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Question: The equivalent weight of \(\text{C}{{\text{u}}_{2}}\text{S}\)(mol. Wt. = M) in the following reactio...

The equivalent weight of Cu2S\text{C}{{\text{u}}_{2}}\text{S}(mol. Wt. = M) in the following reaction is:
Cu2S + 4MnO4 → Cu2+ + SO2 + Mn2+\text{C}{{\text{u}}_{2}}\text{S + 4Mn}{{\text{O}}_{4}}\text{ }\to \text{ C}{{\text{u}}^{2+}}\text{ + S}{{\text{O}}_{2}}\text{ + M}{{\text{n}}^{2+}}
A. M/2
B. M/6
C. M/8
D. M/4

Explanation

Solution

Equivalent of the substance is defined as the ratio of the molecular weight or mass of the compound to the n - factor of the compound or the acidity or basicity. N - factor can be calculated by determining the change in the oxidation state.

Complete Step-by-Step Answer:
- In the given question, we have to calculate the equivalent weight of the copper sulfide from the given reaction.
- It is given in the question that the molecular weight of the copper sulfide is M.
- Now, we have to calculate the n - factor. Firstly, we have to calculate the change in the oxidation state of the copper and sulphur.
- So, change in the oxidation state of copper is:
Cu2S Cu2+ 2x - 2 = 0 x = +2 x = +1 \begin{aligned} & \text{C}{{\text{u}}_{2}}\text{S C}{{\text{u}}^{2+}} \\\ & 2\text{x - 2 = 0 x = +2} \\\ & \text{x = +1} \\\ \end{aligned}
- So, here the oxidation of copper takes place from +1 to +2 and the change in the oxidation state is (2-1) = 1.
- Now, change in the oxidation state of the sulphur will be:
Cu2S SO2 1 × 2 + x = 0 x - 2 × 2 = 0 2 + x = 0 x - 4 = 0 x = -2 x = 4 \begin{aligned} & \text{C}{{\text{u}}_{2}}\text{S S}{{\text{O}}_{2}} \\\ & \text{1 }\times \text{ 2 + x = 0 x - 2 }\times \text{ 2 = 0} \\\ & \text{2 + x = 0 x - 4 = 0} \\\ & \text{x = -2 x = 4} \\\ \end{aligned}
- So, the Oxidation of the sulphur also take place from -2 to +4 and the change in the oxidation state of sulphur is (4-(-2)) = 4+2 = 6.
- So, the n - factor will be 6 + 2(1) = 8.
- The equivalent weight of the copper (I) sulphide is:
Equivalent weight = Molecular weightn - factor\text{Equivalent weight = }\dfrac{\text{Molecular weight}}{\text{n - factor}}
Equivalent weight = M8\text{Equivalent weight = }\dfrac{\text{M}}{8}
Therefore, option C is the correct answer.

Note: The oxidation state of the single element is zero but if it has any charge present on it then it is considered as n - factor. Whereas the basicity is defined for the acidic substance and the acidity is defined for the basic substance.