We determine the oxidation number of the metal in each compound based on the rules:

• HAuCl₄: $\text{H} (+1) + \text{Au} (x) + 4 \times \text{Cl} (-1) = 0 \implies 1 + x - 4 = 0 \implies x = +3$. Representation: HAu(III)Cl₄.
• Tl₂O: $2 \times \text{Tl} (x) + \text{O} (-2) = 0 \implies 2x - 2 = 0 \implies x = +1$. Representation: Tl₂(I)O.
• FeO: $\text{Fe} (x) + \text{O} (-2) = 0 \implies x - 2 = 0 \implies x = +2$. Representation: Fe(II)O.
• Fe₂O₃: $2 \times \text{Fe} (x) + 3 \times \text{O} (-2) = 0 \implies 2x - 6 = 0 \implies x = +3$. Representation: Fe₂(III)O₃.
• CuI: $\text{Cu} (x) + \text{I} (-1) = 0 \implies x - 1 = 0 \implies x = +1$. Representation: Cu(I)I.
• CuO: $\text{Cu} (x) + \text{O} (-2) = 0 \implies x - 2 = 0 \implies x = +2$. Representation: Cu(II)O.
• MnO: $\text{Mn} (x) + \text{O} (-2) = 0 \implies x - 2 = 0 \implies x = +2$. Representation: Mn(II)O.
• MnO₂: $\text{Mn} (x) + 2 \times \text{O} (-2) = 0 \implies x - 4 = 0 \implies x = +4$. Representation: Mn(IV)O₂.

First, assign oxidation numbers to all elements in the reactants and products:

$\quad\quad +1\,\, -2 \quad\quad +1\,\, -2 \quad\quad 0 \quad\quad +4\,\, -2$

$2\text{Cu}_2\text{O(s)} + \text{Cu}_2\text{S(s)} \rightarrow 6\text{Cu(s)} + \text{SO}_2\text{(g)}$

Observe the changes in oxidation numbers:

• Copper (Cu) changes from +1 in both $\text{Cu}_2\text{O}$ and $\text{Cu}_2\text{S}$ to 0 in elemental Cu. The oxidation number of Cu decreases (from +1 to 0). Copper is reduced.
• Sulfur (S) changes from -2 in $\text{Cu}_2\text{S}$ to +4 in $\text{SO}_2$. The oxidation number of S increases (from -2 to +4). Sulfur is oxidised.

Since both oxidation and reduction occur (changes in oxidation numbers are observed), this reaction is a redox reaction.

Identify the agents:

• The species containing the element that is reduced is the oxidising agent. Copper is reduced (from +1 to 0). Copper is present in both $\text{Cu}_2\text{O}$ and $\text{Cu}_2\text{S}$. So, $\text{Cu}_2\text{O}$ and $\text{Cu}_2\text{S}$ act as oxidants (or rather, the $\text{Cu(I)}$ ion in these compounds acts as the oxidant accepting electrons).
• The species containing the element that is oxidised is the reducing agent. Sulfur is oxidised (from -2 to +4). Sulfur is present in $\text{Cu}_2\text{S}$. So, $\text{Cu}_2\text{S}$ acts as the reductant (specifically, the $\text{S}^{2-}$ ion).

Note: In this specific reaction, $\text{Cu}_2\text{S}$ acts as *both* oxidant (providing $\text{Cu(I)}$ to be reduced) and reductant (providing $\text{S}^{2-}$ to be oxidized). However, typically, the species that undergoes the *most* significant oxidation/reduction might be labeled as the primary agent. Here, $\text{Cu}_2\text{O}$ is solely an oxidant (Cu is reduced), and $\text{Cu}_2\text{S}$ is both oxidant and reductant (Cu is reduced, S is oxidized). A simpler view is often taken where the entire compound causing reduction is the reductant and the entire compound causing oxidation is the oxidant. In this case, $\text{Cu}_2\text{S}$ is the reductant because it provides the element that gets oxidized. The oxidant is $\text{Cu}_2\text{O}$ which provides copper that gets reduced, and also $\text{Cu(I)}$ in $\text{Cu}_2\text{S}$ which also gets reduced. This reaction is complex because $\text{Cu}$ in $\text{Cu}_2\text{O}$ and $\text{Cu}$ in $\text{Cu}_2\text{S}$ are reduced, while $\text{S}$ in $\text{Cu}_2\text{S}$ is oxidised.

Based on a common interpretation in complex redox reactions: The species containing the element with increasing oxidation number is the reductant. The species containing the element with decreasing oxidation number is the oxidant. Here, S in $\text{Cu}_2\text{S}$ increases, so $\text{Cu}_2\text{S}$ is the reductant. Cu in $\text{Cu}_2\text{O}$ decreases, so $\text{Cu}_2\text{O}$ is the oxidant. Copper in $\text{Cu}_2\text{S}$ also decreases, so $\text{Cu}_2\text{S}$ is also an oxidant, making it a complex reaction.

