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What are the redoxpairs

That is not the way how to do

One redoxpair is CrO₄ 2- and Cr₃+
The second one is S₂- and SO₄ 2-.

In acidic condition the oxidation needs water to produce H+
The reduction consumes H+ and produces water.

In this case for Reduction

CrO₄ 2- + 8 H+ => Cr 3+ + 4 H₂O

Balance Charge gives

CrO₄ 2- + 8 H+ +3 e- => Cr 3+ + 4 H₂O

Oxidation S₂- + 4 H₂O => SO₄ 2- + 8 H+


Balance Charge gives

S₂- + 4 H₂O => SO₄ 2- + 8 H+ + 8 e-

KGV of elektrones is 24

You get

8 CrO₄ 2- + 64 H+ + 24 e- => 8 Cr 3+ + 32 H₂O

3 S₂- + 12 H₂O =>3 SO₄ 2- + 24 H+ + 24 e-


Addition

8 CrO₄ 2- + 40 H+ + 3 S 2- => 8 Cr 3+ + 3 SO₄ 2- +20 H₂O

That is the basic reaction.

Now add of 2 Cr₃+


8 CrO₄ 2- + 40 H+ + Cr₂S₃ => 10 Cr 3+ + 3 SO₄ 2- +20 H₂O

add K+ and SO₄ 2-


8 K₂CrO₄ + 20 H₂SO₄ + Cr₂S₃ => 5 Cr₂ (SO₄)₃ + 8 K₂SO₄ + 20 H₂O

Add 2 more K₂CrO₄ + one H₂SO₄ to get dichromate.


10 K₂CrO₄ + 21 H₂SO₄ + Cr₂S₃ => K₂Cr₂O₇ + 5 Cr₂ (SO₄)₃ + 9 K₂SO₄ + 21 H₂O

finished