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Now all Chlorine on the right is equalized, so there is the coefficient before

HCl.

K 2Cr 2O 7 + 14HCl → 2CrCl 3 + 3Cl 2 + 2KCl + H 2O.

Next step is to equalize Hydrogen, and to check the balance of Oxygen. The

result is a fully compilited and equated equation.

K 2Cr 2O 7 + 14HCl = 2CrCl 3 + 3Cl 2 + 2KCl + 7H 2O.

Let us identify and record the oxidizer and the reducing agent and the redox

processes:

the reducing agent - HCl;

the oxidizer - K 2Cr 2O 7.

–
0
2Cl - 2e = Cl 2 3 the oxidation process
+6 +3
2Cr + 6e = 2Cr 1 the reduction process

For example 2:
+4
+7
+4
+6
KMn O 4+Na 2S O 3+H 2O = Mn O 2+Na 2S O 4+ KOH

the oxidizer +7 +4
Mn + 3ē → Mn 2 the reduction process
KMnO 4

the reducing agent +4 +6
S − 2ē → S 3 the oxidation process
Na 2SO 3

2KMnO 4 + 3Na 2SO 3+H 2O = 2MnO 2+Na 2SO 4+ 2KOH

For example 3:

+2
+4
0
+5
Mg +4HN O 3 = Mg (NO 3) 2+N O 2+H 2O

0 +2
the reducing agent Mg – 2ē → Mg 1 the oxidation process

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