balance the following redox reaction by ion electron method

Cl2O7(g) + H2O2(aq) - ClO2-(aq) + O2(g)   (in basic medium)

balance the following redox reaction by oxidation number method

Cr2O7 2-(aq) + SO2(g) - Cr3+(aq) +SO4 2-(aq)   (in acidic medium)

I . Divide the equation in two halves.

Cr₂O₇^2-→Cr³⁺  (reduction half reaction)

SO₂→SO₄^2-  (OXIDATION HALF REACTION)

II. Balance the atoms other than O and H 

Cr₂O₇^2-→2Cr³⁺  

SO₂→SO₄^2-  

III. Then O and H are balanced using H and H ₂ O

Cr₂O₇^2-+14H⁺→2Cr³⁺  +7H ₂O 

SO₂  + 2H₂O→SO₄^2-  +4H⁺  

IV. Electrons need to be added to balance the charge.

Cr₂O₇^2-+14H⁺  +6e^- →2Cr³⁺  +7H ₂O ^{.........................(1)}

SO₂  + 2H₂O→SO₄^2-  +4H⁺ +2e^{- ...............................(2)}

V. Loss and gain of electrons need to be made equal:

For this, we need to multiply equation (2) by 3

VI. Adding both equations

Cr₂O₇^2-+2H⁺    +3SO₂ →2Cr³⁺  +7H ₂O 

• Similarly, you can do for the second equation by using H+ for balancing the number of H in place of OH^-ion. let me know in case of any difficulty.