Start by assigning oxidation numbers to all the atoms that take part in the reaction
#stackrel(color(blue)(+1))(K_2) stackrel(color(blue)(+6))(Cr) stackrel(color(blue)(-2))(O_4) + stackrel(color(blue)(+1))(Na_2) stackrel(color(blue)(+4))(S) stackrel(color(blue)(-2))(O_3) + stackrel(color(blue)(+1))(H) stackrel(color(blue)(-1))(Cl) -> stackrel(color(blue)(+1))(K) stackrel(color(blue)(-1))(Cl) + stackrel(color(blue)(+1))(Na_2) stackrel(color(blue)(+6))(S) stackrel(color(blue)(-2))(O_4) + stackrel(color(blue)(+3))(Cr) stackrel(color(blue)(-1))(Cl_3) + stackrel(color(blue)(+1))(H_2) stackrel(color(blue)(-2))(O_2)#
Notice that the oxidation state of chromium goes from +6 on the reactants' side, to +3 on the products' side, which means that it is being reduced.
At the same time, the oxidation number of sulfur goes from +4 on the reactants' side, to +6 on the products' side, which means that it is being oxidized.
Since the reaction takes place in acidic solution, you write the net ionic half-reactions like this
#stackrel(color(blue)(+6))(Cr) O_4^(2-) + 3e^(-) -> stackrel(color(blue)(+3))(Cr^(3+))#
Balance the oxygen and hydrogen atoms by adding water molecules to the side of the equation that needs oxygen, and protons, #H^(+)#, to the side of the equation that needs hydrogens.
#8H^(+) + stackrel(color(blue)(+6))(Cr) O_4^(2-) + 3e^(-) -> stackrel(color(blue)(+3))(Cr^(3+)) + 4H_2O#
#stackrel(color(blue)(+4))(S) O_3^(2-) ->stackrel(color(blue)(+6))(S) O_4^(2-) + 2e^(-)#
Once again, use water molecules and protons to balance the oxygen and hydrogen atoms.
#H_2O + stackrel(color(blue)(+4))(S) O_3^(2-) ->stackrel(color(blue)(+6))(S) O_4^(2-) + 2e^(-) + 2H^(+)#
In any redox reaction, the number of electrons gained in the reduction half-reaction must be equal to the number of electrons lost in the oxidation half-reaction.
Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2 to get
#{ (16H^(+) + 2stackrel(color(blue)(+6))(Cr) O_4^(2-) + 6e^(-) -> 2stackrel(color(blue)(+3))(Cr^(3+)) + 8H_2O), (3H_2O + 3stackrel(color(blue)(+4))(S) O_3^(2-) -> 3stackrel(color(blue)(+6))(S) O_4^(2-) + 6e^(-) + 6H^(+)) :}#
Add the two half-reactions to get
#16H^(+) + 2CrO_4^(2-) + 3H_2O + 3SO_3^(2-) + cancel(6e^(-)) -> 2Cr^(3+) + 3SO_4^(2-) + cancel(6e^(-)) + 6H^(+) + 8H_2O#
The balanced chemical equation will look like this
#10H^(+) + 2CrO_4^(2-) + 3SO_3^(2-) -> 2Cr^(3+) + 3SO_4^(2-) + 5H_2O#
Add the spectator ions to get
#2K_2CrO_4 + 3Na_2SO_3 + 10HCl -> KCl + 3Na_2SO_4 + 2CrCl_3 + 5H_2O#
Finally, balance the potassium and chlorine atoms by multiplying #KCl# by 4
#2K_2CrO_4 + 3Na_2SO_3 + 10HCl -> 4KCl + 3Na_2SO_4 + 2CrCl_3 + 5H_2O#
