Question #4e8ae

1 Answer
May 6, 2015

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

#stackrel(color(blue)(+1))(H_2) stackrel(color(blue)(-2))(S_((s))) + stackrel(color(blue)(+5))(N) stackrel(color(blue)(-2))(O_(3(aq))^(-)) -> stackrel(color(blue)(0))(S_(8(s))) + stackrel(color(blue)(+2))(N) stackrel(color(blue)(-2))(O_((g)))#

Notice that sulfur goes from an oxidation state of -2 on the reactants' side, to an oxidation state of zero on the products' side #-># sulfur is being oxidized.

Likewise, nitrogen's oxidation state changes from +5 to +2 #-># nitrogen is being reduced.

The oxidation and reduction half-reactions will look like this

  • Oxidation half-reaction

#H_2 stackrel(color(blue)(-2))(S) -> stackrel(color(blue)(0))(S_8) + 2e^(-)#

Balance the sulfur atoms by multiplying the left side of the reaction by 8

#8H_2 stackrel(color(blue)(-2))(S) -> stackrel(color(blue)(0))(S_8) + 16e^(-)#

Balance the hydrogen atoms by adding 16 protons, #H^(+)#, to the products' side

#8H_2 stackrel(color(blue)(-2))(S) -> stackrel(color(blue)(0))(S_8) + 16e^(-) + 16H^(+)#

  • Reduction half-reaction

#stackrel(color(blue)(+5))(N) O_3^(-) + 3e^(-) -> stackrel(color(blue)(+2))(N) O#

Balance the oxygens by adding 2 water molecules to the products' side

#stackrel(color(blue)(+5))(N) O_3^(-) + 3e^(-) -> stackrel(color(blue)(+2))(N) O + 2H_2O#

Now balance the hydrogens by adding 4 protons to the reactants' side

#4H^(+) + stackrel(color(blue)(+5))(N) O_3^(-) + 3e^(-) -> stackrel(color(blue)(+2))(N) O + 2H_2O#

In a redox reaction, the number of electrons lost during oxidation and gained during reduction must be equal. At this point, you have #3e^(-)# gained in the reduction half-reaction, and #16e^(-)# lost during the oxidation half-reaction.

Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 16 to balance the electrons out

#{ (24H_2 stackrel(color(blue)(-2))(S) -> 3stackrel(color(blue)(0))(S_8) + 48e^(-) + 48H^(+)), (62H^(+) + 16stackrel(color(blue)(+5))(N) O_3^(-) + 48e^(-) -> 16stackrel(color(blue)(+2))(N) O + 32H_2O) :}#

Add the two half-reactions and cancel the species present on both sides of the equation to get

#24H_2S + 16H^(+) + cancel(48e^(-)) + 16NO_3^(-) ->3S_8 + cancel(48e^(-)) + 16NO + 32H_2O#

The balanced equation will thus be

#24H_2S + 16H^(+) + 16NO_3^(-) -> 3S_8 + 16NO + 32H_2O#