Question

question 17 (1 point)
how many electrons must be added to the half-reaction to balance the charge after
the mass has been balanced?
$ce{cn^{-}(aq) -> cno^{-}(s)}$

1 on the left side
1 on the right side
2 on the left side
2 on the right side
no electrons are needed

Explanation:

Step1: Balance mass (atoms)

First, balance the non - O and non - H atoms. Here, C and N are already balanced (1 C and 1 N on each side). Then balance O by adding \(H_2O\) and balance H by adding \(H^+\) (in acidic medium, but since we can also consider basic medium, but let's first do mass balance). The unbalanced half - reaction is \(CN^-(aq) ightarrow CNO^-(s)\). To balance O, we add 1 \(H_2O\) to the left side: \(CN^-(aq)+H_2O(l) ightarrow CNO^-(s)\). Now balance H by adding \(2H^+\) to the right side: \(CN^-(aq)+H_2O(l) ightarrow CNO^-(s)+2H^+\). (If we consider basic medium, we can add \(2OH^-\) to both sides to neutralize \(H^+\): \(CN^-(aq)+H_2O(l)+2OH^-(aq) ightarrow CNO^-(s)+2H_2O(l)\), and then simplify to \(CN^-(aq)+2OH^-(aq) ightarrow CNO^-(s)+H_2O(l)\). But for charge balance, let's first find the charge on each side before electron addition.)

Step2: Calculate charge on each side

For the left side (after mass balance, let's take the acidic medium first, but the charge calculation is similar in basic medium). The left side has \(CN^-\) and \(H_2O\) (charge of \(H_2O\) is 0) and \(H_2O\) is neutral. So charge on left: charge of \(CN^-\) is - 1. The right side has \(CNO^-\) (charge - 1) and \(2H^+\) (each \(H^+\) has charge + 1, so total + 2). So charge on right: \(-1 + 2\times(+1)=+1\).

Now, let's find the difference in charge. The left side has charge - 1, the right side has charge + 1. To balance the charge, we need to add electrons. Electrons have a charge of - 1. Let the number of electrons be \(n\). We want left charge + charge of electrons = right charge. So \(-1+(-n)= + 1\)? Wait, no. Wait, the oxidation state method: Let's find the oxidation state of C in \(CN^-\) and \(CNO^-\). In \(CN^-\), let oxidation state of C be \(x\), N is - 3 (common oxidation state of N in cyanide). So \(x+( - 3)=-1\), so \(x = + 2\). In \(CNO^-\), let oxidation state of C be \(y\), N is - 3, O is - 2. So \(y+( - 3)+( - 2)=-1\), so \(y=+4\). So C is oxidized from + 2 to + 4, so it loses 2 electrons? Wait, no, wait, maybe I made a mistake in oxidation state. Wait, in \(CN^-\), the oxidation state of C: N is - 3, so \(C + N=-1\), \(C-3=-1\), \(C = + 2\). In \(CNO^-\), N is - 3, O is - 2, so \(C+N + O=-1\), \(C-3 - 2=-1\), \(C = + 4\). So the change in oxidation state of C is \(+4-(+2)=+2\), which means it loses 2 electrons? Wait, no, wait the half - reaction is oxidation (C is oxidized), so electrons are lost (on the right side? No, in oxidation, electrons are products, so they are on the right side). Wait, maybe my mass balance was wrong. Let's start over.

Alternative approach: Let's balance the half - reaction for charge without considering H and O first. The half - reaction is \(CN^- ightarrow CNO^-\). The oxidation state of C: in \(CN^-\), C is + 2 (as above), in \(CNO^-\), C is + 4. So the change in oxidation state is \(+4 - + 2=+2\), which means each C atom loses 2 electrons? Wait, no, the number of C atoms is 1. So the number of electrons lost is 2? Wait, no, wait the charge on \(CN^-\) is - 1, charge on \(CNO^-\) is - 1. Wait, maybe I messed up the mass balance. Wait, the correct mass balance for O: to go from \(CN^-\) (no O) to \(CNO^-\) (1 O), we need to add 1 O. In aqueous solution, O comes from \(H_2O\) (in acidic) or \(OH^-\) (in basic). So the correct half - reaction balancing (in basic medium, which is more likely for cyanide reactions):

1. Balance non - O, non - H: \(CN^- ightarrow CNO^-\) (C and N are balanced).
2. Balance O: add 1 \(H_2O\) to the left: \(CN^-+H_2O ightarrow CNO^-\)
3. Balance H: add 2 \(H^+\) to the right: \(CN^-+H_2O ightarrow CNO^-+2H^+\)
4. Now, to make it basic, add 2 \(OH^-\) to both sides: \(CN^-+H_2O + 2OH^- ightarrow CNO^-+2H^++2OH^-\)
5. Simplify: \(CN^-+2OH^- ightarrow CNO^-+H_2O\) (since \(2H^++2OH^- = 2H_2O\), and we have \(H_2O\) on left, so \(H_2O\) on right is \(2H_2O - H_2O=H_2O\))

Now, calculate charge on each side:

