that are
onidized and
reduced in the following
Identify the substances
the substances that are
reactions:
a) Cuo+H₂ -> Cu+H₂O
b) 4 Na+o2-> 2 Na2o
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Answers
Answer:
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.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.
Explanation: