that 1 mole of Zn provides 2 moles of electrons. Therefore, the total moles of electrons from zinc can be expressed as:
- Electrons from Zn = 0.0005 × V1
For metal M, we have:
- Electrons from M = 0.00075 × V1
To find the valence of metal M, we can set up a ratio based on the moles of electrons transferred:
- Valence of M = (Electrons from M) / (Electrons from Zn) × Valence of Zn
- Valence of M = (0.00075 × V1) / (0.0005 × V1) × 2 = (0.75 / 0.5) × 2 = 1.5 × 2 = 3
Final Thoughts
In conclusion, the valence of zinc is +2, and the valence of metal M is +3. This means that zinc can donate two electrons in the reduction of ferric ions, while metal M can donate three electrons. Understanding these relationships helps us grasp the stoichiometry of redox reactions and the behavior of different metals in such processes.
