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what should be the n factor of Cu2S in the reacCu2S -> Cu2+ +SO2

Anureet Sinha , 11 Years ago

Grade

6 Answers

Sunil Kumar FP

Last Activity: 10 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

Last Activity: 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

Last Activity: 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

Last Activity: 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

Kushagra Madhukar

Last Activity: 4 Years ago

Dear student,

Please find the attached solution to your problem below.

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

Last Activity: 4 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

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