Left side: \(CN^-\) (charge - 1) + \(2OH^-\) (each \(OH^-\) has charge - 1, so total - 2). So total charge on left: \(-1+( - 2)=-3\)

Right side: \(CNO^-\) (charge - 1) + \(H_2O\) (charge 0). So total charge on right: - 1

Wait, that can't be right. Wait, no, in the step 4, when we add \(2OH^-\) to both sides, the left side is \(CN^-+H_2O + 2OH^-\), charge of \(CN^-\) is - 1, \(H_2O\) is 0, \(2OH^-\) is - 2, so total - 3. Right side: \(CNO^-\) ( - 1) + \(2H^+\) ( + 2) + \(2OH^-\) ( - 2). \(2H^++2OH^- = 2H_2O\) (charge 0), so right side charge is - 1 (from \(CNO^-\)) + 0 (from \(2H_2O\))=-1.

Wait, I think I made a mistake in oxidation state. Let's recalculate oxidation state of C. In \(CN^-\), the oxidation state of N is - 3 (because in cyanide, N is more electronegative than C), so C + N=-1, so C = + 2. In \(CNO^-\), N is - 3, O is - 2, so C+(-3)+(-2)=-1, so C = + 4. So the change in oxidation state of C is \(+4 - + 2 = + 2\), which means C is oxidized, so it loses 2 electrons. Wait, but when we look at the charge balance, let's do it without considering acid - base first.

Original unbalanced half - reaction: \(CN^-(aq) ightarrow CNO^-(s)\)

Charge on left: - 1 (from \(CN^-\))

Charge on right: - 1 (from \(CNO^-\))

Wait, that's the same? Wait, no, wait I forgot to balance O and H. Wait, the correct way is that when we balance the half - reaction, we must balance O and H first. Let's do it properly for oxidation - reduction (redox) half - reaction.

The half - reaction is an oxidation reaction (C is oxidized from + 2 to + 4).

1. Balance the atoms:

- C and N are balanced (1 C, 1 N on each side).

- To balance O, add 1 \(H_2O\) to the left: \(CN^-+H_2O ightarrow CNO^-\)

- To balance H, add 2 \(H^+\) to the right: \(CN^-+H_2O ightarrow CNO^-+2H^+\) (in acidic solution) or, in basic solution, add 2 \(OH^-\) to both sides: \(CN^-+H_2O + 2OH^- ightarrow CNO^-+2H^++2OH^-\), then simplify to \(CN^-+2OH^- ightarrow CNO^-+H_2O\)

2. Now, calculate the charge on each side (for the acidic solution version after atom balance):

- Left side: \(CN^-\) (charge - 1) + \(H_2O\) (charge 0). So total charge: - 1

- Right side: \(CNO^-\) (charge - 1) + \(2H^+\) (charge + 2). So total charge: \(-1 + 2=+1\)

3. Now, find the number of electrons to add. Electrons have a charge of - 1. Let the number of electrons be \(n\). We want the charge on the left + charge of electrons = charge on the right.

- Left charge: - 1; Right charge: + 1

- So \(-1+(-n)=+1\)? No, wait, electrons are added to balance the charge. In oxidation, electrons are lost (so they are on the right side). Wait, the oxidation state of C increases by 2 (from + 2 to + 4), so the number of electrons lost is 2. Wait, but the charge on the left is - 1, right is + 1. The difference in charge is \(+1-(-1)=+2\). Since each electron has a charge of - 1, to balance the charge, we need to add 2 electrons to the right side? Wait, no. Wait, the left side has charge - 1, right side has charge + 1. The net charge on the right is more positive. So we need to add electrons (negative charge) to the right side to decrease the positive charge. The difference in charge is \(+1-(-1) = + 2\). So we need to add 2 electrons to the right side? Wait, no, let's think again.

Wait, the correct way: the charge on the left (after atom balance) is - 1 (from \(CN^-\)) and the charge on the right is + 1 (from \(CNO^-\) (- 1) and \(2H^+\) (+ 2)). So the right side is more positive by 2 units. To balance the charge, we need to add 2 electrons (each with charge - 1) to the right side. So the reaction becomes \(CN^-+H_2O ightarrow CNO^-+2H^++2e^-\)

Wait, but let's check the oxidation state. C goes from + 2 to + 4, so it loses 2 electrons, which means electrons are on the product (right) side. So the number of electrons to add to balance the charge is 2 on the right side.

Wait, but let's check the basic medium version. After simplifying to \(CN^-+2OH^- ightarrow CNO^-+H_2O\)

Charge on left: \(CN^-\) (- 1) + \(2OH^-\) (- 2) = - 3

Charge on right: \(CNO^-\) (- 1) + \(H_2O\) (0) = - 1

Difference in charge: \(-1-(-3)=+2\). So we need to add 2 electrons to the right side (since right side is more positive by 2, electrons have negative charge) to balance. So \(CN^-+2OH^- ightarrow CNO^-+H_2O + 2e^-\)

Yes, so the number of electrons to add is 2 on the right side.

Answer:

2 on the right side