Question 4.14: Balancing Oxidation–Reduction Reactions Because metals are s...
Balancing Oxidation–Reduction Reactions
Because metals are so reactive, very few are found in pure form in nature. Metallurgy involves reducing the metal ions in ores to the elemental form. The production of manganese from the ore pyrolusite, which contains MnO_{2} , uses aluminum as the reducing agent. Using oxidation states, balance the equation for this process.
MnO_{2} (s) + Al(s) → Mn(s) + Al_{2} O_{3}(s)
First, we assign oxidation states:
MnO_{2} (s) + Al(s) → Mn(s) + Al_{2} O_{3} (s)
\nearrow ↑ ↑ ↑ ↑ \nwarrow
+4 -2 0 0 +3 -2
(each O) (each Al) (each O)
Each Mn atom undergoes a decrease in oxidation state of 4 (from +4 to 0), whereas each Al atom undergoes an increase of 3 (from 0 to +3).
Thus we need three Mn atoms for every four Al atoms to balance the increase and decrease in oxidation states:
Increase = 4(3) = decrease = 3(4)
3 MnO_{2}(s) + 4 Al(s) → products
We balance the rest of the equation by inspection:
3 MnO_{2}(s) + 4 Al(s) → 3 Mn(s) + 2 Al_{2} O_{3} (s)