Cr₂O₇²⁻ and I⁻ in acid
Cr₂O₇²⁻ + 6I⁻ + 14H⁺ → 2Cr³⁺ + 3I₂ + 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⁻+6I⁻+14H⁺→2Cr³⁺+3I₂+7H₂O
- Oxidising agentCr₂O₇²⁻
- Reducing agentI⁻
- Electrons6
OxidationI −1 → 0
6I⁻→3I₂+6e⁻
loses 6 e⁻ReductionCr +6 → +3
Cr₂O₇²⁻+14H⁺+6e⁻→2Cr³⁺+7H₂O
gains 6 e⁻- Write the skeleton in ionic form
Cr₂O₇²⁻ + I⁻ → Cr³⁺ + I₂ - Split into two half reactions
Oxidation: I⁻ → I₂ (I −1 → 0)
Reduction: Cr₂O₇²⁻ → Cr³⁺ (Cr +6 → +3) - Balance atoms other than O and H
Oxidation: 2I⁻ → I₂
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: 2I⁻ → I₂ + 2e⁻
Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O - Make the electrons equal
Oxidation × 3: 6I⁻ → 3I₂ + 6e⁻
Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O - Add the halves and cancel what appears on both sides
Cr₂O₇²⁻ + 6I⁻ + 14H⁺ → 2Cr³⁺ + 3I₂ + 7H₂O - Check
Each side has 6 I, 2 Cr, 7 O, 14 H and a total charge of +6.
- Write the oxidation numbers and find the change
I: −1 in I⁻ → 0 in I₂, a rise of 2 for 2I⁻ → I₂ (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⁻ + 6I⁻ → 2Cr³⁺ + 3I₂ - Balance the charge with 14 H⁺ (acidic solution)
Cr₂O₇²⁻ + 6I⁻ + 14H⁺ → 2Cr³⁺ + 3I₂ - Balance H with 7 H₂O
Cr₂O₇²⁻ + 6I⁻ + 14H⁺ → 2Cr³⁺ + 3I₂ + 7H₂O - Check
Cr₂O₇²⁻ + 6I⁻ + 14H⁺ → 2Cr³⁺ + 3I₂ + 7H₂O
Each side has 6 I, 2 Cr, 7 O, 14 H and a total charge of +6.
Step by step
Ion-electron method
- Write the skeleton in ionic form
Cr₂O₇²⁻ + I⁻ → Cr³⁺ + I₂ - Split into two half reactions
Oxidation: I⁻ → I₂ (I −1 → 0)
Reduction: Cr₂O₇²⁻ → Cr³⁺ (Cr +6 → +3) - Balance atoms other than O and H
Oxidation: 2I⁻ → I₂
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: 2I⁻ → I₂ + 2e⁻
Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O - Make the electrons equal
Oxidation × 3: 6I⁻ → 3I₂ + 6e⁻
Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O - Add the halves and cancel what appears on both sides
Cr₂O₇²⁻ + 6I⁻ + 14H⁺ → 2Cr³⁺ + 3I₂ + 7H₂O - Check
Each side has 6 I, 2 Cr, 7 O, 14 H and a total charge of +6.
Oxidation number method
- Write the oxidation numbers and find the change
I: −1 in I⁻ → 0 in I₂, a rise of 2 for 2I⁻ → I₂ (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⁻ + 6I⁻ → 2Cr³⁺ + 3I₂ - Balance the charge with 14 H⁺ (acidic solution)
Cr₂O₇²⁻ + 6I⁻ + 14H⁺ → 2Cr³⁺ + 3I₂ - Balance H with 7 H₂O
Cr₂O₇²⁻ + 6I⁻ + 14H⁺ → 2Cr³⁺ + 3I₂ + 7H₂O - Check
Cr₂O₇²⁻ + 6I⁻ + 14H⁺ → 2Cr³⁺ + 3I₂ + 7H₂O
Each side has 6 I, 2 Cr, 7 O, 14 H and a total charge of +6.
Oxidising and reducing agents
- I −1 → 0I⁻ to I₂: oxidised
- Cr +6 → +3Cr₂O₇²⁻ to Cr³⁺: reduced
- Cr₂O₇²⁻oxidising agent, gains electrons
- I⁻reducing agent, loses electrons
Cr₂O₇²⁻ and I⁻ in acid: quick answers
What is the balanced equation for dichromate and iodide in acid?
Cr₂O₇²⁻ + 6I⁻ + 14H⁺ → 2Cr³⁺ + 3I₂ + 7H₂O.
Which is the oxidising agent and which is the reducing agent?
I goes from −1 to 0 and Cr goes from +6 to +3. Cr₂O₇²⁻ is the oxidising agent (it is reduced) and I⁻ is the reducing agent (it is oxidised).
How many electrons are transferred?
6 electrons pass from I⁻ to Cr₂O₇²⁻ in the balanced equation as written.