Question

Balance the following redox equations:
1. H2S + Cr2O72- → S + Cr3+ (acidic medium)
2. MnO2 + HCl → Mn2+ + Cl2 + H2O

Answer 1

Answer on Question #44141 - Chemistry - Inorganic Chemistry

Question:

Balance the following redox equations:

1. $\mathrm{H}_2\mathrm{S} + \mathrm{Cr}_2\mathrm{O}_7^{2-} \rightarrow \mathrm{S} + \mathrm{Cr}^{3+}$ (acidic medium)

2. $\mathrm{MnO}_2 + \mathrm{HCl} \rightarrow \mathrm{Mn}^{2+} + \mathrm{Cl}_2 + \mathrm{H}_2\mathrm{O}$

Solution:

1. $\mathrm{H}_2\mathrm{S} + \mathrm{Cr}_2\mathrm{O}_7^{2-} \rightarrow \mathrm{S} + \mathrm{Cr}^{3+}$ (acidic medium)

Step 1: Separate the half-reactions and balance elements other than O and H.

$$\mathrm{H}_2\mathrm{S} \rightarrow \mathrm{S}$$ $$\mathrm{Cr}_2\mathrm{O}_7^{2-} \rightarrow 2\mathrm{Cr}^{3+}$$

Step 2: Add H₂O to balance oxygen. The chromium reaction needs to be balanced by adding 7 H₂O molecules. This yields:

$$\mathrm{H}_2\mathrm{S} \rightarrow \mathrm{S}$$ $$\mathrm{Cr}_2\mathrm{O}_7^{2-} \rightarrow 2\mathrm{Cr}^{3+} + 7\mathrm{H}_2\mathrm{O}$$

Step 3: Balance hydrogen by adding protons $(\mathsf{H}^{+})$ . 14 protons need to be added to the left side of the chromium reaction to balance the 14 (2 per water molecule * 7 water molecules) hydrogens. 2 protons need to be added to the right side of the other reaction.

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

Step 4: Balance the charge of each equation with electrons. The chromium reaction has $(14+) + (2-) = 12+$ on the left side and $(2 * 3+) = 6+$ on the right side. To balance, add 6 electrons (each with a charge of -1) to the left side:

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

For the other reaction, there is no charge on the left and $2+$ charge on the right. So add 2 electrons to the right side:

$$\mathrm{H}_2\mathrm{S} \rightarrow \mathrm{S} + 2\mathrm{H}^{+} + 2\mathrm{e}^{-}$$

Step 5: Scale the reactions so that the electrons are equal. The chromium reaction has 6e- and the other reaction has 2e-, so it should be multiplied by 3. This gives:

$$3^*[H_2S \rightarrow S + 2H^+ + 2e^-] \text{ then}$$ $$3H_2S \rightarrow 3S + 6H^+ + 6e^-$$ $$6e^- + 14H^+ + Cr_2O_7^{2-} \rightarrow 2Cr^{3+} + 7H_2O$$

Step 6: Add the reactions and cancel out common terms.

$$[3H_2S \rightarrow 3S + 6H^+ + 6e^-] +$$ $$[6e^- + 14H^+ + Cr_2O_7^{2-} \rightarrow 2Cr^{3+} + 7H_2O]$$ $$3H_2S + 6e^- + 14H^+ + Cr_2O_7^{2-} \rightarrow 3S + 6H^+ + 6e^- + 2Cr^{3+} + 7H_2O$$

The electrons cancel out as well as 6 protons. This leaves the balanced reaction of:

**Answer:**

$$3H_2S + 8H^+ + Cr_2O_7^{2-} \rightarrow 3S + 2Cr^{3+} + 7H_2O$$

2. $MnO_2 + HCl \rightarrow Mn^{2+} + Cl_2 + H_2O$

$$MnO_2 + 4HCl \rightarrow Mn^{2+} + Cl_2 + 2H_2O$$

**Answer:**

$$MnO_2 + 4HCl \rightarrow MnCl_2 + Cl_2 + 2H_2O$$

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