In balancing the half-reaction: CN - longrightarrow CNO -(skeletal) the number of electrons that must be added is :

Question

In balancing the half-reaction: CN - longrightarrow CNO -(skeletal) the number of electrons that must be added is : (A) 0 (B) 1 on the right (C) 1 on the left (D) 2 on the right

Tardigrade Admin · Accepted Answer

CN- arrow CNO- Oxidation number of nitrogen in CN-4+x=-1 and in CNO-4+x-2=-1 x=-5,x=-3 CN- is thus oxidised by losing two electrons.

Q. In balancing the half-reaction :
$C N^{-} ⟶ CN O^{-}$ (skeletal) the number of electrons that must be added is :

$0$

$1$ on the right

$1$ on the left

$2$ on the right

$C N^{-} \to CN O^{-}$
Oxidation number of nitrogen in $C N^{-} 4 + x = - 1$ and in $CN O^{-} 4 + x - 2 = - 1$
$x = - 5, x = - 3$
$C N^{-}$ is thus oxidised by losing two electrons.

Q. In balancing the half-reaction : CN−⟶CNO−(skeletal) the number of electrons that must be added is :

0

1 on the right

1 on the left

2 on the right

CN−→CNO− Oxidation number of nitrogen in CN−4+x=−1 and in CNO−4+x−2=−1 x=−5,x=−3 CN− is thus oxidised by losing two electrons.

Q. In balancing the half-reaction :
$C N^{-} ⟶ CN O^{-}$ (skeletal) the number of electrons that must be added is :

$0$

$1$ on the right

$1$ on the left

$2$ on the right

$C N^{-} \to CN O^{-}$
Oxidation number of nitrogen in $C N^{-} 4 + x = - 1$ and in $CN O^{-} 4 + x - 2 = - 1$
$x = - 5, x = - 3$
$C N^{-}$ is thus oxidised by losing two electrons.