Question

Using the rules for balancing redox equations either from your textbook or from lecture, balance

KIO_3(aq) + KI_(aq) + H₂SO_4(aq) ----->  I_2(aq) + K₂SO₄

I_2(aq) + Na₂S₂O_3(aq) --------->   NaI_(aq) + Na₂S₄O_6(aq)

. Note that these are molecular equations that use neutral chemical formulas and that the reactions occur in an acidic medium as indicated by the presence of sulfuric acid. It may be easier to “remove” the spectator ions and write the net ionic equation before you balance (Hint: Na⁺ and K⁺ are almost always spectator cations). Once you have a balanced ionic equation, you can “replace” the spectator ions to have a balanced molecular equation.

Answer 1

2 NaI & Naz Su Oo Iz Cag.) + 2 Na₂S₂O3 + KIO₃ + 5 KI + 3 H₂ Soy Redox halves Oxd" - 2k - Red" → 2 K10 - 3 I + 3 K2504 + 3H₂O the- To floe-

Balancing redox halnes - 2 KI + H₂SO4 -> Iz tk₂S0y the IR 2 KIO₃ + H₂SO4 + 100 - I & K₂ 504 4 Balancing the charge and oxygen atom una au qeen 0 2 KI + H₂Soy & Iz & K₂SO4 + 2e +240 R 2 KIO3 + H₂SO4 + loe & 10H I, tk, sou + 6H₂O

o R Balancing election gain and loss equivalent multiplied oxidation half by 5 and adding both halves 10kI + 5 H₂SO4 - 5I + 5K₂S0y + loe + 1048 2 KIO3 + H₂SO4 loe + 10 H - 2 + k₂ soy + 6H₂O TOKI +6 H₂ SO 4 + 2 KIO 6 I2 + 6 K₂SO4 + 6H₂O simplyfying the equation, 5KI + 3 H₂SO4 +KIO 3 I2 + 3 K₂SO4 + 3H₂O

The second reaction can be balanced by balancing the sodium atom in the reactant and the product.

The first one is a redox reaction in acidic medium. It is solved in following steps

1. Divide the recation into oxidation and reduction halves.

2. Balance the atoms in redox halves except O and H.

3.balance the electrons using H+

Balance the oxygen atoms.

4. Balance electron gain and loss by multiplying oxidation half by 5 and add both of the halves.

5 simplify the equation.