$Cu_2S_{(s)}+ O_{2(g)} = 2CU_{(s)} + SO_{2(g)}$. In the above reaction, the oxidation number of Cu changes from ___ to _______?

Analysis

To determine the change in the oxidation number of Copper (Cu), we need to calculate the oxidation state of Cu in the reactant ($Cu_2S$) and compare it to its oxidation state in the product ($Cu$).

Concept: Rules for Oxidation Numbers

1. Elements in their free state: The oxidation number of an atom in its elemental form (like solid Copper, $Cu$) is always 0.
2. Compounds: The sum of oxidation numbers in a neutral compound is 0.
3. Sulfur: In metal sulfides, Sulfur typically has an oxidation number of -2.

Step-by-Step Solution

Step 1: Find the oxidation number of Cu in $Cu_2S$

• Let the oxidation number of Cu be $x$.
• There are two Cu atoms, so we have $2x$.
• The oxidation number of Sulfur (S) is $-2$.
• Since $Cu_2S$ is a neutral compound, the sum is zero: $2x + (-2) = 0$ $2x = 2$ $x = +1$
• Therefore, the oxidation number of Cu in $Cu_2S$ is +1.

Step 2: Find the oxidation number of Cu in the product

• The product is $Cu_{(s)}$.
• According to the rules, any element in its standard elemental state has an oxidation number of 0.

Step 3: Determine the change

• The oxidation number changes from +1 (in $Cu_2S$) to 0 (in $Cu$).

Conclusion

The oxidation number of Copper decreases from +1 to 0. This indicates that the copper ion is being reduced. Comparing this to your options, it matches option A.

Exam Tip: Always remember that metals in compounds are usually positive, but when they stand alone as a solid element (like $Cu$, $Fe$, or $Zn$), their oxidation number is always zero. This is a common trap in JAMB Chemistry questions!