A species can undergo disproportionation if the element experiencing the oxidation and reduction is in an intermediate oxidation state. It must be able to go to both a higher and a lower oxidation state from its current state.

Let's determine the oxidation state of chlorine in each species:

• ClO⁻ (Hypochlorite ion): Cl + (-2 for O) = -1 $\implies$ Cl = +1. Possible oxidation states for Cl are -1, 0, +1, +3, +5, +7. +1 is an intermediate state (can go to -1, 0, +3, +5, +7). Can disproportionate.
• ClO₂⁻ (Chlorite ion): Cl + 2*(-2 for O) = -1 $\implies$ Cl - 4 = -1 $\implies$ Cl = +3. +3 is an intermediate state (can go to -1, 0, +1, +5, +7). Can disproportionate.
• ClO₃⁻ (Chlorate ion): Cl + 3*(-2 for O) = -1 $\implies$ Cl - 6 = -1 $\implies$ Cl = +5. +5 is an intermediate state (can go to -1, 0, +1, +3, +7). Can disproportionate.
• ClO₄⁻ (Perchlorate ion): Cl + 4*(-2 for O) = -1 $\implies$ Cl - 8 = -1 $\implies$ Cl = +7. +7 is the highest possible oxidation state for chlorine. Cannot go to a higher state, so it cannot disproportionate.

ClO₄⁻ does not disproportionate because chlorine is already in its maximum oxidation state (+7).

Disproportionation reactions for the others (often in alkaline medium):

• ClO⁻ (+1): Can go to Cl⁻ (-1, reduced) and ClO₃⁻ (+5, oxidised).
  $3\text{ClO}^-\text{(aq)} \rightarrow 2\text{Cl}^-\text{(aq)} + \text{ClO}_3^-\text{(aq)}$
• ClO₂⁻ (+3): Can go to Cl⁻ (-1, reduced) and ClO₃⁻ (+5, oxidised), or Cl⁻ (-1) and ClO₄⁻ (+7).
  $6\text{ClO}_2^-\text{(aq)} \rightarrow 4\text{ClO}_3^-\text{(aq)} + 2\text{Cl}^-\text{(aq)}$
• ClO₃⁻ (+5): Can go to Cl⁻ (-1, reduced) and ClO₄⁻ (+7, oxidised).
  $4\text{ClO}_3^-\text{(aq)} \rightarrow \text{Cl}^-\text{(aq)} + 3\text{ClO}_4^-\text{(aq)}$

We classify these based on whether substances combine, decompose, displace elements, or an element disproportionates.

• (a) $\text{N}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{ NO (g)}$
  Two elements combine to form a compound. This is a Combination Redox Reaction. (N goes from 0 to +2, O from 0 to -2).
• (b) $2\text{Pb(NO}_3)_2\text{(s)} \rightarrow 2\text{PbO(s)} + 4\text{ NO}_2\text{ (g)} + \text{O}_2\text{ (g)}$
  A single compound breaks down into multiple substances, including an elemental substance ($\text{O}_2$). This is a Decomposition Redox Reaction. (Pb: +2 to +2 (no change), N: +5 to +4, O: -2 to -2 (in PbO and $\text{NO}_2$), and -2 to 0 (in $\text{O}_2$). N is reduced, some O is oxidised from the nitrate).
• (c) $\text{NaH(s)} + \text{H}_2\text{O(l)} \rightarrow \text{NaOH(aq)} + \text{H}_2\text{ (g)}$
  Sodium (Na) displaces hydrogen (H) from water (where H is +1 in H₂O). Specifically, the hydride ion (H⁻) from NaH displaces H⁺ from water. This is a Displacement Redox Reaction (specifically non-metal displacement - hydrogen). (Na: +1 to +1 (no change), H: -1 (in NaH) to 0 (in $\text{H}_2$), H: +1 (in $\text{H}_2\text{O}$) to 0 (in $\text{H}_2$). H⁻ is oxidized, $\text{H}^+$ from water is reduced).
• (d) $2\text{NO}_2\text{(g)} + 2\text{OH}^-\text{(aq)} \rightarrow \text{NO}_2^-\text{(aq)} + \text{NO}_3^-\text{(aq)}+\text{H}_2\text{O(l)}$
  Nitrogen in $\text{NO}_2$ has an oxidation state of +4. In the products, nitrogen is +3 in $\text{NO}_2^-$ and +5 in $\text{NO}_3^-$. Nitrogen in $\text{NO}_2$ is simultaneously reduced (+4 to +3) and oxidised (+4 to +5). This is a Disproportionation Reaction.

