10 KBr + K2Cr2O7 + 7 H2SO4 → Cr2SO4 + 6 K2SO4 + 7 H2O + 5 Br2
This is an oxidation-reduction (redox) reaction:
10 Br-I - 10 e- → 10 Br0 (oxidation)
SVI + 4 e- → SII (reduction)
2 CrVI + 6 e- → 2 CrIII (reduction)
KBr is a reducing agent, H2SO4 is an oxidizing agent, K2Cr2O7 is an oxidizing agent.