Question

identify the substance that are oxidized and the substance that are reduced in the following reaction; 4Na(s)+o2(g)----->2Na2O(s)​

Answer 1

Answer:

Sodium

Explanation:

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.

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