Question

2. Balance the first redox reaction in acidic solution and the second redox reaction in basic solution.

S + NO3- → SO2 + NO

Cr2O72- + Fe2+ -----> Cr3+ + Fe3+

Answer 1

1) we balance each half reactiosn seperately

S --> SO2 , to balance O we add H2O

S + 2H2O ---> SO2 , to balance H we add H+ since acidic

S + 2H2O --> SO2 + 4H+ , fnally we ad de- to balance charge

S + 2H2O ---> SO2 + 4H+ + 4e-   ..............(1)

now NO3- ---> NO

NO3 - + 4H+ + 3e- ----> NO + 2H2O .......(2)

we must add (1) (2) such that total e- canceels out , hence 3 (1) + 4(2) gives

3 S + 6H2O + 4NO3^-(aq) + 16H+ (aq) ------> 3SO2 +12 H+ + 4NO + 8H2O

3S(s) + 4NO3^-(aq) + 4H+ (aq) ---> 3SO2 (g) + 4NO (g) + 2H2O ( g) is final balanced eq

now second reacton

Cr2O72- ---> Cr3+

Cr2O72- + 14 H+ ----> 2Cr3+ + 7H2O ,, now we add same amount of OH- as H+ since we need to balance in basic solution

Cr2O7 ^2- + 14H+ + 14HO- ---> 2Cr3+ + 7H2O + 14OH-         ( H+ + OH- gives H2O )

Cr2O7^2- + 7H2O + 6e- ---> 2Cr3+ + 14OH- ...........(1)

Fe2+ ---> Fe3+ + e- ....................(2)

now ( 1) + 6(2) gives

Cr2O7^2-(aq) + 7H2O (l) + 6 Fe2+(aq) ----> 2Cr3+(aq) + 6 Fe3+(aq) + 14 OH- (aq)