Balance these redox reaction 1 I-T ClO-3 equal I-3 TCL 2 SO32- t Lui (OH) −3 equal Lui t SO2 4

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Reaction 1: I^− + ClO₃^− → I₃^− + Cl^− (acidic medium) Step 1: Identify oxidation and reduction half-reactions. Oxidation: 3I^− → I₃^− Reduction: ClO₃^− → Cl^− Step 2: Balance oxidation half-reaction for I. 3I^− → I₃^− (already balanced) Step 3: Balance charge in oxidation half-reaction. Left: charge −3, right: charge −1 Add 2e^− to right: 3I^− → I₃^− + 2e^− Step 4: Balance reduction half-reaction for Cl. ClO₃^− → Cl^− Step 5: Balance O atoms with H₂O. ClO₃^− → Cl^− + 3H₂O Step 6: Balance H atoms with H^+. ClO₃^− + 6H^+ → Cl^− + 3H₂O Step 7: Balance charge in reduction half-reaction. Left: charge ( −1 + 6 ) = +5, right: charge −1 Add 6e^− to left: ClO₃^− + 6H^+ + 6e^− → Cl^− + 3H₂O Step 8: Equalize electrons (LCM of 2 and 6 is 6). Multiply oxidation by 3: 3(3I^− → I₃^− + 2e^−) ⇒ 9I^− → 3I₃^− + 6e^− Reduction: ClO₃^− + 6H^+ + 6e^− → Cl^− + 3H₂O Step 9: Add half-reactions. 9I^− + ClO₃^− + 6H^+ + 6e^− → 3I₃^− + Cl^− + 3H₂O + 6e^− Step 10: Simplify (cancel 6e^−). 9I^− + ClO₃^− + 6H^+ → 3I₃^− + Cl^− + 3H₂O 9I^− + ClO₃^− + 6H^+ → 3I₃^− + Cl^− + 3H₂O Reaction 2: SO₃²⁻ + Lu(OH)₃ → Lu + SO₄²⁻ (basic medium) Step 1: Identify oxidation and reduction half-reactions. Oxidation: SO₃²⁻ → SO₄²⁻ Reduction: Lu(OH)₃ → Lu Step 2: Balance oxidation half-reaction for S. SO₃²⁻ → SO₄²⁻ Step 3: Balance O atoms with OH^−. SO₃²⁻ + 2OH^− → SO₄²⁻ + H₂O Step 4: Balance charge in oxidation half-reaction. Left: charge ( −2 −2 ) = −4, right: charge ( −2 −0 ) = −2 Add 2e^− to right: SO₃²⁻ + 2OH^− → SO₄²⁻ + H₂O + 2e^− Step 5: Balance reduction half-reaction for Lu. Lu(OH)₃ → Lu Step 6: Balance O atoms with OH^−. Lu(OH)₃ → Lu + 3OH^− Step 7: Balance charge in reduction half-reaction. Left: charge −3, right: charge ( 0 −3 ) = −3 (balanced for atoms, now electrons) Lu from +3 to 0, add 3e^− to left: Lu(OH)₃ + 3e^− → Lu + 3OH^− Step 8: Equalize electrons (LCM of 2 and 3 is 6). Multiply oxidation by 3: 3(SO₃²⁻ + 2OH^− → SO₄²⁻ + H₂O + 2e^−) ⇒ 3SO₃²⁻ + 6OH^− → 3SO₄²⁻ + 3H₂O + 6e^− Multiply reduction by 2: 2(Lu(OH)₃ + 3e^− → Lu + 3OH^−) ⇒ 2Lu(OH)₃ + 6e^− → 2Lu + 6OH^− Step 9: Add half-reactions. 3SO₃²⁻ + 6OH^− + 2Lu(OH)₃ + 6e^− → 3SO₄²⁻ + 3H₂O + 2Lu + 6OH^− + 6e^− Step 10: Simplify (cancel 6OH^− and 6e^−). 3SO₃²⁻ + 2Lu(OH)₃ → 3SO₄²⁻ + 3H₂O + 2Lu 3SO₃²⁻ + 2Lu(OH)₃ → 3SO₄²⁻ + 3H₂O + 2Lu

Balance these redox reaction 1 I-T ClO-3 equal I-3 TCL 2 SO32- t Lui (OH) -3 equal Lui t SO2 4

Answer

Step-by-step solution

Identify oxidation and reduction half-reactions.

Balance oxidation half-reaction for I.

Balance charge in oxidation half-reaction.

Balance reduction half-reaction for Cl.

Balance O atoms with $H_2O$ .

Balance H atoms with $H^+$ .

Balance charge in reduction half-reaction.

Equalize electrons (LCM of 2 and 6 is 6).

Add half-reactions.

Simplify (cancel $6e^-$ ).

Identify oxidation and reduction half-reactions.

Balance oxidation half-reaction for S.

Balance O atoms with $OH^-$ .

Balance charge in oxidation half-reaction.

Balance reduction half-reaction for Lu.

Balance O atoms with $OH^-$ .

Balance charge in reduction half-reaction.

Equalize electrons (LCM of 2 and 3 is 6).

Add half-reactions.

Simplify (cancel $6OH^-$ and $6e^-$ ).

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Balance these redox reaction 1 I-T ClO-3 equal I-3 TCL 2 SO32- t Lui (OH) -3 equal Lui t SO2 4

Answer

Step-by-step solution

Identify oxidation and reduction half-reactions.

Balance oxidation half-reaction for I.

Balance charge in oxidation half-reaction.

Balance reduction half-reaction for Cl.

Balance O atoms with H2OH_2OH2​O.

Balance H atoms with H+H^+H+.

Balance charge in reduction half-reaction.

Equalize electrons (LCM of 2 and 6 is 6).

Add half-reactions.

Simplify (cancel 6e−6e^-6e−).

Identify oxidation and reduction half-reactions.

Balance oxidation half-reaction for S.

Balance O atoms with OH−OH^-OH−.

Balance charge in oxidation half-reaction.

Balance reduction half-reaction for Lu.

Balance O atoms with OH−OH^-OH−.

Balance charge in reduction half-reaction.

Equalize electrons (LCM of 2 and 3 is 6).

Add half-reactions.

Simplify (cancel 6OH−6OH^-6OH− and 6e−6e^-6e−).

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More Chemistry Questions

Balance O atoms with $H_2O$ .

Balance H atoms with $H^+$ .

Simplify (cancel $6e^-$ ).

Balance O atoms with $OH^-$ .

Balance O atoms with $OH^-$ .

Simplify (cancel $6OH^-$ and $6e^-$ ).