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

Reaction 1: $I^- + ClO_3^- \to I_3^- + Cl^-$ (acidic medium)

Step 1: Identify oxidation and reduction half-reactions.
Oxidation: $3I^- \to I_3^-$
Reduction: $ClO_3^- \to Cl^-$

Step 2: Balance oxidation half-reaction for I.
$3I^- \to I_3^-$ (already balanced)

Step 3: Balance charge in oxidation half-reaction.
Left: charge $-3$, right: charge $-1$
Add $2e^-$ to right: $3I^- \to I_3^- + 2e^-$

Step 4: Balance reduction half-reaction for Cl.
$ClO_3^- \to Cl^-$

Step 5: Balance O atoms with $H_2O$.
$ClO_3^- \to Cl^- + 3H_2O$

Step 6: Balance H atoms with $H^+$.
$ClO_3^- + 6H^+ \to Cl^- + 3H_2O$

Step 7: Balance charge in reduction half-reaction.
Left: charge $( -1 + 6 ) = +5$, right: charge $-1$
Add $6e^-$ to left: $ClO_3^- + 6H^+ + 6e^- \to Cl^- + 3H_2O$

Step 8: Equalize electrons (LCM of 2 and 6 is 6).
Multiply oxidation by 3: $3(3I^- \to I_3^- + 2e^-) \implies 9I^- \to 3I_3^- + 6e^-$
Reduction: $ClO_3^- + 6H^+ + 6e^- \to Cl^- + 3H_2O$

Step 9: Add half-reactions.
$9I^- + ClO_3^- + 6H^+ + 6e^- \to 3I_3^- + Cl^- + 3H_2O + 6e^-$

Step 10: Simplify (cancel $6e^-$).
$9I^- + ClO_3^- + 6H^+ \to 3I_3^- + Cl^- + 3H_2O$

$9I^- + ClO_3^- + 6H^+ \to 3I_3^- + Cl^- + 3H_2O$

Reaction 2: $SO_3^{2-} + Lu(OH)_3 \to Lu + SO_4^{2-}$ (basic medium)

Step 1: Identify oxidation and reduction half-reactions.
Oxidation: $SO_3^{2-} \to SO_4^{2-}$
Reduction: $Lu(OH)_3 \to Lu$

Step 2: Balance oxidation half-reaction for S.
$SO_3^{2-} \to SO_4^{2-}$

Step 3: Balance O atoms with $OH^-$.
$SO_3^{2-} + 2OH^- \to SO_4^{2-} + H_2O$

Step 4: Balance charge in oxidation half-reaction.
Left: charge $( -2 -2 ) = -4$, right: charge $( -2 -0 ) = -2$
Add $2e^-$ to right: $SO_3^{2-} + 2OH^- \to SO_4^{2-} + H_2O + 2e^-$

Step 5: Balance reduction half-reaction for Lu.
$Lu(OH)_3 \to Lu$

Step 6: Balance O atoms with $OH^-$.
$Lu(OH)_3 \to Lu + 3OH^-$

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: $Lu(OH)_3 + 3e^- \to Lu + 3OH^-$

Step 8: Equalize electrons (LCM of 2 and 3 is 6).
Multiply oxidation by 3: $3(SO_3^{2-} + 2OH^- \to SO_4^{2-} + H_2O + 2e^-) \implies 3SO_3^{2-} + 6OH^- \to 3SO_4^{2-} + 3H_2O + 6e^-$
Multiply reduction by 2: $2(Lu(OH)_3 + 3e^- \to Lu + 3OH^-) \implies 2Lu(OH)_3 + 6e^- \to 2Lu + 6OH^-$

Step 9: Add half-reactions.
$3SO_3^{2-} + 6OH^- + 2Lu(OH)_3 + 6e^- \to 3SO_4^{2-} + 3H_2O + 2Lu + 6OH^- + 6e^-$

Step 10: Simplify (cancel $6OH^-$ and $6e^-$).
$3SO_3^{2-} + 2Lu(OH)_3 \to 3SO_4^{2-} + 3H_2O + 2Lu$

3SO_3^{2- + 2Lu(OH)_3 \to 3SO_4^{2-} + 3H_2O + 2Lu}