Question

question 16 (1 point)
how many hydrogen ions must be added to the half - reaction to balance the mass of hydrogen atoms?
$\ce{cn^{-}(aq) -> cno^{-}(s)}$
1 on the left side
1 on the right side
2 on the left side
2 on the right side
no hydrogen ions are needed

Explanation:

Step1: Analyze hydrogen atoms

In the half - reaction $\ce{CN^{-}(aq) -> CNO^{-}(s)}$, we first check the number of hydrogen atoms on each side. The left - hand side (reactant side) has no hydrogen atoms, and the right - hand side (product side) also has no hydrogen atoms from the given species. But we need to balance the oxygen first (usually in redox half - reaction balancing, we balance O by adding $\ce{H2O}$ and then H by adding $\ce{H+}$). Wait, let's correct that. First, balance the oxygen. The left side has 0 O, the right side has 1 O. So we add 1 $\ce{H2O}$ to the left side: $\ce{CN^{-}(aq) + H2O(l) -> CNO^{-}(s)}$. Now, the left side has 2 H atoms (from $\ce{H2O}$) and the right side has 0 H atoms. To balance the H atoms, we add 2 $\ce{H+}$ to the right side? Wait, no, wait. Wait, the original question is about balancing the mass of hydrogen atoms. Wait, maybe I made a mistake. Wait, the initial half - reaction: $\ce{CN^{-}}$ has C, N; $\ce{CNO^{-}}$ has C, N, O. There are no H atoms in either the reactant or the product of the original half - reaction (before adding $\ce{H2O}$ or $\ce{H+}$). Wait, but when we balance the oxygen, we add $\ce{H2O}$. Let's do the correct balancing steps for this half - reaction (oxidation half - reaction, since C is going from +2 in $\ce{CN^{-}}$ to +4 in $\ce{CNO^{-}}$).

1. Balance the non - H, non - O atoms: C and N are already balanced (1 C and 1 N on each side).
2. Balance O: Add $\ce{H2O}$ to the side with less O. Left side: 0 O, right side: 1 O. So add 1 $\ce{H2O}$ to the left: $\ce{CN^{-}(aq) + H2O(l) -> CNO^{-}(s)}$
3. Balance H: Now, left side has 2 H (from $\ce{H2O}$), right side has 0 H. So we add 2 $\ce{H+}$ to the right side: $\ce{CN^{-}(aq) + H2O(l) -> CNO^{-}(s) + 2H^{+}(aq)}$

Wait, but the question is "How many hydrogen ions must be added to the half - reaction to balance the mass of hydrogen atoms?". Wait, maybe the question is a bit simplified. Wait, in the original half - reaction (without considering O balancing first), there are no H atoms in $\ce{CN^{-}}$ or $\ce{CNO^{-}}$. But when we balance O, we introduce $\ce{H2O}$, which introduces H. But maybe the question is considering a different approach. Wait, no, maybe the question has a typo or is simplified. Wait, let's re - examine the half - reaction. The reactant is $\ce{CN^{-}}$, product is $\ce{CNO^{-}}$. There are no H atoms in either. So when we balance the oxygen, we add $\ce{H2O}$ to the left (1 $\ce{H2O}$), which gives 2 H on the left. Then to balance H, we add 2 $\ce{H+}$ to the right. So the number of $\ce{H+}$ added is 2 on the right side.

Step2: Match with options

Looking at the options, the option "2 on the right side" is the correct one.

Answer:

2 on the right side