We start with the unbalanced equation:
#"K" + "B"_2"O"_3 → "K"_2"O" + "B"#
Step 1. Identify the atoms that change oxidation number
#stackrelcolor(blue)(0)("K") + stackrelcolor(blue)(+3)("B")_2stackrelcolor(blue)("-2")("O")_3 → stackrelcolor(blue)(+1)("K")_2stackrelcolor(blue)("-2")("O") + stackrelcolor(blue)(0)("B")#
The changes in oxidation number are:
#"K: 0 → +1; Change =+1 (oxidation)"#
#"B: +3 → 0; Change = -3 (reduction)"#
Step 2. Equalize the changes in oxidation number
We need 3 atoms of #"K"# for every 1 atom of #"B"# or 6 atoms of #"K"# for every 2 atoms of #"B"#. This gives us total changes of +6 and -6.
Step 3. Insert coefficients to get these numbers
#color(red)(6)"K" + color(red)(1)"B"_2"O"_3 → color(red)(3)"K"_2"O" + color(red)(2)"B"#
Every formula now has a coefficient. The equation should be balanced.
Step 4. Check that all atoms are balanced.
#bb"On the left"color(white)(l)bb "On the right"#
#color(white)(mm)"6 K"color(white)(mmmm) "6 K"#
#color(white)(mm)"2 B"color(white)(mmmm) "2 B"#
#color(white)(mm)"3 O"color(white)(mmmm) "3 O"#
The balanced equation is
#color(red)(6"K" + "B"_2"O"_3 → 3"K"_2"O" + 2"B"#
