question 16 (1 point)
how many hydrogen ions must be added to the half - reaction to balance the mass of
hydrogen atoms?
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
question 17 (1 point)
how many electrons must be added to the half - reaction to balance the charge after
the mass has been balanced?
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 electrons are needed
question 18 (1 point)

Question

Accepted Answer

### Question 16
# Explanation:
## Step1: Analyze reactant and product
Reactant: $\ce{CN^- (aq)}$ (contains C, N, no H)
Product: $\ce{CNO^- (s)}$ (contains C, N, O, no H)
## Step2: Check H atoms
Both sides have 0 H atoms. So no H⁺ needed.
# Answer: no hydrogen ions are needed

### Question 17
# Explanation:
## Step1: Balance mass (already balanced for H, C, N, O)
Reaction: $\ce{CN^- (aq) -> CNO^- (s)}$
## Step2: Balance charge
Left charge: -1
Right charge: -1
Wait, no—wait, let's recheck. Wait, maybe I missed oxidation state? Wait, CN⁻ to CNO⁻: C goes from +2 to +4 (loses 2 e⁻? No, wait, no—wait, let's balance the reaction properly. Wait, first, balance O: add H₂O? Wait, no, the original reaction is CN⁻(aq) → CNO⁻(s). Wait, maybe the O comes from water? Wait, no, the problem says "after mass has been balanced"—maybe the mass balance here is just C, N, and O? Wait, CN⁻ has C, N; CNO⁻ has C, N, O. So to balance O, we need to add H₂O? Wait, no, the problem might have a typo, but assuming mass is balanced (maybe O is balanced by some means). Wait, no, let's check charge. Left: CN⁻ is -1. Right: CNO⁻ is -1. Wait, that can't be. Wait, maybe the oxidation: CN⁻ to CNO⁻: N is -3 in CN⁻, in CNO⁻, N is -3? Wait, no, C in CN⁻ is +2, in CNO⁻ is +4 (since O is -2, N is -3: +4 + (-3) + (-2) = -1, which matches CNO⁻ charge). So C is oxidized (loses 2 e⁻). Wait, but the charge on left is -1, right is -1. Wait, that's a problem. Wait, maybe the reaction is in basic or acidic? Wait, no, the problem says "after mass has been balanced"—maybe the mass balance includes adding H⁺ or H₂O. Wait, maybe I made a mistake. Wait, let's do it properly:

1. Balance C, N: CN⁻ → CNO⁻
2. Balance O: add H₂O to left? No, CNO⁻ has O, so add H₂O to left: CN⁻ + H₂O → CNO⁻
3. Balance H: add H⁺ to right: CN⁻ + H₂O → CNO⁻ + 2 H⁺
Now, charge:
Left: -1 (CN⁻) + 0 (H₂O) = -1
Right: -1 (CNO⁻) + 2(+1) (H⁺) = +1
So to balance charge, we need to add 2 e⁻ to right (since left is -1, right is +1; difference is +2, so add 2 e⁻ to right to make right charge -1: -1 + 2(+1) - 2 = -1). Wait, no: left charge -1, right charge (CNO⁻: -1) + (2 H⁺: +2) = +1. So to balance, we need to add 2 e⁻ to the right (reduction) or left (oxidation)? Wait, C is oxidized (loses e⁻), so electrons are on the right. Wait, the problem says "how many electrons must be added to the half-reaction to balance the charge". So left charge: -1 (CN⁻) + 0 (H₂O) = -1. Right charge: -1 (CNO⁻) + 2(+1) (H⁺) = +1. So the difference is +2 (right is more positive). So to balance, we need to add 2 e⁻ to the right (so that right charge is +1 - 2 = -1, same as left). Wait, but the options are 1 or 2 on left/right. Wait, maybe the mass balance was done without H₂O? Maybe the problem is simpler. Wait, the original reaction: CN⁻ → CNO⁻. Charge left: -1, right: -1. But that's only if O is balanced. Wait, maybe the problem assumes that mass is balanced (C, N) and O is already balanced (maybe from the environment). But that's not correct. Wait, the problem says "after the mass has been balanced"—so maybe in the mass balance, we only consider C and N, and O is already there. Then charge: left -1, right -1. But that would mean no electrons needed. But that contradicts oxidation. Wait, maybe the problem has an error, but looking at the options, "no electrons are needed" is an option. Wait, no—wait, let's recalculate. CN⁻: charge -1. CNO⁻: charge -1. So charge is balanced. So no electrons needed? But that can't be, because C is oxidized. Wait, maybe the O is balanced by adding O₂? No, the problem is about half-reaction. Wait, maybe the mass balance is just C and N, and O is already present. So charge is -1 on both sides. So no electrons needed.
Wait, but that's confusing. Alternatively, maybe the reaction is in basic solution. Let's try basic:
CN⁻ + H₂O → CNO⁻ + 2 H⁺ + 2 e⁻ (acidic)
In basic, add 2 OH⁻:
CN⁻ + H₂O + 2 OH⁻ → CNO⁻ + 2 H₂O + 2 e⁻
Simplify: CN⁻ + 2 OH⁻ → CNO⁻ + H₂O + 2 e⁻
Charge left: -1 + 2(-1) = -3
Charge right: -1 + 0 + 2(-1) = -3. Wait, no, electrons are -1 each. Wait, no, electrons are added to balance charge. Wait, I think I messed up. Let's start over.

