Worksheet for balancing redox reactions in acidic and basic solutions.

To balance these redox reactions, we use the half-reaction method.

General Steps for Acidic Solutions:
1. Split the reaction into two half-reactions (Oxidation and Reduction).
2. Balance all atoms except Hydrogen (H) and Oxygen (O).
3. Balance Oxygen (O) by adding water ($H_2O$).
4. Balance Hydrogen (H) by adding hydrogen ions ($H^+$).
5. Balance the charge by adding electrons ($e^-$).
6. Multiply the half-reactions so the number of electrons lost equals the number gained.
7. Add them together and simplify.

General Steps for Basic Solutions:
Follow the steps for acidic solutions first. Then, to convert to basic:
1. Add hydroxide ions ($OH^-$) to both sides to neutralize the $H^+$.
2. Combine $H^+$ and $OH^-$ to make water ($H_2O$).
3. Simplify the water molecules.

Here are the solutions:

(1) $ClO_3^- + Cl^- \rightarrow Cl_2 + ClO_2$
* Oxidation: $Cl^- \rightarrow Cl_2$
* Reduction: $ClO_3^- \rightarrow ClO_2$
* Balanced: $2Cl^- + 2ClO_3^- \rightarrow Cl_2 + 2ClO_2$

(2) $P + Cu^{2+} \rightarrow H_2PO_4^- + Cu$
* Oxidation: $P \rightarrow H_2PO_4^-$
* Reduction: $Cu^{2+} \rightarrow Cu$
* Balanced: $P + 4H_2O + 5Cu^{2+} \rightarrow H_2PO_4^- + 8H^+ + 5Cu$

(3) $NO_2 \rightarrow NO_3^- + NO$
* Oxidation: $NO_2 \rightarrow NO_3^-$
* Reduction: $NO_2 \rightarrow NO$
* Balanced: $2NO_2 + H_2O \rightarrow NO_3^- + NO + 2H^+$

(4) $H_2O_2 + Cr_2O_7^{2-} \rightarrow O_2 + Cr^{3+}$
* Oxidation: $H_2O_2 \rightarrow O_2$
* Reduction: $Cr_2O_7^{2-} \rightarrow Cr^{3+}$
* Balanced: $3H_2O_2 + 8H^+ + Cr_2O_7^{2-} \rightarrow 3O_2 + 2Cr^{3+} + 7H_2O$

(5) $TeO_3^{2-} + N_2O_4 \rightarrow Te + NO_3^-$
* Oxidation: $N_2O_4 \rightarrow NO_3^-$
* Reduction: $TeO_3^{2-} \rightarrow Te$
* Balanced: $TeO_3^{2-} + 2N_2O_4 + 3H_2O \rightarrow Te + 4NO_3^- + 6H^+$

(6) $N_2H_4 + Cu(OH)_2 \rightarrow N_2 + Cu$ (Basic)
* Balanced: $N_2H_4 + 2Cu(OH)_2 \rightarrow N_2 + 2Cu + 4H_2O$

(7) $Cu(NH_3)_4^{2+} + S_2O_4^{2-} \rightarrow SO_3^{2-} + Cu + NH_3$ (Basic)
* Balanced: $2Cu(NH_3)_4^{2+} + S_2O_4^{2-} + 4OH^- \rightarrow 2SO_3^{2-} + 2Cu + 8NH_3 + 2H_2O$

(8) $Pb^{2+} + IO_3^- \rightarrow PbO_2 + I_2$ (Basic)
* Balanced: $2Pb^{2+} + 2IO_3^- + 4H_2O + 4OH^- \rightarrow 2PbO_2 + I_2 + 4H_2O$
* *Simplified:* $2Pb^{2+} + 2IO_3^- + 4OH^- \rightarrow 2PbO_2 + I_2 + 2H_2O$

(9) $CH_3OH + MnO_4^- \rightarrow MnO_4^{2-} + CO_3^{2-}$ (Basic)
* Balanced: $CH_3OH + 6MnO_4^- + 8OH^- \rightarrow 6MnO_4^{2-} + CO_3^{2-} + 6H_2O$

(10) $MnO_4^- + SO_3^{2-} \rightarrow SO_4^{2-} + MnO_2$ (Basic)
* Balanced: $2MnO_4^- + 3SO_3^{2-} + H_2O \rightarrow 3SO_4^{2-} + 2MnO_2 + 2OH^-$

Final Answer:
(1) $2Cl^- + 2ClO_3^- \rightarrow Cl_2 + 2ClO_2$
(2) $P + 4H_2O + 5Cu^{2+} \rightarrow H_2PO_4^- + 8H^+ + 5Cu$
(3) $2NO_2 + H_2O \rightarrow NO_3^- + NO + 2H^+$
(4) $3H_2O_2 + 8H^+ + Cr_2O_7^{2-} \rightarrow 3O_2 + 2Cr^{3+} + 7H_2O$
(5) $TeO_3^{2-} + 2N_2O_4 + 3H_2O \rightarrow Te + 4NO_3^- + 6H^+$
(6) $N_2H_4 + 2Cu(OH)_2 \rightarrow N_2 + 2Cu + 4H_2O$
(7) $2Cu(NH_3)_4^{2+} + S_2O_4^{2-} + 4OH^- \rightarrow 2SO_3^{2-} + 2Cu + 8NH_3 + 2H_2O$
(8) $2Pb^{2+} + 2IO_3^- + 4OH^- \rightarrow 2PbO_2 + I_2 + 2H_2O$
(9) $CH_3OH + 6MnO_4^- + 8OH^- \rightarrow 6MnO_4^{2-} + CO_3^{2-} + 6H_2O$
(10) $2MnO_4^- + 3SO_3^{2-} + H_2O \rightarrow 3SO_4^{2-} + 2MnO_2 + 2OH^-$