10. (a) Balance the following redox equations: (i) $$ \mathrm{H}_{2} \mathrm{~S}+\mathrm{Cl}_{2} \mathrm{O}_{7}^{5} \rightarrow \mathrm{~S}+\mathrm{Cr}^{3+} $$ (acidic medium) (ii) $$ \mathrm{MnO}_{2}+\mathrm{HCl} \rightarrow \mathrm{Cr}^{2+}+\mathrm{Cl}_{2}+\mathrm{H}_{2} \mathrm{O} $$

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To balance the given redox equations, we will follow the half-reaction method. This involves separating the oxidation and reduction half-reactions, balancing them individually, and then combining them.

### (a)(i) Balance the equation: 
$$
\mathrm{H}_{2} \mathrm{~S} + \mathrm{Cl}_{2} \mathrm{O}_{7}^{5} \rightarrow \mathrm{~S} + \mathrm{Cr}^{3+}
$$

**Step 1: Identify oxidation and reduction.**
- **Oxidation:** $$ \mathrm{H}_{2} \mathrm{S} $$ is oxidized to $$ \mathrm{S} $$.
- **Reduction:** $$ \mathrm{Cl}_{2} \mathrm{O}_{7}^{5} $$ is reduced to $$ \mathrm{Cr}^{3+} $$.

**Step 2: Write half-reactions.**
- Oxidation half-reaction: 
  $$
  \mathrm{H}_{2} \mathrm{S} \rightarrow \mathrm{S}
  $$
- Reduction half-reaction: 
  $$
  \mathrm{Cl}_{2} \mathrm{O}_{7}^{5} \rightarrow \mathrm{Cr}^{3+}
  $$

**Step 3: Balance each half-reaction.**
1. **Oxidation half-reaction:**
   $$
   \mathrm{H}_{2} \mathrm{S} \rightarrow \mathrm{S} + 2 \mathrm{H}^{+} + 2 e^{-}
   $$

2. **Reduction half-reaction:**
   $$
   \mathrm{Cl}_{2} \mathrm{O}_{7}^{5} + 6 e^{-} + 14 \mathrm{H}^{+} \rightarrow 2 \mathrm{Cr}^{3+} + 7 \mathrm{H}_{2} \mathrm{O}
   $$

**Step 4: Equalize electrons and combine.**
- Multiply the oxidation half-reaction by 3 to equalize the electrons:
  $$
  3 \mathrm{H}_{2} \mathrm{S} \rightarrow 3 \mathrm{S} + 6 \mathrm{H}^{+} + 6 e^{-}
  $$

**Step 5: Combine the half-reactions:**
$$
3 \mathrm{H}_{2} \mathrm{S} + \mathrm{Cl}_{2} \mathrm{O}_{7}^{5} + 14 \mathrm{H}^{+} \rightarrow 3 \mathrm{S} + 2 \mathrm{Cr}^{3+} + 7 \mathrm{H}_{2} \mathrm{O}
$$

### (a)(ii) Balance the equation:
$$
\mathrm{MnO}_{2} + \mathrm{HCl} \rightarrow \mathrm{Cr}^{2+} + \mathrm{Cl}_{2} + \mathrm{H}_{2} \mathrm{O}
$$

**Step 1: Identify oxidation and reduction.**
- **Oxidation:** $$ \mathrm{Cl}^{-} $$ in $$ \mathrm{HCl} $$ is oxidized to $$ \mathrm{Cl}_{2} $$.
- **Reduction:** $$ \mathrm{MnO}_{2} $$ is reduced to $$ \mathrm{Mn}^{2+} $$.

**Step 2: Write half-reactions.**
- Oxidation half-reaction: 
  $$
  2 \mathrm{Cl}^{-} \rightarrow \mathrm{Cl}_{2} + 2 e^{-}
  $$
- Reduction half-reaction: 
  $$
  \mathrm{MnO}_{2} + 4 e^{-} + 4 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+} + 2 \mathrm{H}_{2} \mathrm{O}
  $$

**Step 3: Balance each half-reaction.**
1. **Oxidation half-reaction:**
   $$
   2 \mathrm{Cl}^{-} \rightarrow \mathrm{Cl}_{2} + 2 e^{-}
   $$

2. **Reduction half-reaction:**
   $$
   \mathrm{MnO}_{2} + 4 e^{-} + 4 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+} + 2 \mathrm{H}_{2} \mathrm{O}
   $$

**Step 4: Equalize electrons and combine.**
- Multiply the oxidation half-reaction by 2:
  $$
  4 \mathrm{Cl}^{-} \rightarrow 2 \mathrm{Cl}_{2} + 4 e^{-}
  $$

**Step 5: Combine the half-reactions:**
$$
\mathrm{MnO}_{2} + 4 \mathrm{Cl}^{-} + 4 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+} + 2 \mathrm{Cl}_{2} + 2 \mathrm{H}_{2} \mathrm{O}
$$

### Final Balanced Equations:
1. $$ 3 \mathrm{H}_{2} \mathrm{S} + \mathrm{Cl}_{2} \mathrm{O}_{7}^{5} + 14 \mathrm{H}^{+} \rightarrow 3 \mathrm{S} + 2 \mathrm{Cr}^{3+} + 7 \mathrm{H}_{2} \mathrm{O} $$
2. $$ \mathrm{MnO}_{2} + 4 \mathrm{Cl}^{-} + 4 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+} + 2 \mathrm{Cl}_{2} + 2 \mathrm{H}_{2} \mathrm{O} $$
The balanced equations are: 1. $$ 3 \mathrm{H}_{2} \mathrm{S} + \mathrm{Cl}_{2} \mathrm{O}_{7}^{5} + 14 \mathrm{H}^{+} \rightarrow 3 \mathrm{S} + 2 \mathrm{Cr}^{3+} + 7 \mathrm{H}_{2} \mathrm{O} $$ 2. $$ \mathrm{MnO}_{2} + 4 \mathrm{Cl}^{-} + 4 \mathrm{H}^{+} \rightarrow \mathrm{Mn}^{2+} + 2 \mathrm{Cl}_{2} + 2 \mathrm{H}_{2} \mathrm{O} $$

(a) Balance the following redox equations: (i) (acidic medium)
(ii)

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(a)(i) Balance the equation:

(a)(ii) Balance the equation:

Final Balanced Equations:

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