Question

Sa se egalizeze ecuatiile, precizand reactiile redox partiale, care sunt oxidantii si care sunt reducatorii:
- In mediu acid: 1. Br– (aq) + MnO4 – (aq) --> Br2(l) + Mn2+(aq);
2. Cr2O7 2- (aq) + H2O2(aq) --> Cr3+(aq) + H2O(l) + O2(g)
- In mediu bazic: 1. Br2(aq) + OH- (aq) --> BrO3 - (aq) + Br- (aq) + H2O;
2. Cr(OH)4 - (aq) + OH- (aq) --> CrO4 2- (aq) + H2O

Answer 1

in acest caz vom introduce H⁺, ca mediu acid

1.

Br- (aq) + H⁺ + MnO4-(aq) --> Br2(l) + Mn2+(aq)

5x  2Br⁻__-2e-_> Br⁰2 oxidare (ag. red, Br⁻)

2x  Mn⁷⁺O4 __+5e-__> Mn²⁺ + 4H2O reducere (ag. ox, MnO4⁻)

10Br-(aq) + H⁺ + 2MnO4-(aq) --> 5Br2(l) + 2Mn2+(aq) + 8H2O

2.

Cr2O7²⁻(aq) + H⁺ + H2O2(aq) --> Cr3+(aq) + H2O(l) + O2(g)

3x  H2O2⁻ __-2e-__> O2⁰ reducere (ag. ox)

     Cr2O7²⁻ __+6e-__> 2Cr³⁺ + 7H2O oxidare (ag. red (Cr2O7)²⁻)

Cr2O7²⁻(aq) + 8H⁺ + 3H2O2(aq) --> 2Cr3+(aq) + 7H2O(l) + 3O2(g)

in acest caz vom introduce HO⁻, ca mediu bazic

1.

Br2(aq) + OH- (aq) --> Br⁵⁺O3- (aq) + Br- (aq) + H2O

    Br2⁰__-10e__> 2Br⁵⁺ oxidare (ag. red. Br2) + 3H2O

5x Br2⁰__+2e-__> 2Br- reducere (ag. ox Br2)

3Br2(aq) + 6OH- (aq) --> Br⁵⁺O3- (aq) + 5Br- (aq) + 3H2O

2.

Cr(OH)4-(aq) + OH-(aq) --> CrO4²⁻(aq) + H2O

cred ca lipseste oxidantul....?????

verifica.....!!