Redox reactions can be written as two half-reactions, focusing on the gain or loss of electrons by one of the chemical substances. one half-reaction shows the oxidation while the other shows the reduction. when the two half-reactions are combined, the overall reaction is obtained.

K2Cr2O7 + HI + HClO4  → Cr(ClO4)3  +  KClO4  + I2 + H2O

                                          ↓

8HClO4 + K2Cr2O7 + 6HI  →  3I₂ +  2Cr(ClO4)3 + 2KClO4  +  7H2O

Explanation:

I put and example of what you said.

Potassium dichromate and iodide acid react with perchloric acid to generate chromium perchlorate, potassium chlorate, iodine and water.

First of all think all the oxidation number of each element. One ox. number will increase and the other decrease, so those will be our half reactions (one of reduction, the other of oxidation).

In the case above, I in HI acts with -1 and I2 has 0 (all elements in ground state has 0 as oxidation number). (Increase) - Oxidation

Cr in K2Cr2O7 acts with +6, in Cr(ClO4)3 is 3+ (Decrease) -  Reduction

So, the first half reaction is:

2I⁻  →  I₂ + 2e⁻           (OXIDATION)

I have to put 2 iodides to ballance, so the total charge is 2-. It has to release 2 electrons.

Cr2O7²⁻  →  2Cr³⁺

In products side, I have to add 2 chromes but we don't have the charges ballanced. At the main equation, we have acids so this redox occurs in an acidic medium. In the acidic medium we add water, the same as oxygen we have, so:

Cr2O7²⁻  →  2Cr³⁺  +  7H2O

and in reactant side, we add protons the same as hydrogen, we have, in this case like this

14H⁺  +  Cr2O7²⁻  →  2Cr³⁺  +  7H2O

finally, we add the electrons. Chrome to decrease +6 to +3 had to lose 3 electrons but, we have 2 Cr, so 6 in total. These are final the 2 half reaction

14H⁺  +  Cr2O7²⁻  + 6e-  →  2Cr³⁺  +  7H2O  (REDUCTION)

2I⁻  →  I₂ + 2e⁻ (OXIDATION)

Electrons are not ballanced, we have to multiply by a minimum common multiple. For 2 and 6, this number is 12 so:

(14H⁺  +  Cr2O7²⁻  + 6e-  →  2Cr³⁺  +  7H2O) .2

(2I⁻  →  I₂ + 2e⁻) .6

Afterwards, we can sum the reactions:

28H⁺ + 2Cr2O7²⁻ + 12e- + 12I⁻  →  6I₂ + 12e⁻ + 4Cr³⁺  +  14H2O

As we have 12e- in both sides, we cancel them

28H⁺ + 2Cr2O7²⁻ + 12I⁻  →  6I₂ +  4Cr³⁺  +  14H2O (still balanced)

Look that all the stoichiometry is even, so we can /2.

14H⁺ + Cr2O7²⁻ + 6I⁻  →  3I₂ +  2Cr³⁺  +  7H2O

So the final reaction is:

8HClO4 + K2Cr2O7 + 6HI  →  3I₂ +  2Cr(ClO4)3 + 2KClO4  +  7H2O

We have in total 14H+, so 6 protons are for HI and 8 for the HClO4.

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