Original half-reaction: CN⁻(aq) → CNO⁻(s)

1. Balance C and N: already balanced (1 C, 1 N on each side).
2. Balance O: add 1 H₂O to the left (since product has 1 O):
CN⁻ + H₂O → CNO⁻
3. Balance H: add 2 H⁺ to the right (since left has 2 H from H₂O):
CN⁻ + H₂O → CNO⁻ + 2 H⁺
4. Balance charge:
Left charge: -1 (CN⁻) + 0 (H₂O) = -1
Right charge: -1 (CNO⁻) + 2(+1) (H⁺) = +1
Difference: +2 (right is more positive). So we need to add 2 electrons to the right (reduction) or left (oxidation)? Since C is oxidized (loses electrons), electrons are on the right:
CN⁻ + H₂O → CNO⁻ + 2 H⁺ + 2 e⁻
Now, charge: left -1, right: -1 + 2(+1) + 2(-1) = -1 + 2 - 2 = -1. Balanced.

But the problem says "after the mass has been balanced"—maybe the mass balance here is just C and N, and O is considered balanced (maybe the problem assumes O is already there). But according to the steps, after mass balance (including O, H), we need to add 2 e⁻ to the right. But the options are 1 on left, 1 on right, 2 on left, 2 on right, or no electrons. Wait, maybe the problem has a different approach. Wait, the original reaction: CN⁻ → CNO⁻. Let's check oxidation states:

- In CN⁻: C is +2, N is -3.
- In CNO⁻: C is +4, N is -3, O is -2.

So C is oxidized from +2 to +4, so it loses 2 electrons. So the half-reaction is oxidation: CN⁻ → CNO⁻ + 2 e⁻. Now, check charge:

Left: -1
Right: -1 + 2(-1) = -3. Wait, that's not balanced. Wait, no—if we add 2 e⁻ to the right, charge is -1 (CNO⁻) + 2(-1) (e⁻) = -3. Left is -1. So we need to add H⁺ or OH⁻. Ah, here's the mistake: the mass balance includes H and O. So we need to balance H and O first.

So correct steps:

1. Balance O: add H₂O to left (1 O in CNO⁻, so 1 H₂O on left):
CN⁻ + H₂O → CNO⁻
2. Balance H: add 2 H⁺ to right (2 H in H₂O):
CN⁻ + H₂O → CNO⁻ + 2 H⁺
3. Balance charge:
Left charge: -1 (CN⁻) + 0 (H₂O) = -1
Right charge: -1 (CNO⁻) + 2(+1) (H⁺) = +1
Difference: +2 (right is more positive). So to balance, add 2 e⁻ to the right (since electrons are negative, adding 2 e⁻ to right: +1 + 2(-1) = -1, which matches left charge -1).

So the half-reaction is: $\ce{CN^- (aq) + H2O (l) -> CNO^- (s) + 2 H^+ (aq) + 2 e^-}$

Now, after mass balance (steps 1 and 2), we add 2 e⁻ to the right. So the answer is 2 on the right side.

Wait, but the options include "2 on the right side". So that's the answer.

# Explanation:
## Step1: Balance mass (add H₂O and H⁺)
$\ce{CN^- + H2O -> CNO^- + 2 H^+}$
## Step2: Balance charge
Left charge: -1
Right charge: -1 + 2(+1) = +1
Need to add 2 e⁻ to right (charge becomes +1 + 2(-1) = -1, matching left)
# Answer: 2 on the right side

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QUESTION IMAGE

Question

Explanation:

Question 16

Step1: Analyze reactant and product

Step2: Check H atoms

Step1: Balance mass (already balanced for H, C, N, O)

Step2: Balance charge

Step1: Balance mass (add H₂O and H⁺)

Step2: Balance charge

Answer:

Question 17