Balance the following redox reaction if it occurs in acidic solution. what are the coefficients in front of h+ and fe3+ in the balanced reaction? fe2+(aq) + mno4-(aq) → fe3+(aq) + mn2+(aq) balance the following redox reaction if it occurs in acidic solution. what are the coefficients in front of h+ and fe3+ in the balanced reaction? fe2+(aq) + mno4-(aq) → fe3+(aq) + mn2+(aq) h+ = 8, fe3+ = 1 h+ = 8, fe3+ = 5 h+ = 2, fe3+ = 3 h+ = 3, fe3+ = 2 h+ = 5, fe3+ = 1

H⁺ = 8; Fe³⁺ = 5

Explanation:

In order to balance a redox reaction, we use the ion-electron method.

Step 1: Identify both half-reactions.

Reduction: MnO₄⁻(aq) → Mn²⁺(aq)

Oxidation: Fe²⁺(aq) → Fe³⁺(aq)

Step 2: Balance the masses using H⁺ and H₂O where appropiate.

8 H⁺(aq) + MnO₄⁻(aq) → Mn²⁺(aq) + 4 H₂O(l)

Fe²⁺(aq) → Fe³⁺(aq)

Step 3: Balance the charges adding electrons where appropiate.

8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l)

Fe²⁺(aq) → Fe³⁺(aq) + 1 e⁻

Step 4: Multiply both half-reactions by numbers that assure that the number of electrons gained is equal to the number of electrons lost.

1 × [8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l)]

5 × [Fe²⁺(aq) → Fe³⁺(aq) + 1 e⁻]

Step 5: Add both half-reactions

8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ + 5 Fe²⁺(aq) → Mn²⁺(aq) + 4 H₂O(l) + 5 Fe³⁺(aq) + 5 e⁻

8 H⁺(aq) + MnO₄⁻(aq) + 5 Fe²⁺(aq) → Mn²⁺(aq) + 4 H₂O(l) + 5 Fe³⁺(aq)

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

the volume increases