Igualar las siguientes ecuaciones por el método del ion-electrón (medio ácido).
1) NO3 + Mg → N₂+ Mg2+
2) Fe + SO42- Fe³+ + SO2
3) 101 + 103"
4) MnO2 + Cr³+ →→ Mn2+ + Cr2O72
5) NO3 + Al → Al2O3 + NO
6) SO2 + S→ SO2
7) Mn2+ + MnO4
8) H2C2O4+ MnO4 →MnO2 CO₂+ Mn2+
9) 12+ S2032-1+ S4082
10) H₂O → H₂+ O₂
Respuestas a la pregunta
Respuesta:
The unbalanced chemical equation is:
MnO
4
−
(aq)+SO
2
(g)→Mn
2+
(aq)+HSO
4
−
(aq)
The oxidation half reaction is
SO
2
(g)+2H
2
O(l)→HSO
4
−
(aq)+3H
+
(aq)+2e
−
(aq).
The reduction half reaction is
MnO
4
−
(aq)→Mn
(
2+)(aq).
In the reduction half reaction, the oxidation number of Mn changes from +7 to +2. Hence, 5 electrons are added to LHS of the reaction.
MnO
4
−
(aq)+5e
−
→Mn
2+
(aq)
Charge is balanced in the reduction half reaction by adding 8 hydrogen ions to LHS.
MnO
4
−
(aq)+5e
−
+8H
+
(aq)→Mn
2+
(aq)
To balance O atoms, 4 water molecules are added on RHS.
MnO
4
−
(aq)+5e
−
+8H
+
(aq)→Mn
2+
(aq)+4H
2
O(l)
To equalize the number of electrons, the oxidation half reaction is multiplied by 5 and the reduction half reaction is multiplied by 2.
5SO
2
(g)+10H
2
O(l)→5HSO
4
−
(aq)+15H
+
(aq)+10e
−
(aq)
2MnO
4
−
(aq)+10e
−
+16H
+
(aq)→2Mn
2+
(aq)
Two half cell reactions are added to obtain a balanced equation.
2MnO
4
−
(aq)+5SO
2
(g)+2H
2
O(l)→2Mn
2+
(aq)+5HSO
4
−
(aq)