The difference in reactions is due to the nature of $\text{Pb}_3\text{O}_4$ and the oxidizing strength of the acids involved.

$\text{Pb}_3\text{O}_4$ is a mixed oxide, correctly represented as $2\text{PbO} \cdot \text{PbO}_2$. It contains lead in two oxidation states: +2 in $\text{PbO}$ and +4 in $\text{PbO}_2$.

• In $\text{PbO}$, lead is in the +2 oxidation state, which is a stable oxidation state for lead. $\text{PbO}$ is a basic oxide.
• In $\text{PbO}_2$, lead is in the +4 oxidation state. $\text{Pb(IV)}$ is less stable than $\text{Pb(II)}$ and $\text{PbO}_2$ acts as a strong oxidizing agent (lead can be reduced from +4 to +2 or 0).

Now consider the reactions with the acids:

Reaction with HCl: $\text{Pb}_3\text{O}_4 + 8\text{HCl} \rightarrow 3\text{PbCl}_2 + \text{Cl}_2 + 4\text{H}_2\text{O}$

This reaction can be viewed as two separate reactions happening simultaneously:

1. Acid-base reaction between $\text{PbO}$ (basic oxide) and $\text{HCl}$ (acid):
$2\text{PbO} + 4\text{HCl} \rightarrow 2\text{PbCl}_2 + 2\text{H}_2\text{O}$ (Lead oxidation state remains +2)

2. Redox reaction between $\text{PbO}_2$ (oxidizing agent) and $\text{HCl}$ (here $\text{Cl}^-$ acts as a reducing agent):
$+4 \quad -1 \quad\quad +2 \quad\quad 0$
$\text{PbO}_2 + 4\text{HCl} \rightarrow \text{PbCl}_2 + \text{Cl}_2 + 2\text{H}_2\text{O}$ (Lead is reduced from +4 to +2, Chlorine is oxidized from -1 to 0)

Adding these two reactions ($2\text{PbO} + \text{PbO}_2 = \text{Pb}_3\text{O}_4$ and $4\text{HCl} + 4\text{HCl} = 8\text{HCl}$) gives the overall reaction with HCl.

Reaction with HNO₃: $\text{Pb}_3\text{O}_4 + 4\text{HNO}_3 \rightarrow 2\text{Pb(NO}_3)_2 + \text{PbO}_2 + 2\text{H}_2\text{O}$

Here too, the reaction can be viewed in terms of $\text{PbO}$ and $\text{PbO}_2$ components:

1. Acid-base reaction between $\text{PbO}$ and $\text{HNO}_3$:
$2\text{PbO} + 4\text{HNO}_3 \rightarrow 2\text{Pb(NO}_3)_2 + 2\text{H}_2\text{O}$ (Lead oxidation state remains +2)

2. Reaction of $\text{PbO}_2$ with $\text{HNO}_3$: Unlike HCl, nitric acid ($\text{HNO}_3$) is itself a strong oxidizing agent. Therefore, it cannot be oxidized by $\text{PbO}_2$. Instead, $\text{PbO}_2$ is unreactive towards $\text{HNO}_3$ under these conditions and remains as a product.

Adding these two reactions gives the overall reaction with $\text{HNO}_3$. The difference arises because $\text{HCl}$ is a reducing acid (due to $\text{Cl}^-$), allowing $\text{PbO}_2$ to act as an oxidant and be reduced to $\text{Pb}^{2+}$ products, whereas $\text{HNO}_3$ is an oxidizing acid, preventing $\text{PbO}_2$ from being reduced and causing it to appear as a product.

The reactants and products in ionic form are $\text{Cr}_2\text{O}_7^{2-}$, $\text{SO}_3^{2-}$, $\text{Cr}^{3+}$, and $\text{SO}_4^{2-}$. The medium is acidic (implies $\text{H}^+$ and $\text{H}_2\text{O}$ are involved).

Step 1: Skeletal ionic equation:
$\text{Cr}_2\text{O}_7^{2-}\text{(aq)} + \text{SO}_3^{2-}\text{(aq)} \rightarrow \text{Cr}^{3+}\text{(aq)} + \text{SO}_4^{2-}\text{(aq)}$

Step 2: Assign oxidation numbers to atoms changing state:
$+6 \quad\quad +4 \quad\quad +3 \quad\quad +6$
$\text{Cr}_2\text{O}_7^{2-} + \text{SO}_3^{2-} \rightarrow \text{Cr}^{3+} + \text{SO}_4^{2-}$

Step 3: Calculate total increase/decrease and balance atoms:
Cr changes from +6 to +3 (decrease of 3 per atom). Since there are 2 Cr atoms in $\text{Cr}_2\text{O}_7^{2-}$, total decrease is $3 \times 2 = 6$.
S changes from +4 to +6 (increase of 2 per atom). There is 1 S atom in $\text{SO}_3^{2-}$ and $\text{SO}_4^{2-}$. Total increase is 2.

