# 8.21  The $Mn^{3+}$  ion is unstable in solution and undergoes disproportionation to give $Mn^{2+}$ , $MnO_{2}$ and $H^{+}$ ion. Write a balanced ionic equation for the reaction

The base equation

$Mn^{3+}\rightarrow Mn^{2+}+MnO_{2}+H^{+}$

write oxidation half with their oxidation state

$Mn^{3+}(+3)\rightarrow MnO_{2}(+4)$

Balance the charge on $Mn$ by adding 1 $e^{-}$ on RHS side. To balance charge add $H^{+}$ ions on RHS side and then for oxygen balance add $H_{2}O$ molecule on LHS side.

$Mn^{3+}+2H_{2}O\rightarrow MnO_{2}+1e^{-}+4H^{+}$

reduction half

$Mn^{3+}(+3)\rightarrow Mn^{2+}(+2)$

balancing the reduction half by adding 1$e^{-}$ on LHS side

$Mn^{3+}+1e^{-}\rightarrow Mn^{2+}$

Add both balanced reduction half and oxidation half

$2Mn^{3+}+2H_{2}O\rightarrow Mn^{2+}+MnO_{2}+4H^{+}$

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