Question

Complete and balance the following reaction:

${Ag}^{+}+As{H}_3⟶Ag\downarrow +H_{3}As{O}_3+H^{+}$

• A. $6{Ag}^{+}+As{H}_3+3H_{2}O\to 6Ag\downarrow +6H^{+}+H_{3}As{O}_3$
• B. $6{Ag}^{+}+2As{H}_3+5H_{2}O\to 6Ag\downarrow +10H^{+}+H_{3}A{s}_2{O}_3$
• C. $6{Ag}^{+}+2As{H}_3+3H_{2}O\to 6Ag\downarrow +6H^{+}+H_{3}A{s}_2{O}_3$
• D. None of these

Answer 1

The correct option is A $6{Ag}^{+}+As{H}_3+3H_{2}O\to 6Ag\downarrow +6H^{+}+H_{3}As{O}_3$

1. $As{H}_3\left(g\right)\to H_{3}As{O}_3\left(aq\right)+\left(n\right)e^{-}$
The number of electrons involved in this one is not so obvious, and we need to figure out the oxidation state of arsenic in $As{H}_3$ and $H_{3}As{O}_3$ in order to deduce the number of electrons involved in this half-cell reaction.
$As{H}_3\to A{s}^{3-}$ (each H is treated as $H^{+}$ in this molecule)
$H_{3}As{O}_3\to A{s}^{3+}$ (the anion is $As{O}_3{)}^{3-}$
This half-cell reaction can then be written as follows:
$A{s}^{3-}\to A{s}^{3+}+6e^{-}$
2. We can see that the silver half-cell needs to be multiplied by $6$ to balance the number of electrons gained and lost:
$6A{g}^{+}\left(aq\right)+6e^{-}\to 6Ag\left(s\right)$
3. We use this information on the two half-cells to start entering the balancing coefficients into the original equation,
$6A{g}^{+}\left(aq\right)+As{H}_3\left(g\right)+H_{2}O\to 6Ag\left(s\right)+H_{3}As{O}_3\left(aq\right)+H^{+}\left(aq\right)$
4. The left side of the equation has a net charge of $+6$ and so, the right side must also have a net charge of $+6$. That means that the $6A{g}^{+}$ on the left side must be balanced by $6$ $H^{+}$ to make the net charge equal on both side.
$6A{g}^{+}\left(aq\right)+As{H}_3\left(g\right)+H_{2}O\to 6Ag\left(s\right)+H_{3}As{O}_3\left(aq\right)+6H^{+}\left(aq\right)$
5. Now balance the oxygens and hydrogens by putting $3$ in front of the $H_{2}O$ on the right side.
$6A{g}^{+}\left(aq\right)+As{H}_3\left(g\right)+3H_{2}O\to 6Ag\left(s\right)+H_{3}As{O}_3\left(aq\right)+6H^{+}\left(aq\right)$

Hence, the correct option is $\left(A\right)$.