To make total increase equal to total decrease (6), multiply $\text{SO}_3^{2-}$ and $\text{SO}_4^{2-}$ by 3, and $\text{Cr}^{3+}$ by 2 (already done implicitly when balancing Cr). Balance Cr atoms (already done):

$\text{Cr}_2\text{O}_7^{2-} + 3\text{SO}_3^{2-} \rightarrow 2\text{Cr}^{3+} + 3\text{SO}_4^{2-}$
Total decrease in ON for Cr: $2 \times (6 \rightarrow 3) = 2 \times (-3) = -6$.
Total increase in ON for S: $3 \times (4 \rightarrow 6) = 3 \times (+2) = +6$. (Total change is balanced).

Step 4: Balance charge using H⁺ (acidic medium):
Left side total charge: $-2 + 3(-2) = -2 - 6 = -8$.
Right side total charge: $2(+3) + 3(-2) = +6 - 6 = 0$.

The left side has a charge of -8, the right side has 0. Add $\text{H}^+$ to the left to balance charge ($+8\text{H}^+$):
$\text{Cr}_2\text{O}_7^{2-}\text{(aq)} + 3\text{SO}_3^{2-}\text{(aq)} + 8\text{H}^+\text{(aq)} \rightarrow 2\text{Cr}^{3+}\text{(aq)} + 3\text{SO}_4^{2-}\text{(aq)}$
Left side charge: $-2 + 3(-2) + 8(+1) = -2 - 6 + 8 = 0$. Right side charge: $2(+3) + 3(-2) = +6 - 6 = 0$. (Charge is balanced).

Step 5: Balance O and H using H₂O:
Left side: 7 O (in $\text{Cr}_2\text{O}_7^{2-}$) + 3*3 = 9 O (in $\text{SO}_3^{2-}$) = 16 O atoms. 8 H atoms.
Right side: 3*4 = 12 O (in $\text{SO}_4^{2-}$) + 0 H atoms.

Add $\text{H}_2\text{O}$ to the side deficient in oxygen (right side). Need 16 - 12 = 4 O atoms. Add $4\text{H}_2\text{O}$. This also adds $4 \times 2 = 8$ H atoms.
$\text{Cr}_2\text{O}_7^{2-}\text{(aq)} + 3\text{SO}_3^{2-}\text{(aq)} + 8\text{H}^+\text{(aq)} \rightarrow 2\text{Cr}^{3+}\text{(aq)} + 3\text{SO}_4^{2-}\text{(aq)} + 4\text{H}_2\text{O(l)}$

Check O atoms: Left (7 + 9 = 16), Right (12 + 4 = 16). Balanced.
Check H atoms: Left (8), Right (4*2 = 8). Balanced.

The balanced ionic equation is: $\text{Cr}_2\text{O}_7^{2-}\text{(aq)} + 3\text{SO}_3^{2-}\text{(aq)} + 8\text{H}^+\text{(aq)} \rightarrow 2\text{Cr}^{3+}\text{(aq)} + 3\text{SO}_4^{2-}\text{(aq)} + 4\text{H}_2\text{O(l)}$.

Reactants: Permanganate ion ($\text{MnO}_4^-$) and bromide ion ($\text{Br}^-$). Products: Manganese dioxide ($\text{MnO}_2$) and bromate ion ($\text{BrO}_3^-$). Medium is basic (implies $\text{OH}^-$ and $\text{H}_2\text{O}$ are involved).

Step 1: Skeletal ionic equation:
$\text{MnO}_4^-\text{(aq)} + \text{Br}^-\text{(aq)} \rightarrow \text{MnO}_2\text{(s)} + \text{BrO}_3^-\text{(aq)}$

Step 2: Assign oxidation numbers to atoms changing state:
$+7 \quad -1 \quad\quad +4 \quad\quad +5$
$\text{MnO}_4^- + \text{Br}^- \rightarrow \text{MnO}_2 + \text{BrO}_3^-$

Step 3: Calculate total increase/decrease and balance atoms:
Mn changes from +7 to +4 (decrease of 3). Multiply $\text{MnO}_2$ by 1 (implicit).
Br changes from -1 to +5 (increase of 6). Multiply $\text{Br}^-$ and $\text{BrO}_3^-$ by 1 (implicit).

To make total increase equal to total decrease (LCM of 3 and 6 is 6), multiply Mn species by 2:
$\text{2MnO}_4^- + \text{Br}^- \rightarrow \text{2MnO}_2 + \text{BrO}_3^-$
Total decrease in ON for Mn: $2 \times (7 \rightarrow 4) = 2 \times (-3) = -6$.
Total increase in ON for Br: $1 \times (-1 \rightarrow 5) = 1 \times (+6) = +6$. (Total change is balanced).

