Balance the following redox reactions by ion – electron method

$$Cr_{ 2 }^{ 2- }O_{ 7 \; (aq) } \; + \; SO_{ 2 \; (g) } \; \rightarrow \; Cr^{ 3+ }_{ (aq) } \; + \; SO^{ 2- }_{ (aq) }$$

Asked by Pragya Singh | 1 year ago |  89

##### Solution :-

$$Cr_{ 2 }^{ 2- }O_{ 7 \; (aq) } \; + \; SO_{ 2 \; (g) } \; \rightarrow \; Cr^{ 3+ }_{ (aq) } \; + \; SO^{ 2- }_{ (aq) }$$

Step 1

Similar to (i), oxidation half reaction is:

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

Step 2

Reduction half reaction is:

$$Cr_{ 2 }O^{ 2- }_{ 7 \; (aq) } \; + \; 14 \; H^{ + }_{ (aq) } \; + \; 6 \; e^{ – } \;$$

$$\rightarrow \; 2 \; Cr_{ (aq) }^{ 3+ } \; + \; 7 \; H_{ 2 }O_{(l)}$$

Step 3

Multiply the oxidation half reaction with 2 then add it to the reduction half reaction. We get the balanced reaction as given below:

$$Cr_{ 2 }^{ 2- }O_{ 7 \; (aq) } \; + \;3 \; SO_{ 2 \; (g) } \; + \; 2 \; H^{ + }_{ (aq) } \;$$

$$\rightarrow \; 2 \; Cr_{ (aq) }^{ 3+ } \; + \;3 \; SO^{ 2- }_{ 4 \; (aq) } \;+ \; H_{ 2 }O_{(l)}$$

Answered by Abhisek | 1 year ago

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