Question

the redox reaction given below occurs in a basic solution. \\(\ce{hpo^{2-}_{3} + n_{2}h_{4} -> h_{2}po^{-}_{2} + n_{2}}\\) balance the \\(\ce{hpo^{2-}_{3}}\\) half - reaction. how many electrons are transferred?

Explanation:

Step1: Balance atoms other than O and H

The half - reaction for $\ce{HPO_3^{2 - }}$ to $\ce{H_2PO_2^-}$:
First, balance the P atoms. There is 1 P atom on each side, so P is balanced.

Step2: Balance O atoms by adding $\ce{H_2O}$

For $\ce{HPO_3^{2 - }}\to\ce{H_2PO_2^-}$:
Left side O: 3, Right side O: 2. So we add 1 $\ce{H_2O}$ to the right side to balance O.
$\ce{HPO_3^{2 - }}\to\ce{H_2PO_2^- + H_2O}$

Step3: Balance H atoms by adding $\ce{H^+}$ (then adjust for basic solution later)

Left side H: 1, Right side H: 2 + 2 (from $\ce{H_2O}$) = 4. So we add 3 $\ce{H^+}$ to the left side.
$\ce{3H^+ + HPO_3^{2 - }}\to\ce{H_2PO_2^- + H_2O}$

Step4: Adjust for basic solution by adding $\ce{OH^-}$ to both sides

Since the solution is basic, we add 3 $\ce{OH^-}$ to both sides to neutralize $\ce{H^+}$.
$\ce{3OH^- + 3H^+ + HPO_3^{2 - }}\to\ce{H_2PO_2^- + H_2O + 3OH^-}$
The $\ce{H^+}$ and $\ce{OH^-}$ on the left combine to form $\ce{H_2O}$.
$\ce{3H_2O + HPO_3^{2 - }}\to\ce{H_2PO_2^- + H_2O + 3OH^-}$
Simplify the $\ce{H_2O}$: Subtract 1 $\ce{H_2O}$ from both sides.
$\ce{2H_2O + HPO_3^{2 - }}\to\ce{H_2PO_2^- + 3OH^-}$

Step5: Balance charge by adding electrons

Left side charge: - 2 (from $\ce{HPO_3^{2 - }}$) + 0 (from $\ce{H_2O}$) = - 2
Right side charge: - 1 (from $\ce{H_2PO_2^-}$) + 3*(- 1) (from $\ce{OH^-}$) = - 4
To balance the charge, we need to gain 2 electrons (since charge goes from - 2 to - 4, we add 2 electrons to the left side to make the charge equal).
$\ce{2e^- + 2H_2O + HPO_3^{2 - }}\to\ce{H_2PO_2^- + 3OH^-}$

Now, let's check the oxidation state of P to confirm the electron transfer.
In $\ce{HPO_3^{2 - }}$: Let oxidation state of P be $x$. $1 + x+3*(-2)= - 2\Rightarrow x = + 3$
In $\ce{H_2PO_2^-}$: Let oxidation state of P be $y$. $2*1 + y+2*(-2)= - 1\Rightarrow y = + 1$
The change in oxidation state of P is from + 3 to + 1, a decrease of 2 per P atom. Since there is 1 P atom, the number of electrons gained (transferred) is 2.

Answer:

The number of electrons transferred in the $\ce{HPO_3^{2 - }}$ half - reaction is 2. The balanced half - reaction is $\ce{2e^- + 2H_2O + HPO_3^{2 - } = H_2PO_2^- + 3OH^-}$ (the number of electrons transferred is 2).