# what should be the n factor of Cu2S in the reac Cu2S -> Cu2+  +    SO2

Sunil Kumar FP
9 years ago
n factor of oxidising and reducing agent is equal to the change in oxidation number per molecule
for the reaction
Cu2S----->Cu+2 +SO2
Change in oxidation number of Cu+1 =2*1 -2=0
change in oxidation number of S=-2 -4 =!-6!=6
Therefore the n factor of the Cu2S is equal to the sum of the change in oxidation number =0+6 =6
ROHAN THE KING
27 Points
7 years ago
Since in Cu2S, Cu changes to Cu+2, Cu loses 2 electrons and S changes to SO2, sulphur gains 6 electrons, the n factor is 6+2+2 = 8
ROHAN THE KING
27 Points
7 years ago
Since in Cu2S, Cu changes to Cu+2 Cu loses 2 electrons and S changes to SO2, sulphur gains 6 electrons , the n factor is 6+2=8
manvir
13 Points
5 years ago
Firstly it is half cell reaction as both the specis are oxidised Cu+1 changes to Cu+2 and S-2 changes to SO4 2- so n factor of Cu2S is 2*1+6=8
4 years ago
Dear student,

n factor of oxidising and reducing agent is equal to the change in oxidation number per molecule
for the reaction
Cu2S → 2Cu+2 +SO2
Change in oxidation number of Cu =2 x (2 – 1) = 2
change in oxidation number of S = 4 – ( – 2) = 6
Therefore the n factor of the Cu2S is equal to the sum of the change in oxidation number = 2+6 = 8

Hope it helps.
Thanks and regards,
Kushagra
Anurag Bhattacharjee
15 Points
3 years ago
n factor of oxidising and reducing agent is equal to the change in oxidation number per molecule
for the reaction
Cu2S → 2Cu+2 +SO2
Change in oxidation number of Cu =2 x (2 – 1) = 2
change in oxidation number of S = 4 – ( – 2) = 6
Therefore the n factor of the Cu2S is equal to the sum of the change in oxidation number = 2+6 = 8