acidified k2cr2o7 solution turns green when sodium sulphite is added to it . explain

 Potassium Dichromate (K2Cr2O7) is Oxidizer whilst Soodium Sulphite (Na2SO3) acts as Reductants.
I start from half-reaction about OXIDIZER : I balance Oxidation Number of atom changing it, I refer to Chromium atoms
K2Cr2O7 + 6e ---> 2 K+ + 2 Cr+++ + 7 O--
now I add hydrogen ions to convert oxygen to water molecules
K2Cr2O7 + 6e + 14 H+ ---> 2 K+ + 2 Cr+++ + 7 H2O
and I finished since electrical charge are balanced already.
Now, I come to half-reaction about REDUCTANT : I balance Oxidation Number of atom changing it, I refer to Sulphur atoms
Na2SO3 ---> 2 Na+ + SO4-- + 2e
now I add water molecules to balance oxygen atoms
Na2SO3 + H2O ---> 2 Na+ + SO4-- + 2e
so I add hydrogen ions to balance it
Na2SO3 + H2O ---> 2 Na+ + SO4-- + 2 H+ + 2e
and I finished one more time since electrical charge are balanced already.
I have to sum two writings looking to erase electrons : how to do it?
I multiply for 3 (6e / 2e = 3) the second writing and I sum it to obtain the final formula

K2Cr2O7 + 4 H2SO4 + 3 Na2SO3 --> 
---> K2SO4 + Cr2(SO4)3 + 3 Na2SO4 + 4 H2O

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