Cr2O72- + Fe2+ → Cr3+ + Fe3+

14 H⁺ + Cr₂O₇²⁻ + 6 Fe²⁺ ⇒ 2 Cr³⁺ + 7 H₂O + 6 Fe³⁺

Explanation:

The question is missing but I guess it must be about balancing the redox reaction.

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⁺

the answer is c. nacl