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Balance the following redox reactions by ion electron method

8.18   Balance the following redox reactions by ion – electron method 

(a) MnO_{4}^{-}(aq)+ I^{-}(aq)\rightarrow MnO_{2}(s)+ I_{2}(s)  (In basic medium)

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reduction half reaction

MnO_{4}^{-}\rightarrow MnO_{2}  (+7 to +4) 

Add 3 electron on LHS side and after that to balance charge add OH ions. And to balance O atom add water molecule on whichever side it needed

balance it 

MnO_{4}^{-}+2H_{2}O+3e^{-}\rightarrow MnO_{2} +4OH^{-}

oxidation half

I\rightarrow I_{2}

balance it 

2I^{-}\rightarrow I_{2}+2e^{-}

equalising the no. of electrons by multiplying the oxidation half by 3 and reduction half by 2 and then add it.

6I^{-}\rightarrow 3I_{2}+ 6e^{-}\\&\ 2MnO_{4}^{-}+4H_{2}O+6e^{-}\rightarrow 2MnO_{2}+8OH^{-}

final answer-

2MnO_{4}^{-}+4H_{2}O+6I^{-}\rightarrow 2MnO_{2}+3I_{2}+8OH^{-}

 

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