Q.Balance the following ionic equations

The key idea for balancing these ionic equations is Oxidation-Reduction (Redox) balancing, specifically using the half-reaction method in acidic medium. This involves separating the overall reaction into oxidation and reduction half-reactions, balancing atoms (other than O and H), then balancing O with H$_2$O, H with H$^+$, and finally charges with electrons ($e^-$). The electrons are then equalised and the half-reactions are added.

Here are the balanced equations:

(i) Cr$_2$O$_7^{2-}$ + H$^+$ + I$^-$ → Cr$^{3+}$ + I$_2$ + H$_2$O

1. Identify oxidation/reduction: Cr goes from $+6$ to $+3$ (reduction). I goes from $-1$ to $0$ (oxidation).
2. Balance half-reactions:
   • Reduction: Cr$_2$O$_7^{2-}$ + 14H$^+$ + 6e$^-$ → 2Cr$^{3+}$ + 7H$_2$O
   • Oxidation: 2I$^-$ → I$_2$ + 2e$^-$
3. Equalise electrons and add: Multiply oxidation half-reaction by 3. …

Balancing redox reactions is a fundamental skill in chemistry, crucial for understanding stoichiometry and reaction mechanisms. The most common and reliable method for balancing ionic redox equations is the ion-electron method (also known as the half-reaction method). This method systematically accounts for the conservation of mass (atoms) and charge by splitting the overall reaction into two half-reactions: one for oxidation and one for reduction.

General Approach for Balancing Redox Reactions in Acidic Medium

1. Assign Oxidation States: Determine the oxidation states of all elements to identify which species are oxidised and which are reduced.
2. Split into Half-Reactions: Write separate unbalanced half-reactions for oxidation and reduction.
3. Balance Atoms (excluding O and H): Balance all atoms other than oxygen and hydrogen in each half-reaction.
4. Balance Oxygen Atoms: Add water molecules ($H_2O$) to the side deficient in oxygen.
5. Balance Hydrogen Atoms: Add hydrogen ions ($H^+$) to the side deficient in hydrogen (since the reactions are in acidic medium).
6. Balance Charges: Add electrons ($e^-$) to the more positive side of each half-reaction to balance the charges.
7. Equalise Electrons: Multiply each half-reaction by an appropriate integer so that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction.
8. Combine Half-Reactions: Add the two balanced half-reactions and cancel out any common species (electrons, $H^+$, $H_2O$) appearing on both sides.
9. Verify: Double-check that all atoms and charges are balanced.

(i) Cr2O7^2- + H^+ + I^- → Cr^3+ + I2 + H2O

1. Identify Oxidation States:

   • In $Cr_2O_7^{2-}$, Cr is in the $+6$ oxidation state ($2x + 7(-2) = -2 \Rightarrow 2x = 12 \Rightarrow x = +6$).
   • In $I^-$, I is in the $-1$ oxidation state.
   • In $Cr^{3+}$, Cr is in the $+3$ oxidation state.
   • In $I_2$, I is in the $0$ oxidation state.
   • Chromium is reduced from $+6$ to $+3$. Iodide is oxidised from $-1$ to $0$.

2. Write Half-Reactions:

   • Reduction: $Cr_2O_7^{2-} \rightarrow Cr^{3+}$
   • Oxidation: $I^- \rightarrow I_2$

3. Balance Atoms (excluding O and H):

   • Reduction: $Cr_2O_7^{2-} \rightarrow 2Cr^{3+}$ (Balance Cr atoms)
   • Oxidation: $2I^- \rightarrow I_2$ (Balance I atoms)

4. Balance Oxygen Atoms (using $H_2O$):

   • Reduction: $Cr_2O_7^{2-} \rightarrow 2Cr^{3+} + 7H_2O$ (Add $7H_2O$ to the right to balance 7 oxygen atoms on the left)
   • Oxidation: $2I^- \rightarrow I_2$ (No oxygen atoms)

5. Balance Hydrogen Atoms (using $H^+$):

   • Reduction: $14H^+ + Cr_2O_7^{2-} \rightarrow 2Cr^{3+} + 7H_2O$ (Add $14H^+$ to the left to balance 14 hydrogen atoms in $7H_2O$ on the right)
   • Oxidation: $2I^- \rightarrow I_2$ (No hydrogen atoms)

6. Balance Charges (using $e^-$):

   • Reduction: $14H^+ + Cr_2O_7^{2-} \rightarrow 2Cr^{3+} + 7H_2O$
     • Left side charge: $14(+1) + (-2) = +12$
     • Right side charge: $2(+3) = +6$
     • Add $6e^-$ to the left side to balance the charge: $6e^- + 14H^+ + Cr_2O_7^{2-} \rightarrow 2Cr^{3+} + 7H_2O$ (Net charge on both sides is $+6$)
   • Oxidation: $2I^- \rightarrow I_2$
     • Left side charge: $2(-1) = -2$
     • Right side charge: $0$
     • Add $2e^-$ to the right side: $2I^- \rightarrow I_2 + 2e^-$ (Net charge on both sides is $-2$)

7. Equalise Electrons:

   • The reduction half-reaction involves $6e^-$.
   • The oxidation half-reaction involves $2e^-$.
   • Multiply the oxidation half-reaction by 3 to get $6e^-$: $3 \times (2I^- \rightarrow I_2 + 2e^-) \Rightarrow 6I^- \rightarrow 3I_2 + 6e^-$

8. Combine Half-Reactions:

   • $6e^- + 14H^+ + Cr_2O_7^{2-} \rightarrow 2Cr^{3+} + 7H_2O$
   • $6I^- \rightarrow 3I_2 + 6e^-$
   • Adding them: $14H^+ + Cr_2O_7^{2-} + 6I^- + 6e^- \rightarrow 2Cr^{3+} + 3I_2 + 7H_2O + 6e^-$
   • Cancel $6e^-$ from both sides: $14H^+ + Cr_2O_7^{2-} + 6I^- \rightarrow 2Cr^{3+} + 3I_2 + 7H_2O$

9. Verify:

   • Atoms: Cr (2), O (7), H (14), I (6) are balanced on both sides. …