Separate into oxidation and reduction half-reactions. Balance electrons.

Balance Cr₂O₇²⁻ + Fe²⁺ in Acidic Medium — Half Reaction Method

Question

Balance the following redox reaction in acidic medium using the half-reaction method:

\text{Cr}_2\text{O}_7^{2-} + \text{Fe}^{2+} \rightarrow \text{Cr}^{3+} + \text{Fe}^{3+}

Solution — Step by Step

Chromium in $\text{Cr}_2\text{O}_7^{2-}$ goes from +6 → +3 (gains electrons — reduction). Iron goes from +2 → +3 (loses electrons — oxidation).

We split the reaction into two half-reactions based on this.

\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-

Atoms balance. Charge: left side is +2, right side is +3 − 1 = +2. Balanced.

Start with just chromium:

\text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+}

There are 7 oxygen atoms on the left. In acidic medium, we balance oxygen using $\text{H}_2\text{O}$ and hydrogen using $\text{H}^+$ :

\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}

Now balance charge. Left: −2 + 14 = +12. Right: +6. Difference is 6, so add 6 electrons to the left:

\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}

Verify: left charge = −2 + 14 − 6 = +6. Right charge = +6. ✓

Oxidation half-reaction transfers 1 electron, reduction needs 6 electrons. Multiply the oxidation half by 6:

6\text{Fe}^{2+} \rightarrow 6\text{Fe}^{3+} + 6e^-

\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- + 6\text{Fe}^{2+} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} + 6\text{Fe}^{3+} + 6e^-

Cancel $6e^-$ from both sides:

\boxed{\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Fe}^{2+} \rightarrow 2\text{Cr}^{3+} + 6\text{Fe}^{3+} + 7\text{H}_2\text{O}}

Why This Works

The half-reaction method works because it forces us to conserve both mass and charge separately in each half — which is harder to mess up than balancing the full equation in one shot.

In acidic medium, oxygen is always balanced by adding $\text{H}_2\text{O}$ to the oxygen-deficient side, then hydrogen is balanced using $\text{H}^+$ . This sequence is fixed — oxygen first, then hydrogen. Flipping the order creates a mess.

The electron equalisation step is where the stoichiometry of the final equation comes from. Here, dichromate is a 6-electron oxidising agent per formula unit — that’s why it can oxidise 6 $\text{Fe}^{2+}$ ions in one go. This ratio (1:6) is a favourite data point in MCQ options designed to trap students.

Alternative Method — Oxidation Number Method

Assign oxidation numbers and track the total change:

Each Cr: +6 → +3 (change = 3). Two Cr atoms → total decrease = 6
Each Fe: +2 → +3 (change = 1). Need 6 Fe atoms → total increase = 6

So the ratio of $\text{Cr}_2\text{O}_7^{2-}$ : $\text{Fe}^{2+}$ is 1 : 6.

Place these coefficients, then balance $\text{H}^+$ and $\text{H}_2\text{O}$ by inspection. You’ll land on the same equation. The half-reaction method is preferred for complex ions, but the oxidation number method is faster for MCQs where you just need the ratio.

In JEE MCQs, they often just ask for the coefficient of $\text{H}^+$ or the ratio of oxidant to reductant. The answer here: 14 H⁺ and 1:6 ratio. Memorise these for this specific reaction — it appeared in JEE Main 2024.

Common Mistake

The most common error is adding electrons to the wrong side of the reduction half-reaction. Students write $\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+$ on the left (charge = +12) and $2\text{Cr}^{3+}$ (charge = +6) on the right, then add 6 electrons to the right to “balance” it — completely backwards. Electrons are added to the more positive side to reduce the charge difference. In reduction, electrons always go on the left (reactant side). If you find yourself adding electrons to the product side, stop and recheck which species is being reduced.

Question

Solution — Step by Step

Why This Works

Alternative Method — Oxidation Number Method

Common Mistake

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