We start with the unbalanced equation:
#"Cu" + "H"_2"SO"_4 → "CuSO"_4 + "SO"_2 + "H"_2"O"#
Step 1. Identify the atoms that change oxidation number
#stackrelcolor(blue)(0)("Cu") + stackrelcolor(blue)(+1)("H")_2stackrelcolor(blue)(+6)("S")stackrelcolor(blue)("-2")("O")_4 → stackrelcolor(blue)(+2)("Cu")stackrelcolor(blue)(+6)("S")stackrelcolor(blue)(-2)("O")_4 + stackrelcolor(blue)(+4)("S")stackrelcolor(blue)("-2")("O")_2 + stackrelcolor(blue)(+1)("H")_2stackrelcolor(blue)("-2")("O")#
The changes in oxidation number are:
#"Cu: 0 → +2; Change =+2" (bb "oxidation")#
#"S: +6 → +4 in SO"_2; "Change = -2" (bb"reduction")#
#"Cu"# is oxidized; #"S"# is reduced.
Step 2. Equalize the changes in oxidation number
We need 1 atom of #"Cu"# for every 1 atom of #"S"# in #"SO"_2"#. This gives us total changes of +2 and -2.
Step 3. Insert coefficients to get these numbers
#color(red)(1)"Cu" + "H"_2"SO"_4 → color(red)(1)"CuSO"_4 + color(red)(1)"SO"_2 + "H"_2"O"#
Step 4. Balance #"S"#
We have fixed 2 #"S"# atoms on the right, so we need 2 #"S"# atoms on the left. Put a 2 before #"H"_2"SO"_4#.
#color(red)(1)"Cu" + color(blue)(2)"H"_2"SO"_4 → color(red)(1)"CuSO"_4 + color(red)(1)"SO"_2 + "H"_2"O"#
Step 5. Balance #"O"#.
We have fixed 8 #"O"# atoms on the left and 6 #"O"# atoms on the right.
We need 2 more #"O"# atoms on the right. Put a 2 before #"H"_2"O"#.
#color(red)(1)"Cu" + color(blue)(2)"H"_2"SO"_4 → color(red)(1)"CuSO"_4 + color(red)(1)"SO"_2 + color(green)(2)"H"_2"O"#
Every formula now has a coefficient. The equation should be balanced.
Step 6. Check that all atoms are balanced.
#bb"On the left"color(white)(l)bb "On the right"#
#color(white)(mm)"1 Cu"color(white)(mmmm) "1 Cu"#
#color(white)(mm)"4 H"color(white)(mmmmll) "4 H"#
#color(white)(mm)"2 S"color(white)(mmmmm) "2 S"#
#color(white)(mm)"8 O"color(white)(mmmmll) "8 O"#
The balanced equation is
#color(red)("Cu"color(white)(l) + "2H"_2"SO"_4 → "CuSO"_4 + "SO"_2 +2"H"_2"O")#
