How do you balance this redox reaction using the oxidation number method? Al(s) + H2SO4(aq) → Al2(SO4)3(aq) + H2(g)

May 13, 2014

$2 A l$ + $3 {H}_{2} S {O}_{4}$ --> $A {l}_{2} {\left(S {O}_{4}\right)}_{3}$ + $3 {H}_{2}$

Step 1. Write the oxidation numbers for each atom:

$A l$ + ${H}_{2} S {O}_{4}$ --> $A {l}_{2} {\left(S {O}_{4}\right)}_{3}$ + ${H}_{2}$
0 +1 +6 -2-2 +3 +6 -2-2 0
+1 -2-2 +3 -2-2

Step 2. Identify what has been oxidised and what as been reduced, and balance the equation ONLY for these species

Al has been oxidised 0 to +3. We need 2Al on the LHS as we have 2 on the right.

Hydrogen has been reduced from +1 to 0. We are balanced for H, so the equation at this stage looks like:

$2 A l$ + ${H}_{2} S {O}_{4}$ --> $A {l}_{2} {\left(S {O}_{4}\right)}_{3}$ + ${H}_{2}$

Step 3. Calculate the total increase in oxidation numbers due to the oxidation taking place. We are oxidising 2 x Al (oxidation state 0) to 2 x Al3+ (oxidation state +3) so the total increase is +6.

Do the same for the reduction. We reduce 2 x H+ (+1) to 2 x H (0) so the total decrease is -2

Step 4. Multiply the the species being oxidised and/or reduced so that the total increase in oxidation number = total decrease in oxidation number. In this example we need to multiply the species involved in the reduction of hydrogen x3 to get a total decrease of -6. This means we need 3H2SO4 and we get 3H2.

Final equation: $2 A l$ + $3 {H}_{2} S {O}_{4}$ --> $A {l}_{2} {\left(S {O}_{4}\right)}_{3}$ + $3 {H}_{2}$ as above.