Write a balanced ionic equation to represent the reaction between sulfur dioxide, ##SO_2##, and the dichromate ion, ##Cr_2O_7^(2-)## in acidic solution to yield the sulfate ion and chromium (III) ion, ##Cr^(3+)##?

Start by assigning to all the atoms that take part in the reaction.

##stackrel(color(blue)(+4))(S) stackrel(color(blue)(-2))(O_2) + stackrel(color(blue)(+6))(Cr_2) stackrel(color(blue)(-2))(O_7^(2-)) -> stackrel(color(blue)(+6))(S) stackrel(color(blue)(-2))(O_4^(2-)) + stackrel(color(blue)(+3))(Cr^(3+))##

Notice that oxidation state of sulfur goes from +4 on the reactants' side, to +6 on the products' side, which means that it is being oxidized.

At the same time, 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.

The oxidation and reduction half-reactions will look like this

• Oxidation half-reaction

##stackrel(color(blue)(+4))(S) O_2 -> stackrel(color(blue)(+6))(S) O_4^(2-) + 2e^(-)##

Balance the oxygen and hydrogen atoms by adding water molecules to the side that needs oxygen, and protons, ##H^(+)##, to the side that needs hydrogen.

##2H_2O + stackrel(color(blue)(+4))(S) O_2 -> stackrel(color(blue)(+6))(S) O_4^(2-) + 2e^(-) + 4H^(+)##

• Reduction half-reaction

##stackrel(color(blue)(+6))(Cr_2) O_7^(2-) + 3e^(-) -> stackrel(color(blue)(+3))(Cr^(3+))##

Notice that you have two cxhromium atoms on the reactants' side, and only one on the products' side. Multiply the product by 2 to get

##stackrel(color(blue)(+6))(Cr_2) O_7^(2-) + 6e^(-) -> 2stackrel(color(blue)(+3))(Cr^(3+))##

Notice that two chromium atoms lose a total of 6 electrons to go from an oxidation state of +6 to an aoxidation state of +3 per atom.

Once again, balance the oxygen and hydrogens atoms by adding water and protons.

##14H^(+) + stackrel(color(blue)(+6))(Cr_2) O_7^(2-) + 6e^(-) -> 2stackrel(color(blue)(+3))(Cr^(3+)) + 7H_2O##

In , 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 to balance the number of transferred electrons.

##{ (6H_2O + 3stackrel(color(blue)(+4))(S) O_2 -> 3stackrel(color(blue)(+6))(S) O_4^(2-) + 6e^(-) + 4H^(+)), (14H^(+) + stackrel(color(blue)(+6))(Cr_2) O_7^(2-) + 6e^(-) -> 2stackrel(color(blue)(+3))(Cr^(3+)) + 7H_2O) :}##

Add the two half-reactions together to get

##6H_2O + 3SO_2 + 14H^(+) + Cr_2O_7^(2-) + cancel(6e^(-)) -> 3SO_4^(2-) + 2Cr^(3+) + 4H^(+) + 7H_2O + cancel(6e^(-))##

The balanced net ionic equation will be

##H_2O_((l)) + 3SO_(2(g)) + Cr_2O_(7(aq))^(2-) -> 3SO_(4(aq))^(2-) + 2Cr_((aq))^(3+) + 2H_((aq))^(+)##