Question

Balance the following reaction:

$Mn{O}_4^{-}+C{N}^{-}+H_{2}O\to Mn{O}_2+CN{O}^{-}+O{H}^{-}$

Answer 1

$Mn{O}_4^{-}+C{N}^{-}⟶Mn{O}_2+CN{O}^{-}$

In basic medium, half reactions are:

$Mn{O}_4^{-}⟶Mn{O}_2$

$C{N}^{-}⟶CN{O}^{-}$

Now, $Mn{O}_4^{-}\stackrel{H^{+}}{⟶}Mn{O}_2+2H_{2}O$

Also, $4H^{+}+Mn{O}_4^{-}+4\left(O{H}^{-}\right)⟶Mn{O}_2+2H_{2}O+4\left(O{H}^{-}\right)$

From the $2^{nd}$ we get,

$Mn{O}_4^{-}+4H_{2}O⟶Mn{O}_2+2H_{2}O+4\left(O{H}^{-}\right)$

$⟹Mn{O}_4^{-}+2H_{2}O⟶Mn{O}_2+4\left(O{H}^{-}\right)$ equation $01$

Now, $3e^{-}+2H_{2}O+Mn{O}_4^{-}⟶Mn{O}_2+4O{H}^{-}$

Also,

$H_{2}O+C{N}^{-}⟶CN{O}^{-}$

$H_{2}O+C{N}^{-}⟶CN{O}^{-}+2H^{+}$

$2\left(O{H}^{-}\right)+H_{2}O+C{N}^{-}⟶CN{O}^{-}+2H^{+}+2O{H}^{-}$

$2\left(O{H}^{-}\right)+H_{2}O+C{N}^{-}⟶CN{O}^{-}+H_{2}O$

$2\left(O{H}^{-}\right)+C{N}^{-}⟶CN{O}^{-}+H_{2}O$

$2\left(O{H}^{-}\right)+C{N}^{-}⟶CN{O}^{-}+H_{2}O+2e^{-}$ equation $02$

Now adding $01$ and $02$

$2\left(H_{2}O\right)+\left(Mn{O}_4^{-}\right)+3\left(e^{-}\right)⟶\left(Mn{O}_2\right)+4\left(O{H}^{-}\right)$

$2\left(O{H}^{-}\right)+\left(C{N}^{-}\right)⟶\left(CN{O}^{-}\right)+\left(H_{2}O\right)+2\left(e^{-}\right)$

Multiplying equation $01$ by $02$ and equation $02$ by $03$, we get,.

$4\left(H_{2}O\right)+2\left(Mn{O}_4^{-}\right)+6\left(e^{-}\right)⟶2\left(Mn{O}_2\right)+8\left(O{H}^{-}\right)$

$6\left(O{H}^{-}\right)+3\left(C{N}^{-}\right)⟶3\left(CN{O}^{-}\right)+3\left(H_{2}O\right)+6\left(e^{-}\right)$

Now, adding we get,

$H_{2}O+2Mn{O}_4^{-}+3C{N}^{-}⟶2Mn{O}_2+2\left(O{H}^{-}\right)+3CN{O}^{-}$