Question

the redox reaction given below occurs in basic solution.\\(\ce{teo^{2-}_{3} + n_{2}o_{4} \
ightarrow te + no^{-}_{3}}\\)\
balance the half - reaction:\\(\ce{n_{2}o_{4} \
ightarrow no^{-}_{3}}\\)\
how many electrons are transferred?\\(\ce{?e^{-}}\\)

Explanation:

Step1: Balance N atoms

On the left, we have 2 N atoms in $\ce{N2O4}$, and on the right, we have 1 N atom in $\ce{NO^-3}$. So we multiply $\ce{NO^-3}$ by 2 to balance N:
$\ce{N2O4 -> 2NO^-3}$

Step2: Balance O atoms by adding $\ce{H2O}$ (in basic solution, we can also use $\ce{OH^-}$ and $\ce{H2O}$, but first balance O with $\ce{H2O}$)

Left O: 4, Right O: $2\times3 = 6$. So we need to add $2\ce{H2O}$ to the left to balance O? Wait, no, right has more O. Wait, let's do it properly. After balancing N: $\ce{N2O4 -> 2NO^-3}$
Left O: 4, Right O: 6. So we need to add $2\ce{H2O}$ to the left? No, wait, in basic solution, we can add $\ce{OH^-}$ and $\ce{H2O}$. Wait, let's use the method for basic solution.

First, balance N: $\ce{N2O4 -> 2NO^-3}$

Now balance O: Left O: 4, Right O: 6. So we need to add $2\ce{H2O}$ to the left? No, wait, to balance O, we can add $\ce{H2O}$ to the side with less O. Left has 4 O, right has 6 O. So we need to add $2\ce{H2O}$ to the left? Wait, no, that would make left O: 4 + 2*1 = 6? No, $\ce{H2O}$ has 1 O. Wait, 2 $\ce{H2O}$ has 2 O. So left O: 4 + 2 = 6, which matches right O (2*3=6). So add $2\ce{H2O}$ to left:
$\ce{N2O4 + 2H2O -> 2NO^-3}$

Now balance H atoms by adding $\ce{H^+}$, but since it's basic solution, we will later convert $\ce{H^+}$ to $\ce{H2O}$ by adding $\ce{OH^-}$. Left H: $2\times2 = 4$, right H: 0. So add $4\ce{H^+}$ to right:
$\ce{N2O4 + 2H2O -> 2NO^-3 + 4H^+}$

Now, since it's basic solution, add $4\ce{OH^-}$ to both sides to neutralize $\ce{H^+}$:
$\ce{N2O4 + 2H2O + 4OH^- -> 2NO^-3 + 4H^+ + 4OH^-}$

The $\ce{H^+}$ and $\ce{OH^-}$ on the right form $4\ce{H2O}$:
$\ce{N2O4 + 2H2O + 4OH^- -> 2NO^-3 + 4H2O}$

Now, subtract $2\ce{H2O}$ from both sides (since left has 2 $\ce{H2O}$, right has 4 $\ce{H2O}$):
$\ce{N2O4 + 4OH^- -> 2NO^-3 + 2H2O}$

Now balance charge. Let's calculate the charge on each side.

Left: $\ce{N2O4}$ is neutral, $4\ce{OH^-}$ is -4. So total left charge: -4.

Right: $2\ce{NO^-3}$ is 2*(-1) = -2, $2\ce{H2O}$ is neutral. So total right charge: -2.

We need to balance charge by adding electrons. Left charge: -4, right charge: -2. So we need to lose 2 electrons (since oxidation: N in $\ce{N2O4}$ is +4, in $\ce{NO^-3}$ is +5. Each N loses 1 electron, 2 N atoms lose 2 electrons). So add $2e^-$ to the right:

$\ce{N2O4 + 4OH^- -> 2NO^-3 + 2H2O + 2e^-}$

Wait, let's check the oxidation state of N. In $\ce{N2O4}$, each N is +4 (since O is -2, 2x + 4*(-2) = 0 => 2x = 8 => x=+4). In $\ce{NO^-3}$, N is +5 (x + 3*(-2) = -1 => x = +5). So each N atom is oxidized from +4 to +5, losing 1 electron per N. There are 2 N atoms, so total electrons lost: 2*1 = 2. So the number of electrons transferred is 2.

Wait, but let's check the balanced half-reaction. Let's verify all steps:

1. N: 2 on left, 2 on right (2*1). Good.

2. O: Left: 4 (from $\ce{N2O4}$) + 4 (from $\ce{4OH^-}$) = 8? Wait no, wait the half-reaction after balancing O: Wait, no, earlier step: when we added $2\ce{H2O}$ to left: $\ce{N2O4 + 2H2O -> 2NO^-3}$. Then O: left 4 + 2 = 6, right 2*3=6. Good. Then H: left 2*2=4, so we add $4\ce{H^+}$ to right: $\ce{N2O4 + 2H2O -> 2NO^-3 + 4H^+}$. Then in basic solution, add $4\ce{OH^-}$ to both sides: $\ce{N2O4 + 2H2O + 4OH^- -> 2NO^-3 + 4H^+ + 4OH^-}$. Then $\ce{4H^+ + 4OH^- -> 4H2O}$, so: $\ce{N2O4 + 2H2O + 4OH^- -> 2NO^-3 + 4H2O}$. Then subtract $2\ce{H2O}$ from both sides: $\ce{N2O4 + 4OH^- -> 2NO^-3 + 2H2O}$. Now charge: left: $\ce{N2O4}$ (0) + $4\ce{OH^-}$ (-4) = -4. Right: $2\ce{NO^-3}$ (-2) + $2\ce{H2O}$ (0) = -2. So the difference is +2 (right is more positive). So to balance charge, we need to add 2 electrons to the right (since oxidation, electrons are lost, so they appear on the right). So: $\ce{N2O4 + 4OH^- -> 2NO^-3 + 2H2O + 2e^-}$. Now check charge: left: -4, right: -2 + 2*(-1) = -4? Wait no, electrons have charge -1. So 2 electrons: -2. So right charge: -2 (from $\ce{2NO^-3}$) + 0 (from $\ce{2H2O}$) + (-2) (from $2e^-$) = -4. Which matches left charge (-4). Good.

Now, the number of electrons transferred: each N atom goes from +4 to +5, losing 1 electron. There are 2 N atoms, so total electrons lost: 2. So the number of electrons transferred is 2.

Answer:

2