Question

Identifying redox reactions under cidic and basic conditions

Cr2O7^2-(aq) + NH4^+(aq) == Cr2O3(s) + N2(g)

Please help me to balance this equation in acidic and basic conditions.

How to find the oxidation numbers for EU(NO2)3 .

Please explain Thank you so much.

Answer 1

Cr₂O₇^2-(aq) + NH₄⁺(aq) ---> Cr₂O₃(s) + N₂(g)

a) In acidic medium

Oxidation half reaction-- 2NH₄⁺ → NO₂ + 6e ^- + 8H⁺

Reduction half reaction-- Cr₂O₇^2- + 6e^- + 8H⁺ → Cr₂O₃ + 4H₂O

Now, add the half reactions together, we will get,

2NH₄⁺ + Cr₂O₇^2- + 6e- + 8H⁺ → NO₂ + Cr₂O₃ + 6e- + 4H₂O + 8H⁺

=> 2NH₄⁺ + Cr₂O₇^2- → N₂ + Cr₂O₃ + 4H₂O

b) In basic medium

Oxidation half reaction--- 2NH₄⁺ + 8OH^- → NO₂ + 6e^- + 8H₂O

Reduction half reaction-- Cr₂O₇^2- + 6^e- + 4H₂O → Cr₂O₃ + 8OH^-

Now, add both the half reactions together---

2NH₄⁺ + Cr₂O₇^2- + 8OH^- + 6e^- + 4H₂O → NO₂ + Cr₂O₃ + 6e^- + 8OH^- + 8H₂O

=> 2NH₄⁺ + Cr₂O₇^2- → N₂ + Cr₂O₃ + 4H₂O