Waiting for answer This question has not been answered yet. You can hire a professional tutor to get the answer.

QUESTION

# Balancing redox equation? K_2CrO_4+Na_2SO_3+HCl-&gt;KCl+Na_2SO_4+CrCl_3+H_2O

Start by assigning 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

• Reduction half-reaction

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

• Oxidation half-reaction

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 , 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