Grade 12th passGeneral Physics

Find out no. Of moles of KMnO4 needed to oxidise one mole of Cu2S in acidic medium.The reaction. ;: KMnO4 + Cu2S = Mn+2 + Cu+2 + SO2. (please write the balanced reaction of Cu2S.. What should i take the vakency of Cu) ?

Parul Rajput

9 Years agoGrade 12th pass

2 Answers

Vikas TU

Approved Tutor Answer9 Years ago

To do this U need the valemce factor not the oxidation state.

First find the valence factor for Cu in the compund using oxidation no. in both sides from the Reactant and the Product respectively.
Therfore, v.f. for Mn => +5

and v.f. for Cu => +2 as there is no change in its valency we take the v.f as its highest state.
Therfore moles of KMnO4 x 5 = moles Cu2+ * 2

Kushagra Madhukar

5 Years ago

Dear student,

Please find the solution to your problem below.

From law of equivalence,

equivalents of Cu₂S = equivalents of KMnO₄
moles of Cu₂S × valence factor of Cu₂S = moles of KMnO₄× valence factor of KMnO₄.

Now, for Cu₂S we can condense the equation as

Cu⁺¹₂S^-2 → 2Cu²⁺ + S⁺⁴O₄

Hence vf_Cu2S = 2(2 – 1) + 1(4 – ( – 2)) = 2 + 6 = 8

For KMnO₄ we can write

KMn⁺⁷O₄ → Mn⁺² (here we haven’t considered O and K because their oxidation state remains unchanged in the reaction)

Hence vf_KMnO4 = 7 – 2 = 5

Moles of Cu₂S = 1

Therefore,

1 × 8 = moles of KMnO₄ × 5

⇒ moles of KMnO₄ = 8/5 = 1.6 moles

Hope it helps.

Thanks for the question.

Regards,

Kushagra