Answer:
14 H⁺ + Cr₂O₇²⁻ + 6 Fe²⁺ ⇒ 2 Cr³⁺ + 7 H₂O + 6 Fe³⁺
Explanation:
<em>The question is missing but I guess it must be about balancing the redox reaction.</em>
We will balance the redox reaction using the ion-electron method.
Step 1: Identify both half-reactions
Reduction: Cr₂O₇²⁻ ⇒ Cr³⁺
Oxidation: Fe²⁺ ⇒ Fe³⁺
Step 2: Perform the mass balance adding H⁺ and H₂O where appropriate (this reaction takes place in acid medium)
14 H⁺ + Cr₂O₇²⁻ ⇒ 2 Cr³⁺ + 7 H₂O
Fe²⁺ ⇒ Fe³⁺
Step 3: Perform the charge balance adding electrons where appropriate
14 H⁺ + Cr₂O₇²⁻ + 6 e⁻ ⇒ 2 Cr³⁺ + 7 H₂O
Fe²⁺ ⇒ Fe³⁺ + 1 e⁻
Step 4: Multiply half-reactions by numbers that make that the electrons gained and lost are equal
1 × [14 H⁺ + Cr₂O₇²⁻ + 6 e⁻ ⇒ 2 Cr³⁺ + 7 H₂O]
6 × [Fe²⁺ ⇒ Fe³⁺ + 1 e⁻]
Step 5: Add both half-reactions and cancel what is repeated
14 H⁺ + Cr₂O₇²⁻ + 6 Fe²⁺ ⇒ 2 Cr³⁺ + 7 H₂O + 6 Fe³⁺
