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JAMB Chemistry 2014 Paper

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Question : 20

Total: 45

Zn _(s) + CuSO ₄(aq) → ZnSO ₄(aq) + Cu _(s)

In the reaction above, the oxidation number of the reducing agent changes from

0 to + 4
0 to + 2
+ 1 to + 2
+ 1 to + 3

Solution:

Concept:

The reducing agent is oxidised, so its oxidation number increases.

Explanation:

In the reaction,

\mathrm{Zn(s)}

is the reducing agent.

Oxidation number of

\mathrm{Zn(s)} = 0

.

In

\mathrm{ZnSO}_4

, zinc has oxidation number

+2

.

Therefore, the oxidation number changes from

0

to

+2

.

Answer:

B.

0

to

+2

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