a certain volume of ferric sulphate solution is reduced by zinc. the resulting solution could be reoxidised by V₁ml of 0.1 N KMnO_4... when same volume of ferric sulphate solution is reduced by another metal,M the resulting solution required V₂ ml of 0.1 N KMnO₄ solution for reoxidation. if V₁:V₂ = 1:1.5..determine valences of metal Zn & M

To tackle this problem, we need to analyze the reactions involved and how they relate to the stoichiometry of the redox processes. The key here is to understand how the valences of the metals affect the amount of potassium permanganate (KMnO4) required for reoxidation after reduction of ferric sulfate (Fe2(SO4)3).

Understanding the Reactions

When ferric sulfate is reduced by zinc, the zinc donates electrons, reducing Fe³⁺ ions to Fe²⁺ ions. The balanced half-reaction for this process can be represented as:

• Fe³⁺ + e^- → Fe²⁺

For every mole of Fe³⁺ reduced, one mole of electrons is transferred. Zinc, being a more reactive metal, can be oxidized to Zn²⁺ as follows:

• Zn → Zn²⁺ + 2e^-

Reduction by Zinc

In this case, one mole of Zn will reduce two moles of Fe³⁺ ions because it releases two electrons. Therefore, if V₁ ml of 0.1 N KMnO₄ is needed to reoxidize the Fe²⁺ back to Fe³⁺, it indicates that the amount of Fe²⁺ produced is equivalent to the moles of KMnO₄ used.

Analyzing the Second Metal

Now, when the same volume of ferric sulfate solution is reduced by another metal, M, we know that it requires V₂ ml of 0.1 N KMnO₄ for reoxidation. Given that V₁:V₂ = 1:1.5, we can express this relationship mathematically:

• V₂ = 1.5 * V₁

This implies that the amount of Fe²⁺ produced by metal M is greater than that produced by zinc. To find the valence of metal M, we need to consider how many electrons it donates during the reduction of Fe³⁺.

Relating the Valences

Let’s denote the valence of metal M as 'n'. If M donates 'n' electrons, then the stoichiometry of the reaction will be:

• Fe³⁺ + n e^- → Fe²⁺

For every mole of M, it will reduce n moles of Fe³⁺. The relationship between the moles of Fe²⁺ produced by Zn and M can be expressed as:

• 2 moles of Fe²⁺ from 1 mole of Zn
• n moles of Fe²⁺ from 1 mole of M

Setting Up the Equation

Since V₁ corresponds to the amount of Fe²⁺ produced by Zn and V₂ corresponds to the amount produced by M, we can set up the following ratio based on the volumes and the number of electrons transferred:

• V₁ : V₂ = 2 : n

Substituting the known ratio:

• 1 : 1.5 = 2 : n

Solving for n

Cross-multiplying gives us:

• 1 * n = 2 * 1.5
• n = 3

This means that metal M has a valence of 3. Now, we can summarize the valences:

• Valence of Zn = 2
• Valence of M = 3

Final Thoughts

In summary, zinc donates 2 electrons per atom, while metal M donates 3 electrons per atom during the reduction of ferric sulfate. This stoichiometric relationship helps us understand the different reactivity and electron transfer capabilities of these metals in redox reactions.