For a particular redox reaction, Cr is oxidized to CrO 2 − 4 and Fe 3 + is reduced to Fe 2 + . Complete and balance the equation for this reaction in basic solution. The phases are optional. balanced reaction: Cr + Fe^{3+} -> CrO_{4}^{2} + Fe^{2+} Cr + Fe 3 + ⟶ CrO 2 4 + Fe 2 +

8OH⁻ + Cr³⁺  + 3Fe³⁺ → 3Fe²⁺  +  CrO₄²⁻  +  4H₂O

is the balanced reaction

Explanation:

Cr³⁺ is oxidized to CrO₄²⁻ and Fe³⁺ is reduced to Fe²⁺

In basic medium, you complete with water where you have the highest value of oxygen. In the opposite side we add OH⁻ as many oxygens, we have.

8OH⁻ + Cr³⁺  →  CrO₄²⁻  +  4H₂O + 3e⁻

Chromium increased the oxidation state from +3 to +6. It released 3 electrons.

Fe³⁺  + 1e⁻ → Fe²⁺

Iron decreased the oxidation state from +3 to +2, so it won an electron.

We multiply the second half reaction x3, to balance the e⁻

(Fe³⁺  + 1e⁻ → Fe²⁺)x3 → 3Fe³⁺  + 3e⁻ → 3Fe²⁺

We mix the half reactions:

8OH⁻ + Cr³⁺  + 3Fe³⁺  + 3e⁻  → 3Fe²⁺  +  CrO₄²⁻  +  4H₂O + 3e⁻

We cancel the electrons, so the balance reaction is:

8OH⁻ + Cr³⁺  + 3Fe³⁺ → 3Fe²⁺  +  CrO₄²⁻  +  4H₂O

it's all four options.