Cu\(_2\)S(g) + O\(_2\)(g) → 2Cu + SO\(_2\)(g) What is the change in the oxidation number...

Chemistry

JAMB 2017

Cu\(_2\)S(g) + O\(_2\)(g) → 2Cu + SO\(_2\)(g)
What is the change in the oxidation number of copper in the reaction?

A. 0 to + 2
B. 0 to + 1
C. + 1 to 0
D. + 2 to + 1

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Correct Answer: Option C

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

There is an explanation video available below.

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