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

1 Answer

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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