Urgent reply needed

Calculate the n-factor of K₂MnO₄ in acidic medium

5 Answers

someone plzz help me within the next 3 hrs.

MnO4^â€“ + 8H⁺ + 5e^â€“ â†’ Mn²⁺ + 4H₂O.

So the n factor is 5

r u sure mentor sir ?

i thought this way.
this is the general equation.

3 K₂MnO₄ + 2 H₂O â†’ 2 KMnO₄ + MnO₂ + 4 KOH

now, here Mn (+6) changes to Mn(+7) and Mn(+4). So, change in oxidation state is 1 and 2 respec.

so , equivalent weight = M / 1 + M / 2 = 3M / 2 = M / (2/3).

thus, the n-factor is 2/3.

ur right aieee.....

mentor ur answer is for KMnO₄

ohh ya sorry i wrote it for KMNO₄

√ √ || {} [] () → ~ ^ x x x → → ln log lg lim lim cos sin tan cot arcsin arccos arctan

∫ ∫ ∫ ∫ ∫ ∬ ∬ ∬ ∬ ∬ ∭ ∭ ∭ ∭ ∭ ⨌ ⨌ ⨌ ⨌ ⨌

∑ ∑ ∑ ∑ ∑ ∮ ∮ ∮ ∮ ∮ ∏ ∏ ∏ ∏ ∏ ∐ ∐ ∐ ∐ ∐

+ - × ÷ ± ∓ = ∪ ∩ ≠ ∼ ≈ ∅ ∀ ∃ ∈ ∋ ∝ ≤ ≥ ≃ ≅ ≡ < > ≪ ≫ ⊂ ∉ ⊃ ⊆ ⊇ ⇒ ⇐ ⇔ ⇒ ⇐ ⇔ ⇑ ⇓ ⇕ → ← ↔ → ← ↔ ↑ ↓ ↕ ↖ ↗ ↘ ↙ ↪ ↩ ⇀ ↼ ⇁ ↽

° ∂ ∞ α β γ δ θ λ μ ν ξ π ρ σ φ ω ε ƒ κ τ υ Γ Δ Λ Σ Π Ω