Step 4: Balance charge using OH⁻ (basic medium):
Left side total charge: $2(-1) + (-1) = -3$.
Right side total charge: $2(0) + (-1) = -1$.

The left side has a charge of -3, the right side has -1. Add $\text{OH}^-$ to the side deficient in negative charge (right side). Need $-3 = -1 + (\text{charge from OH}^-)$, so add $2\text{OH}^-$ to the right:
$2\text{MnO}_4^-\text{(aq)} + \text{Br}^-\text{(aq)} \rightarrow 2\text{MnO}_2\text{(s)} + \text{BrO}_3^-\text{(aq)} + 2\text{OH}^-\text{(aq)}$
Left side charge: $2(-1) + (-1) = -3$. Right side charge: $2(0) + (-1) + 2(-1) = 0 - 1 - 2 = -3$. (Charge is balanced).

Step 5: Balance O and H using H₂O:
Left side: $2 \times 4 = 8$ O atoms.
Right side: $2 \times 2 = 4$ O (in $\text{MnO}_2$) + 3 O (in $\text{BrO}_3^-$) + 2 O (in $\text{OH}^-$) = 9 O atoms. 2 H atoms (in $\text{OH}^-$).

Need to add H₂O to the left to balance O atoms (deficit of 1 O). Add $1\text{H}_2\text{O}$. This also adds $1 \times 2 = 2$ H atoms.
$2\text{MnO}_4^-\text{(aq)} + \text{Br}^-\text{(aq)} + \text{H}_2\text{O(l)} \rightarrow 2\text{MnO}_2\text{(s)} + \text{BrO}_3^-\text{(aq)} + 2\text{OH}^-\text{(aq)}$

Check O atoms: Left ($8 + 1 = 9$), Right ($4 + 3 + 2 = 9$). Balanced.
Check H atoms: Left (2), Right (2). Balanced.

The balanced ionic equation is: $2\text{MnO}_4^-\text{(aq)} + \text{Br}^-\text{(aq)} + \text{H}_2\text{O(l)} \rightarrow 2\text{MnO}_2\text{(s)} + \text{BrO}_3^-\text{(aq)} + 2\text{OH}^-\text{(aq)}$.

Reactants: $\text{MnO}_4^-$ (Permanganate(VII), Mn is +7), $\text{I}^-$ (Iodide, I is -1). Products: $\text{MnO}_2$ (Manganese(IV) oxide, Mn is +4), $\text{I}_2$ (Molecular iodine, I is 0). Medium is basic.

Step 1: Skeletal ionic equation:
$\text{MnO}_4^-\text{(aq)} + \text{I}^-\text{(aq)} \rightarrow \text{MnO}_2\text{(s)} + \text{I}_2\text{(s)}$

Step 2: Separate into half-reactions:
Oxidation half: $\text{I}^-\text{(aq)} \rightarrow \text{I}_2\text{(s)}$
Reduction half: $\text{MnO}_4^-\text{(aq)} \rightarrow \text{MnO}_2\text{(s)}$

Step 3: Balance atoms (other than O and H) in half-reactions:
Oxidation half: $2\text{I}^-\text{(aq)} \rightarrow \text{I}_2\text{(s)}$ (Balance I atoms by adding coefficient 2 to $\text{I}^-$)
Reduction half: $\text{MnO}_4^-\text{(aq)} \rightarrow \text{MnO}_2\text{(s)}$ (Mn atoms are already balanced)

Step 4: Balance O and H (basic medium):
Oxidation half: No O or H.
Reduction half: $\text{MnO}_4^-$ has 4 O, $\text{MnO}_2$ has 2 O. Need to balance O atoms. In basic medium, add $\text{H}_2\text{O}$ to the side with *excess* oxygen, and add $\text{OH}^-$ to the opposite side. Excess oxygen is on the left (4 vs 2). Add $2\text{H}_2\text{O}$ to the right (difference of 2 O) and $4\text{OH}^-$ to the left.
$\text{MnO}_4^-\text{(aq)} + 4\text{OH}^-\text{(aq)} \rightarrow \text{MnO}_2\text{(s)} + 2\text{H}_2\text{O(l)}$ (Check O: 4 + 4 = 8 on left; 2 + 2 = 4 on right. Incorrect method used above. Let's use the standard method for basic media: add $\text{H}_2\text{O}$ to balance O and $\text{OH}^-$ to balance H after that.)

Correct approach for balancing O and H in basic medium:

Reduction half: $\text{MnO}_4^-\text{(aq)} \rightarrow \text{MnO}_2\text{(s)}$ (Deficient in O on the right by 2 atoms).

