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Balance the following redox reactions by ion-electron method

Akash , 6 Years ago

Grade 11

3 Answers

Vishi

Reduction half -

MnO4- --> Mn2+

MnO4- --> Mn2+ + 4H2O

MnO4- + 8H+ --> Mn2+ + 4H2O

MnO4- + 8H+ + 5e --> Mn2+ + 4H2O

2MnO4- + 16H+ + 10e --> 2Mn2+ + 8H2O

Oxidation half -

SO2 --> HSO4-

SO2 + 2H2O --> HSO4-

SO2 + 2H2O --> HSO4- + 3H+

SO2 + 2H2O --> HSO4- + 3H+ + 2e

5SO2 + 10H2O --> 5HSO4- + 15H+ + 10e

Adding both eq ,

2MnO4- + 16H+ + 10e + 5SO2 + 10H2O -----> 2Mn2+ + 8H2O + 5HSO4- + 15H+ + 10e

2MnO4- + H+ + 5SO2 + 2H2O -----> 2Mn2+ + 5HSO4-

Last Activity: 6 Years ago

Isra Abass

Reduction half -

MnO4- --> Mn2+

MnO4- --> Mn2+ + 4H2O

MnO4- + 8H+ --> Mn2+ + 4H2O

MnO4- + 8H+ + 5e --> Mn2+ + 4H2O

2MnO4- + 16H+ + 10e --> 2Mn2+ + 8H2O

Oxidation half -

SO2 --> HSO4-

SO2 + 2H2O --> HSO4-

SO2 + 2H2O --> HSO4- + 3H+

SO2 + 2H2O --> HSO4- + 3H+ + 2e

5SO2 + 10H2O --> 5HSO4- + 15H+ + 10e

Adding both eq ,

2MnO4- + 16H+ + 10e + 5SO2 + 10H2O -----> 2Mn2+ + 8H2O + 5HSO4- + 15H+ + 10e

2MnO4- + H+ + 5SO2 + 2H2O -----> 2Mn2+ + 5HSO4-

Last Activity: 4 Years ago

Isra Abass

Manganese goes from oxidation state +7 to +2. A 5-electron change and a reduction.
Sulfur goes from an oxidation state of +4 to +6. A 2-electron change and an oxidation.

5e- + MnO4- = Mn2+

SO2 = HSO4- + 2e-

Balance the charge with H+

8 H+ + 5e- + MnO4- = Mn2+

SO2 = HSO4- + 2e- + 3 H+

Now balance the H using H2O
If done correctly, the O should be balanced
Multiply each half reaction by a number that will give the same number of electrons in each—in this case TEN electrons and then add the two half reactions.

Last Activity: 4 Years ago