Balance the redox reactions by ion – electron method :

$$MnO^{ – }_{ 4 \; (aq) } \; + \; SO_{ 2 \; (g) } \; \rightarrow \; Mn^{ 2+ }_{ (aq) } \; + \; H_{ 2 }SO_{ 4 }^{ – }$$

Asked by Pragya Singh | 1 year ago |  84

#### 1 Answer

##### Solution :-

$$MnO^{ – }_{ 4 \; (aq) } \; + \; SO_{ 2 \; (g) } \; \rightarrow \; Mn^{ 2+ }_{ (aq) } \; + \; H_{ 2 }SO_{ 4 }^{ – }$$

Step 1

Similar to (i), oxidation half reaction is:

$$SO_{ 2 \; (g) } \; + \; 2\; H_{ 2 }O_{ (l) } \; \rightarrow \; HSO^{ – }_{ 4 \; (aq) } \; + \; 3 \; H^{ + }_{ (aq) } \; + \; 2 \; e^{ – }_{ (aq) }$$

Step 2

Reduction half reaction is:

$$MnO^{ – }_{ 4 \; (aq) } \; + \; 8\; H^{ + }_{ (aq) } \; + \; 5 \; e^{ – } \; \rightarrow \; Mn^{ 2+ }_{ (aq) } \; + \; 4 \; H_{ 2 }O_{ (l) }$$

Step 3

Multiply the oxidation half reaction with 5 and the reduction half reaction with 2, then add them. We get the balanced reaction as given below:

$$2 \; MnO^{ – }_{ 4 \; (aq) } \; + \; 5 \; SO_{ 2 \; (g) } \; + \; 2\; H_{ 2 }O_{ (l) } \; +\; H^{ + }_{ (aq) } \;$$

$$\rightarrow \; 2 \; Mn^{ 2+ }_{ (aq) } \; + \; 5 \; HSO^{ – }_{ 4 \; (aq) }$$

Answered by Abhisek | 1 year ago

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