Cr₂O₇²⁻ and Fe²⁺ in acid
Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O
Redox balancer
Type the reactants and products with an arrow, without numbers: MnO4- + Fe2+ -> Mn2+ + Fe3+. Leave out H+, OH- and water; they are added for you. Salts and acids such as KMnO4 or HNO3 are fine.
Cr₂O₇²⁻+6Fe²⁺+14H⁺→2Cr³⁺+6Fe³⁺+7H₂O
- Oxidising agentCr₂O₇²⁻
- Reducing agentFe²⁺
- Electrons6
OxidationFe +2 → +3
6Fe²⁺→6Fe³⁺+6e⁻
loses 6 e⁻ReductionCr +6 → +3
Cr₂O₇²⁻+14H⁺+6e⁻→2Cr³⁺+7H₂O
gains 6 e⁻- Write the skeleton in ionic form
Cr₂O₇²⁻ + Fe²⁺ → Cr³⁺ + Fe³⁺ - Split into two half reactions
Oxidation: Fe²⁺ → Fe³⁺ (Fe +2 → +3)
Reduction: Cr₂O₇²⁻ → Cr³⁺ (Cr +6 → +3) - Balance atoms other than O and H
Oxidation: Fe²⁺ → Fe³⁺
Reduction: Cr₂O₇²⁻ → 2Cr³⁺ - Balance O by adding H₂O
Reduction: Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O - Balance H by adding H⁺
Reduction: Cr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O - Balance the charge with electrons
Oxidation: Fe²⁺ → Fe³⁺ + e⁻
Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O - Make the electrons equal
Oxidation × 6: 6Fe²⁺ → 6Fe³⁺ + 6e⁻
Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O - Add the halves and cancel what appears on both sides
Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O - Check
Each side has 6 Fe, 2 Cr, 7 O, 14 H and a total charge of +24.
- Write the oxidation numbers and find the change
Fe: +2 in Fe²⁺ → +3 in Fe³⁺, a rise of 1 for Fe²⁺ → Fe³⁺ (oxidation)
Cr: +6 in Cr₂O₇²⁻ → +3 in Cr³⁺, a fall of 6 for Cr₂O₇²⁻ → 2Cr³⁺ (reduction) - Make the total rise equal the total fall (6 each)
Cr₂O₇²⁻ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ - Balance the charge with 14 H⁺ (acidic solution)
Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ - Balance H with 7 H₂O
Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O - Check
Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O
Each side has 6 Fe, 2 Cr, 7 O, 14 H and a total charge of +24.
Step by step
Ion-electron method
- Write the skeleton in ionic form
Cr₂O₇²⁻ + Fe²⁺ → Cr³⁺ + Fe³⁺ - Split into two half reactions
Oxidation: Fe²⁺ → Fe³⁺ (Fe +2 → +3)
Reduction: Cr₂O₇²⁻ → Cr³⁺ (Cr +6 → +3) - Balance atoms other than O and H
Oxidation: Fe²⁺ → Fe³⁺
Reduction: Cr₂O₇²⁻ → 2Cr³⁺ - Balance O by adding H₂O
Reduction: Cr₂O₇²⁻ → 2Cr³⁺ + 7H₂O - Balance H by adding H⁺
Reduction: Cr₂O₇²⁻ + 14H⁺ → 2Cr³⁺ + 7H₂O - Balance the charge with electrons
Oxidation: Fe²⁺ → Fe³⁺ + e⁻
Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O - Make the electrons equal
Oxidation × 6: 6Fe²⁺ → 6Fe³⁺ + 6e⁻
Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O - Add the halves and cancel what appears on both sides
Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O - Check
Each side has 6 Fe, 2 Cr, 7 O, 14 H and a total charge of +24.
Oxidation number method
- Write the oxidation numbers and find the change
Fe: +2 in Fe²⁺ → +3 in Fe³⁺, a rise of 1 for Fe²⁺ → Fe³⁺ (oxidation)
Cr: +6 in Cr₂O₇²⁻ → +3 in Cr³⁺, a fall of 6 for Cr₂O₇²⁻ → 2Cr³⁺ (reduction) - Make the total rise equal the total fall (6 each)
Cr₂O₇²⁻ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ - Balance the charge with 14 H⁺ (acidic solution)
Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ - Balance H with 7 H₂O
Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O - Check
Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O
Each side has 6 Fe, 2 Cr, 7 O, 14 H and a total charge of +24.
Oxidising and reducing agents
- Fe +2 → +3Fe²⁺ to Fe³⁺: oxidised
- Cr +6 → +3Cr₂O₇²⁻ to Cr³⁺: reduced
- Cr₂O₇²⁻oxidising agent, gains electrons
- Fe²⁺reducing agent, loses electrons
Cr₂O₇²⁻ and Fe²⁺ in acid: quick answers
What is the balanced equation for dichromate and iron(II) in acid?
Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O.
Which is the oxidising agent and which is the reducing agent?
Fe goes from +2 to +3 and Cr goes from +6 to +3. Cr₂O₇²⁻ is the oxidising agent (it is reduced) and Fe²⁺ is the reducing agent (it is oxidised).
How many electrons are transferred?
6 electrons pass from Fe²⁺ to Cr₂O₇²⁻ in the balanced equation as written.