Identify the substance oxidized and the substance reduced in the following reaction oxidized and reduced
1.H2(g) +Cl2 (g) --- 2HCL (g)
2.H2(g)+Cuo(g) --Cu(s)+H2o(l)
3.H2s(g)+So2(g)--S(s)+H2o(l)
Identify the substance oxidized and the substance reduced in the following reaction oxidized and reduced
1.H2(g) +Cl2 (g) --- 2HCL (g)
2.H2(g)+Cuo(g) --Cu(s)+H2o(l)
3.H2s(g)+So2(g)--S(s)+H2o(l)
Reaction-1
H 2(g) +Cl2 (g) → 2HCl (g)
Cl2 get reduced by H2 from 0 oxidation state of hydrogen to -1 oxidation state of hydrogen in the product. Thus H2 is oxidised
Reaction-2
H2(g)+ CuO(g) → Cu(s)+H2O (l)
CuO get reduced by the gain of two electron from+2 oxidation state of Cu to zero oxidation state of Cu in solid form in product and H2 get oxidised by the addition of oxygen atom
Reaction-3
H2S(g)+SO2(g) → S(s)+H2O(l)
Sulphur is in +4 oxidation state in SO2 and get reduced to zero oxidation state by the gain of electron in solid sulphur in product. Hence, SO2 get reduced and H2S get oxidised.
Reaction-1
H 2(g) +Cl2 (g) → 2HCl (g)
Cl2 get reduced by H2 from 0 oxidation state of hydrogen to -1 oxidation state of hydrogen in the product. Thus H2 is oxidised
Reaction-2
H2(g)+ CuO(g) → Cu(s)+H2O (l)
CuO get reduced by the gain of two electron from+2 oxidation state of Cu to zero oxidation state of Cu in solid form in product and H2 get oxidised by the addition of oxygen atom
Reaction-3
H2S(g)+SO2(g) → S(s)+H2O(l)
Sulphur is in +4 oxidation state in SO2 and get reduced to zero oxidation state by the gain of electron in solid sulphur in product. Hence, SO2 get reduced and H2S get oxidised.
