JAMB - Chemistry (2017 - No. 29)
Cu\(_2\)S(g) + O\(_2\)(g) → 2Cu + SO\(_2\)(g)
What is the change in the oxidation number of copper in the reaction?
What is the change in the oxidation number of copper in the reaction?
0 to + 2
0 to + 1
+ 1 to 0
+ 2 to + 1
Explanation
In the reactant;
Cu\(_2\)S
2 Cu - 2(1) = 0
2 Cu = 2
Cu = \(\frac{2}{2}\)
Cu = +1
In the product, Cu
Cu = O (Oxidation state of an element in its free state is 0)
The oxidation number of Cu in Cu\(_2\)S and Cu respectively is +1 and 0 respectively
Cu\(_2\)S
2 Cu - 2(1) = 0
2 Cu = 2
Cu = \(\frac{2}{2}\)
Cu = +1
In the product, Cu
Cu = O (Oxidation state of an element in its free state is 0)
The oxidation number of Cu in Cu\(_2\)S and Cu respectively is +1 and 0 respectively
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