07 - Balancing Redox in Acid or Basic | PDF | Redox | Chemistry - Free Printable

Let's continue solving the redox reaction balancing problem step-by-step, based on the image you provided. The goal is to balance the redox reaction in acidic solution:

Skeleton Equation:

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

We are given the first three steps of the process. Let's complete it fully.

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✔ Step 1: Divide into Half-Reactions

Oxidation half-reaction (Fe²⁺ → Fe³⁺):
\[
\text{Fe}^{2+} \rightarrow \text{Fe}^{3+}
\]

Reduction half-reaction (Cr₂O₇²⁻ → Cr³⁺):
\[
\text{Cr}_2\text{O}_7^{2-} \rightarrow \text{Cr}^{3+}
\]

---

✔ Step 2: Balance atoms other than H and O

For Cr:
There are 2 Cr atoms on the left, so we need 2 Cr³⁺ on the right:
\[
\text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+}
\]

For Fe:
Already balanced:
\[
\text{Fe}^{2+} \rightarrow \text{Fe}^{3+}
\]

---

✔ Step 3: Balance oxygen by adding H₂O

The Cr₂O₇²⁻ has 7 O atoms, but the right side has none. Add 7 H₂O molecules to the right:
\[
\text{Cr}_2\text{O}_7^{2-} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}
\]

Now, oxygen is balanced.

---

✔ Step 4: Balance hydrogen by adding H⁺ (since it's acidic solution)

Left side has no H; right has 14 H atoms (from 7 H₂O). So add 14 H⁺ to the left:
\[
\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}
\]

---

✔ Step 5: Balance charge using electrons (e⁻)

Now balance the charges on both sides.

Left side charge:
- Cr₂O₇²⁻: -2
- 14 H⁺: +14
- Total: (-2) + (+14) = +12

Right side charge:
- 2 Cr³⁺: 2 × (+3) = +6
- 7 H₂O: neutral
- Total: +6

To balance, add 6 electrons (e⁻) to the left side (since reduction occurs here — Cr is gaining electrons):
\[
\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}
\]

✔ Reduction half-reaction is now balanced.

---

Now, oxidation half-reaction: Fe²⁺ → Fe³⁺

Balance charge:
- Left: Fe²⁺ → charge = +2
- Right: Fe³⁺ → charge = +3

Add 1 electron to the right side:
\[
\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-
\]

But this produces only 1 electron per Fe atom. We need to match the number of electrons from the reduction half-reaction.

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✔ Step 6: Equalize electrons in both half-reactions

Reduction: gains 6 electrons
Oxidation: loses 1 electron per Fe²⁺

So multiply oxidation half-reaction by 6:
\[
6\text{Fe}^{2+} \rightarrow 6\text{Fe}^{3+} + 6e^-
\]

---

✔ Step 7: Add the two half-reactions together

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

Oxidation:
\[
6\text{Fe}^{2+} \rightarrow 6\text{Fe}^{3+} + 6e^-
\]

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

Cancel out the 6e⁻ on both sides:
\[
\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{Fe}^{2+} \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O} + 6\text{Fe}^{3+}
\]

---

✔ Final Balanced Equation (in acidic solution):

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

---

🔍 Verification:

| Atom | Left Side | Right Side |
|------|-----------|------------|
| Cr | 2 | 2 |
| O | 7 | 7 (in H₂O) |
| Fe | 6 | 6 |
| H | 14 | 14 (in 7 H₂O) |
| Charge: | (-2) + 6×(+2) + 14×(+1) = -2 + 12 + 14 = +24 | 2×(+3) + 6×(+3) = 6 + 18 = +24 |

✔ All atoms and charges are balanced!

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Summary of Steps for Acidic Solution:

1. Split into half-reactions.
2. Balance atoms other than H and O.
3. Balance O with H₂O.
4. Balance H with H⁺.
5. Balance charge with e⁻.
6. Equalize electrons in both half-reactions.
7. Add and simplify.

This method works for any redox reaction in acidic solution. For basic solutions, you would add OH⁻ to both sides after balancing in acidic conditions to neutralize H⁺.

Let me know if you'd like the basic version too!