Balance O by adding $\text{H}_2\text{O}$ to the deficient side (right), then H by adding $\text{H}^+$ to the deficient side (left) as if in acidic medium, and finally convert to basic by adding $\text{OH}^-$ equal to $\text{H}^+$ to both sides.

$\text{MnO}_4^-\text{(aq)} \rightarrow \text{MnO}_2\text{(s)} + 2\text{H}_2\text{O(l)}$ (Balances O by adding $2\text{H}_2\text{O}$)
$\text{MnO}_4^-\text{(aq)} + 4\text{H}^+\text{(aq)} \rightarrow \text{MnO}_2\text{(s)} + 2\text{H}_2\text{O(l)}$ (Balances H by adding $4\text{H}^+$)

Convert to basic medium by adding $4\text{OH}^-$ to both sides:

$\text{MnO}_4^-\text{(aq)} + 4\text{H}^+\text{(aq)} + 4\text{OH}^-\text{(aq)} \rightarrow \text{MnO}_2\text{(s)} + 2\text{H}_2\text{O(l)} + 4\text{OH}^-\text{(aq)}$

Combine $\text{H}^+$ and $\text{OH}^-$ to form $\text{H}_2\text{O}$ on the left ($4\text{H}^+ + 4\text{OH}^- \rightarrow 4\text{H}_2\text{O}$):

$\text{MnO}_4^-\text{(aq)} + 4\text{H}_2\text{O(l)} \rightarrow \text{MnO}_2\text{(s)} + 2\text{H}_2\text{O(l)} + 4\text{OH}^-\text{(aq)}$

Cancel $2\text{H}_2\text{O}$ from both sides:

$\text{MnO}_4^-\text{(aq)} + 2\text{H}_2\text{O(l)} \rightarrow \text{MnO}_2\text{(s)} + 4\text{OH}^-\text{(aq)}$ (Balanced reduction half-reaction in basic medium).

Step 5: Balance charge by adding electrons:
Oxidation half: $2\text{I}^-\text{(aq)} \rightarrow \text{I}_2\text{(s)}$. Charge on left: $2(-1) = -2$. Charge on right: 0. Add 2e⁻ to the right:
$2\text{I}^-\text{(aq)} \rightarrow \text{I}_2\text{(s)} + 2\text{e}^-$

Reduction half: $\text{MnO}_4^-\text{(aq)} + 2\text{H}_2\text{O(l)} \rightarrow \text{MnO}_2\text{(s)} + 4\text{OH}^-\text{(aq)}$. Charge on left: -1. Charge on right: -4. Add 3e⁻ to the left:
$\text{MnO}_4^-\text{(aq)} + 2\text{H}_2\text{O(l)} + 3\text{e}^- \rightarrow \text{MnO}_2\text{(s)} + 4\text{OH}^-\text{(aq)}$

Step 6: Equalize electrons in half-reactions:
Multiply oxidation half by 3: $6\text{I}^-\text{(aq)} \rightarrow 3\text{I}_2\text{(s)} + 6\text{e}^-$
Multiply reduction half by 2: $2\text{MnO}_4^-\text{(aq)} + 4\text{H}_2\text{O(l)} + 6\text{e}^- \rightarrow 2\text{MnO}_2\text{(s)} + 8\text{OH}^-\text{(aq)}$

Step 7: Add balanced half-reactions and cancel electrons:
$6\text{I}^-\text{(aq)} + 2\text{MnO}_4^-\text{(aq)} + 4\text{H}_2\text{O(l)} + 6\text{e}^- \rightarrow 3\text{I}_2\text{(s)} + 2\text{MnO}_2\text{(s)} + 8\text{OH}^-\text{(aq)} + 6\text{e}^-$
$6\text{I}^-\text{(aq)} + 2\text{MnO}_4^-\text{(aq)} + 4\text{H}_2\text{O(l)} \rightarrow 3\text{I}_2\text{(s)} + 2\text{MnO}_2\text{(s)} + 8\text{OH}^-\text{(aq)}$

Step 8: Verify balance:
Left: 6 I, 2 Mn, 8 O, 8 H. Charge: $-6 + 2(-1) = -8$.
Right: 6 I, 2 Mn, 4 O + 8 O = 12 O, 8 H. Charge: $2(0) + 8(-1) = -8$.

Oxygen atoms are not balanced (8 on left, 12 on right). Let's recheck balancing O and H in basic medium in Step 4.

Alternative standard method for balancing O and H in basic medium:

Reduction half: $\text{MnO}_4^-\text{(aq)} \rightarrow \text{MnO}_2\text{(s)}$

Balance O deficit by adding 2 $\text{H}_2\text{O}$ to the right:
$\text{MnO}_4^-\text{(aq)} \rightarrow \text{MnO}_2\text{(s)} + 2\text{H}_2\text{O(l)}$ (O is balanced: 4 on left, 2+2=4 on right)

Balance H excess (4 H on right) by adding $4\text{OH}^-$ to the left:
$\text{MnO}_4^-\text{(aq)} + 4\text{OH}^-\text{(aq)} \rightarrow \text{MnO}_2\text{(s)} + 2\text{H}_2\text{O(l)}$ (H is balanced, O is balanced, but charge is not yet balanced, that's next).

