In balancing the following half-reaction:
${C N}^{-} ⟶ C N O^{-} (s k e l e t a l)$ , the number of electrons that must be added on the right is :

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Solution

${C N}^{-} ⟶ C N O^{-}$
Let the oxidation number of $C$ in $C N^{-}$ be $x$ .
$x - 3 = - 1$
or, $x = + 2$
Let the oxidation number of $C$ in $C N O^{-}$ be $y$ .
$y - 3 - 2 = - 1$
or, $y = 4$ .
So, the oxidation of carbon changes by $+ 2$ .
${C N}^{-} ⟶ C N O^{-} + 2 e$

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