Question #de191

!! LONG ANSWER !!

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

#stackrel(color(blue)(+1))(Cu_2) stackrel(color(blue)(-2))(S) + stackrel(color(blue)(+1))(H) stackrel(color(blue)(+5))(N) stackrel(color(blue)(-2))(O_3) -> stackrel(color(blue)(+2))(Cu) overbrace((NO_3)_2)^(color(blue)(-2)) + stackrel(color(blue)(+4))(S) stackrel(color(blue)(-2))(O_2) + stackrel(color(blue)(+4))(N) stackrel(color(blue)(-2))(O_2) + stackrel(color(blue)(+1))(H_2) stackrel(color(blue)(-2))(O)#

Notice that copper's oxidation state changes from #color(blue)(+1)# on the reactants' side, to #color(blue)(+2)# on the products' side, which means that copper is getting oxidized.

Moreover, sulfur is being oxidized as well, since its oxidation state goes from #color(blue)(-2)# to #color(blue)(+4)#.

On the other hand, nitrogen is going from #color(blue)(+5)# on the reactants' side, to #color(blue)(+4)# on the products' side, which means that nitrogen is getting reduced.

Write the oxidation and reduction half-reactions and balance using #H^(+)# and #H_2O# when needed.

• Oxidation half-reaction

#stackrel(color(blue)(+1))(Cu_2) stackrel(color(blue)(-2))(S) -> 2stackrel(color(blue)(+2))(Cu_2) + stackrel(color(blue)(+4))(S)O_2 + 8e^(-)#

You get #1e^(-)# lost from each of the two copper atoms, and #6e^(-)# lost from the sulfur atom #-># #8e^(-)# are lost altogether.

Balance the oxygen by adding two water molecules on the reactants' side, and the resulting hydrogen by adding four #H^(+)# on the products' side

#2H_2O + Cu_2S -> 2Cu_2^(2+) + SO_2 + 8e^(-) + 4H^(+)#

• Reduction half-reaction

#stackrel(color(blue)(+5))(N)O_3^(-) + 1e^(-) -> stackrel(color(blue)(+4))(N)O_2#

Once again, balance the oxygen by adding one water molecule on the products' side, and two #H^(+)# on the reactants' side

#2H^(+) + NO_3^(-) + 1e^(-) -> NO_2 + H_2O#

Now it's time to balance the electrons lost during oxidation half-reaction and gained during the reduction half-reaction. Notice that you have #8e^(-)# in the oxidation half-reaction, but only #1e^(-)# in the reduction half-reaction.

Multiply the reduction half-reaction by 8 to get

#{ (2H_2O + Cu_2S -> 2Cu_2^(2+) + SO_2 + 8e^(-) + 4H^(+)), (16H^(+) + 8NO_3^(-) + 8e^(-) -> 8NO_2 + 8H_2O) :}#

Add these two equations to get

#2H_2O + Cu_2S + 16H^(+) + 8NO_3^(-) + cancel(8e^(-)) -> 2Cu_2^(2+) + SO_2 + 8NO_2 + cancel(8e^(-)) + 4H^(+) + 8H_2O#

Eliminate compounds that are on both sides of the equation to get

#Cu_2S + 12H^(+) + 8NO_3^(-) -> 2Cu_2^(2+) + SO_2 + 8NO_2 + 6H_2O#

The complete and balanced equation will be

#Cu_2S + 12HNO_3 -> 2Cu_2(NO_3)_2 + SO_2 + 8NO_2 + 6H_2O#

Warning! This is a long answer. The balanced equation is

#"Cu"_2"S" + "12HNO"_3 → "2Cu"("NO"_3)_2 + "SO"_2 + "8NO"_2 + "6H"_2"O"#

We start with the unbalanced equation:

#"Cu"_2"S" + "HNO"_3 → "Cu"("NO"_3)_2 + "SO"_2 + "NO"_2 + "H"_2"O"#

Step 1. Identify the atoms that change oxidation number

# stackrel(color(blue)(+1))("Cu")_2 stackrel(color(blue)(-2))("S") + stackrel(color(blue)(+1))("H") stackrel(color(blue)(+5))("N") stackrel(color(blue)(-2))("O")_3 → stackrel(color(blue)(+2))("Cu")( stackrel(color(blue)(+5))("N") stackrel(color(blue)(-2))("O")_3)_2 + stackrel(color(blue)(+4))("S") stackrel(color(blue)(-2))("O")_2 + stackrel(color(blue)(+4))("N") stackrel(color(blue)(-2))("O")_2 + stackrel(color(blue)(+1))("H")_2 stackrel(color(blue)(-2))("O")#

Left hand side: #"Cu" = +1#; #"S" = -2#; #"H" = +1#; #"N" = +5#; #"O" = -2#
Right hand side: #"Cu" = +2#; #"N in NO"_3 = +5#; #"S" = +4#; #"O" = -2#; #"N in NO"_2 = +4#; #"H" = +1#

The changes in oxidation number are:

#"Cu"#: +1 → +2; Change =+1
#"S"#: -2 → +4; Change = +6
#"N"#: +5 → +4; Change = -1

But #"Cu"_2"S"# behaves as a unit.

The total change for #"Cu"_2"S"# is [2×(+1) + 6] = +8.

Step 2. Equalize the changes in oxidation number

You need 8 atoms of #"N"# that change oxidation number for every 1 unit of #"Cu"_2"S"#. This gives us total changes of -8 and +8.

Step 3. Insert coefficients to get these numbers

# color(red)(1)"Cu"_2"S" + "HNO"_3 → color(red)(2)"Cu"("NO"_3)_2 + color(red)(1)"SO"_2 + color(red)(8)"NO"_2 + "H"_2"O"#

Note: We can't fix a number in front of the #"HNO"_3# because some of the #"N"# atoms go to #"NO"_2# and some remain unchanged as #"NO"_3# ions.

Step 4. Balance #"N"# by adding #"HNO"_3"# molecules to the appropriate side

# color(red)(2)"Cu"_2"S" + color(blue)(12)"HNO"_3 → color(red)(2)"Cu"("NO"_3)_2 + color(red)(1)"SO"_2 + color(red)(8)"NO"_2 + "H"_2"O"#

Step 4. Balance #"O"# by adding #"H"_2"O"# molecules to the appropriate side

# color(red)(2)"Cu"_2"S" + color(blue)(12)"HNO"_3 → color(red)(2)"Cu"("NO"_3)_2 + color(red)(1)"SO"_2 + color(red)(8)"NO"_2 + color(orange)(6)"H"_2"O"#

The balanced equation is

#color(red)("Cu"_2"S" + "12HNO"_3 → "2Cu"("NO"_3)_2 + "SO"_2 + "8NO"_2 + "6H"_2"O")#