Let's go back to Step 5 with this balanced reduction half-reaction:

Oxidation half: $2\text{I}^-\text{(aq)} \rightarrow \text{I}_2\text{(s)} + 2\text{e}^-$ (Charge: -2 on left, 0 + -2 = -2 on right)

Reduction half: $\text{MnO}_4^-\text{(aq)} + 2\text{H}_2\text{O(l)} + 3\text{e}^- \rightarrow \text{MnO}_2\text{(s)} + 4\text{OH}^-\text{(aq)}$ (Charge: -1 + 0 + (-3) = -4 on left, 0 + (-4) = -4 on right). This reduction half-reaction was correctly balanced in Step 5 in terms of atoms and charge *before* multiplying.

So the correctly balanced half-reactions before multiplying for electrons are:
Oxidation: $2\text{I}^-\text{(aq)} \rightarrow \text{I}_2\text{(s)} + 2\text{e}^-$
Reduction: $\text{MnO}_4^-\text{(aq)} + 2\text{H}_2\text{O(l)} + 3\text{e}^- \rightarrow \text{MnO}_2\text{(s)} + 4\text{OH}^-\text{(aq)}$

Multiply oxidation by 3, reduction by 2 to equalize electrons (6e⁻):
$6\text{I}^-\text{(aq)} \rightarrow 3\text{I}_2\text{(s)} + 6\text{e}^-$
$2\text{MnO}_4^-\text{(aq)} + 4\text{H}_2\text{O(l)} + 6\text{e}^- \rightarrow 2\text{MnO}_2\text{(s)} + 8\text{OH}^-\text{(aq)}$

Add the multiplied half-reactions:
$6\text{I}^-\text{(aq)} + 2\text{MnO}_4^-\text{(aq)} + 4\text{H}_2\text{O(l)} \rightarrow 3\text{I}_2\text{(s)} + 2\text{MnO}_2\text{(s)} + 8\text{OH}^-\text{(aq)}$

Final check:
Left: 6 I, 2 Mn, 8 O, 8 H. Charge: $-6 + 2(-1) = -8$.
Right: 6 I, 2 Mn, 4 O + 8 O = 12 O, 8 H. Charge: $3(0) + 2(0) + 8(-1) = -8$.

Still an issue with Oxygen count (8 vs 12). Let's re-examine the balancing of the reduction half-reaction in basic medium carefully.

Reduction half: $\text{MnO}_4^-\text{(aq)} \rightarrow \text{MnO}_2\text{(s)}$ (Basic)

Difference in O atoms = $4 - 2 = 2$ (Left has more O). Add 2 $\text{H}_2\text{O}$ to the side with *less* oxygen (right):
$\text{MnO}_4^-\text{(aq)} \rightarrow \text{MnO}_2\text{(s)} + 2\text{H}_2\text{O(l)}$ (O is balanced)

Balance H atoms (4 H on right). Add $4\text{OH}^-$ to the side with *no* H (left):
$\text{MnO}_4^-\text{(aq)} + 4\text{OH}^-\text{(aq)} \rightarrow \text{MnO}_2\text{(s)} + 2\text{H}_2\text{O(l)}$ (H is balanced)

Check charge: Left: $-1 + 4(-1) = -5$. Right: 0 + 0 = 0. Add 5e⁻ to the right to balance charge:
$\text{MnO}_4^-\text{(aq)} + 4\text{OH}^-\text{(aq)} \rightarrow \text{MnO}_2\text{(s)} + 2\text{H}_2\text{O(l)} + 5\text{e}^-$ (This half-reaction is now balanced in atoms and charge, and is the reduction half in basic medium).

Let's restart Step 6 with the correct balanced half-reactions:
Oxidation half: $2\text{I}^-\text{(aq)} \rightarrow \text{I}_2\text{(s)} + 2\text{e}^-$
Reduction half: $\text{MnO}_4^-\text{(aq)} + 4\text{OH}^-\text{(aq)} \rightarrow \text{MnO}_2\text{(s)} + 2\text{H}_2\text{O(l)} + 5\text{e}^-$

Multiply oxidation by 5 (for 10e⁻) and reduction by 2 (for 10e⁻):
$10\text{I}^-\text{(aq)} \rightarrow 5\text{I}_2\text{(s)} + 10\text{e}^-$
$2\text{MnO}_4^-\text{(aq)} + 8\text{OH}^-\text{(aq)} \rightarrow 2\text{MnO}_2\text{(s)} + 4\text{H}_2\text{O(l)} + 10\text{e}^-$

