Question

Balance the following redox reaction in acid and basic mediums: MnO4^2- + SO3^2- -> Mn^2+ + SO4^2-

Answer 1

In acidic medium

First half cell :

MnO4- --> Mn2+

Add e-'s,

MnO4- + 5e- --> Mn2+

balance O,

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

balance H,

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

IInd half cell :

SO3^2- ---> SO4^2-

Add e-'s

SO3^2- --> SO4^2- + 2e-

balance O,

SO3^2- + H2O --> SO42- + 2e-

balance H,

SO3^2- + H2O ---> SO4^2- + 2H+ + 2e-

Multiply the Ist half cell by 2 and IInd half cell by 5 and add them together,

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

5SO3^2- + 5H2O ---> 5SO4^2- + 10H+ + 10e-

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2MnO4- + 5SO3^2- + 6H+ ----> 2Mn2+ + 5SO4^2- + 3H2O

Is the final balanced equation

Now in basic medium,

First half cell :

MnO4- --> Mn2+

Add e-'s,

MnO4- + 5e- --> Mn2+

balance O,

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

balance H,

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

Add OH- on each side equivalent to number of H+ amd combine to form H2O,

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

IInd half cell :

SO3^2- ---> SO4^2-

Add e-'s

SO3^2- --> SO4^2- + 2e-

balance O,

SO3^2- + H2O --> SO42- + 2e-

balance H,

SO3^2- + H2O ---> SO4^2- + 2H+ + 2e-

Add OH- on each side equivalent to number of H+ amd combine to form H2O,

SO3^2- + 2OH- ---> SO4^2- + H2O + 2e-

Multiply the Ist half cell by 2 and IInd half cell by 5 and add them together,

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

5SO3^2- + 10OH- ---> 5SO4^2- + 5H2O + 10e-

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2MnO4- + 5SO3^2- + 3H2O ----> 2Mn2+ + 5SO4^2- + 6OH-

Is the final belanced equation in basic medium