Chemistry, asked by disha3012, 5 months ago


Identify the substances that are oxidised and the substances
that are reduced in the following reactions.
(i) 4Na(s) + O, (g) → 2Na r'a)
(1) CuO(s) + H, (g) → Cu(s) + 420(1)​

Answers

Answered by Anonymous
1

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.

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

Answered by JennieRocks
6

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.

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

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