Question

Q). Identify the substances that are oxidized and that are reduced in the following equation.

i) 4Na(s) + O2(g) → 2Na2O(s)

ii) CuO(s) + H2(g) → Cu(s) + H2O(l)​

Answer 1

Answer:

Correct option is

A

sodium

Sodium atom to sodium ion: oxidation, e loss, ox. state change is from 0 to +1

2 x Na increases, each from (0) to (+1), losing one electron per Na atom, which is 'electronically' balanced by the oxygen molecule to peroxide ion: reduction, e gain, ox. state change 2 x O decreases, each from (0) to (1), gaining two electrons per O atom (or two electrons per O

2

molecule), and the halfreactions are:

(i) oxidation: Na → Na

+

+ $$e^$$ (formation of sodium ion)

(ii) reduction: O

2

+ $$2e^$$ → O

2

−−

(formation of the peroxide ion)

Adding and balancing 2 x (i) + (ii) gives the full equation

4Na (s) + O

2

(g) → 2Na

2

O(s)

Sodium (Na) is oxidised as it gains oxygen and oxygen gets reduced.

Option A is correct

Answer 2

Answer:

(i)

If a substance gain oxygen during a reaction, it is said to be oxidized. During this reaction, the sodium is gaining oxygen and is being oxidized. Oxygen is getting reduced.

(ii)

During this reaction, copper oxide is losing oxygen and is being reduced. The hydrogen is gaining oxygen and is being oxidized.