Add the multiplied half-reactions:
$10\text{I}^-\text{(aq)} + 2\text{MnO}_4^-\text{(aq)} + 8\text{OH}^-\text{(aq)} \rightarrow 5\text{I}_2\text{(s)} + 2\text{MnO}_2\text{(s)} + 4\text{H}_2\text{O(l)}$

Final check:
Left: 10 I, 2 Mn, 8 O, 8 H. Charge: $-10 + 2(-1) + 8(-1) = -10 - 2 - 8 = -20$. Wait, something is still wrong with charge balance.
Let's recheck charge on multiplied reduction half: $2\text{MnO}_4^-\text{(aq)} + 8\text{OH}^-\text{(aq)} \rightarrow 2\text{MnO}_2\text{(s)} + 4\text{H}_2\text{O(l)} + 10\text{e}^-$. Charge left: $2(-1) + 8(-1) = -2 - 8 = -10$. Charge right: $2(0) + 4(0) + 10(-1) = -10$. Charge is balanced for the multiplied half-reaction.

Adding the reactions:
$10\text{I}^-\text{(aq)} + 2\text{MnO}_4^-\text{(aq)} + 8\text{OH}^-\text{(aq)} \rightarrow 5\text{I}_2\text{(s)} + 2\text{MnO}_2\text{(s)} + 4\text{H}_2\text{O(l)}$

Total charge Left: $-10 + (-2) + (-8) = -20$. Something is very wrong. Let's check the original species charges again.

Reactants: $\text{MnO}_4^-$ (-1), $\text{I}^-$ (-1). Products: $\text{MnO}_2$ (0), $\text{BrO}_3^-$ (-1). This was for the example with Br, not I.

Let's re-read Problem 7.10 carefully. Permanganate(VII) ion, $\text{MnO}_4^-$ (Charge -1) + iodide ion, $\text{I}^-$ (Charge -1) in basic solution gives manganese (IV) oxide, $\text{MnO}_2$ (Charge 0) and molecular iodine ($\text{I}_2$) (Charge 0). Yes, that matches the skeletal equation.

Let's re-do Step 5 charge balancing:
Oxidation half: $2\text{I}^-\text{(aq)} \rightarrow \text{I}_2\text{(s)}$. Charge left: $-2$. Charge right: 0. Add 2e⁻ to right: $2\text{I}^-\text{(aq)} \rightarrow \text{I}_2\text{(s)} + 2\text{e}^-$ (Charge: -2 on both sides). Correct.
Reduction half: $\text{MnO}_4^-\text{(aq)} + 2\text{H}_2\text{O(l)} \rightarrow \text{MnO}_2\text{(s)} + 4\text{OH}^-\text{(aq)}$. Charge left: $-1 + 0 = -1$. Charge right: $0 + (-4) = -4$. Need to add electrons to the left. Add 3e⁻ to left: $\text{MnO}_4^-\text{(aq)} + 2\text{H}_2\text{O(l)} + 3\text{e}^- \rightarrow \text{MnO}_2\text{(s)} + 4\text{OH}^-\text{(aq)}$ (Charge: -1 + 0 + (-3) = -4 on left, -4 on right). Correct.

Steps 6 and 7 (multiplication and addition):
Multiply oxidation half by 3: $6\text{I}^-\text{(aq)} \rightarrow 3\text{I}_2\text{(s)} + 6\text{e}^-$
Multiply reduction half by 2: $2\text{MnO}_4^-\text{(aq)} + 4\text{H}_2\text{O(l)} + 6\text{e}^- \rightarrow 2\text{MnO}_2\text{(s)} + 8\text{OH}^-\text{(aq)}$

Add the half-reactions:
$6\text{I}^-\text{(aq)} + 2\text{MnO}_4^-\text{(aq)} + 4\text{H}_2\text{O(l)} \rightarrow 3\text{I}_2\text{(s)} + 2\text{MnO}_2\text{(s)} + 8\text{OH}^-\text{(aq)}$

Final check:
Left: I: 6, Mn: 2, O: $2 \times 4 + 4 = 12$, H: $4 \times 2 = 8$. Charge: $-6 + 2(-1) + 0 = -8$.
Right: I: $3 \times 2 = 6$, Mn: 2, O: $2 \times 2 + 8 = 12$, H: 8. Charge: $3(0) + 2(0) + 8(-1) = -8$.

Atoms and charge are balanced. The balanced ionic equation is $6\text{I}^-\text{(aq)} + 2\text{MnO}_4^-\text{(aq)} + 4\text{H}_2\text{O(l)} \rightarrow 3\text{I}_2\text{(s)} + 2\text{MnO}_2\text{(s)} + 8\text{OH}^-\text{(aq)}$.