Question

examples of acidic medium in chemistry
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Answer 1

Answer:

Example #1: ClO3¯ + SO2 ---> SO42¯ + Cl¯

Example #1: ClO3¯ + SO2 ---> SO42¯ + Cl¯Solution:

Example #1: ClO3¯ + SO2 ---> SO42¯ + Cl¯Solution:1) Split into unbalanced half-reactions:

Example #1: ClO3¯ + SO2 ---> SO42¯ + Cl¯Solution:1) Split into unbalanced half-reactions:ClO3¯ ---> Cl¯

Example #1: ClO3¯ + SO2 ---> SO42¯ + Cl¯Solution:1) Split into unbalanced half-reactions:ClO3¯ ---> Cl¯SO2 ---> SO42¯

Example #1: ClO3¯ + SO2 ---> SO42¯ + Cl¯Solution:1) Split into unbalanced half-reactions:ClO3¯ ---> Cl¯SO2 ---> SO42¯2) Balance the half-reactions:

Example #1: ClO3¯ + SO2 ---> SO42¯ + Cl¯Solution:1) Split into unbalanced half-reactions:ClO3¯ ---> Cl¯SO2 ---> SO42¯2) Balance the half-reactions:6e¯ + 6H+ + ClO3¯ ---> Cl¯ + 3H2O

Example #1: ClO3¯ + SO2 ---> SO42¯ + Cl¯Solution:1) Split into unbalanced half-reactions:ClO3¯ ---> Cl¯SO2 ---> SO42¯2) Balance the half-reactions:6e¯ + 6H+ + ClO3¯ ---> Cl¯ + 3H2O2H2O + SO2 ---> SO42¯ + 4H+ + 2e¯

Example #1: ClO3¯ + SO2 ---> SO42¯ + Cl¯Solution:1) Split into unbalanced half-reactions:ClO3¯ ---> Cl¯SO2 ---> SO42¯2) Balance the half-reactions:6e¯ + 6H+ + ClO3¯ ---> Cl¯ + 3H2O2H2O + SO2 ---> SO42¯ + 4H+ + 2e¯3) Make the number of electrons equal:

Example #1: ClO3¯ + SO2 ---> SO42¯ + Cl¯Solution:1) Split into unbalanced half-reactions:ClO3¯ ---> Cl¯SO2 ---> SO42¯2) Balance the half-reactions:6e¯ + 6H+ + ClO3¯ ---> Cl¯ + 3H2O2H2O + SO2 ---> SO42¯ + 4H+ + 2e¯3) Make the number of electrons equal:6e¯ + 6H+ + ClO3¯ ---> Cl¯ + 3H2O

Example #1: ClO3¯ + SO2 ---> SO42¯ + Cl¯Solution:1) Split into unbalanced half-reactions:ClO3¯ ---> Cl¯SO2 ---> SO42¯2) Balance the half-reactions:6e¯ + 6H+ + ClO3¯ ---> Cl¯ + 3H2O2H2O + SO2 ---> SO42¯ + 4H+ + 2e¯3) Make the number of electrons equal:6e¯ + 6H+ + ClO3¯ ---> Cl¯ + 3H2O6H2O + 3SO2 ---> 3SO42¯ + 12H+ + 6e¯ <--- multiplied through by a factor of 3

Example #1: ClO3¯ + SO2 ---> SO42¯ + Cl¯Solution:1) Split into unbalanced half-reactions:ClO3¯ ---> Cl¯SO2 ---> SO42¯2) Balance the half-reactions:6e¯ + 6H+ + ClO3¯ ---> Cl¯ + 3H2O2H2O + SO2 ---> SO42¯ + 4H+ + 2e¯3) Make the number of electrons equal:6e¯ + 6H+ + ClO3¯ ---> Cl¯ + 3H2O6H2O + 3SO2 ---> 3SO42¯ + 12H+ + 6e¯ <--- multiplied through by a factor of 34) Add the two half-reactions for the final answer:

Example #1: ClO3¯ + SO2 ---> SO42¯ + Cl¯Solution:1) Split into unbalanced half-reactions:ClO3¯ ---> Cl¯SO2 ---> SO42¯2) Balance the half-reactions:6e¯ + 6H+ + ClO3¯ ---> Cl¯ + 3H2O2H2O + SO2 ---> SO42¯ + 4H+ + 2e¯3) Make the number of electrons equal:6e¯ + 6H+ + ClO3¯ ---> Cl¯ + 3H2O6H2O + 3SO2 ---> 3SO42¯ + 12H+ + 6e¯ <--- multiplied through by a factor of 34) Add the two half-reactions for the final answer:ClO3¯ + 3H2O + 3SO2 ---> 3SO42¯ + Cl¯ + 6H+

Example #1: ClO3¯ + SO2 ---> SO42¯ + Cl¯Solution:1) Split into unbalanced half-reactions:ClO3¯ ---> Cl¯SO2 ---> SO42¯2) Balance the half-reactions:6e¯ + 6H+ + ClO3¯ ---> Cl¯ + 3H2O2H2O + SO2 ---> SO42¯ + 4H+ + 2e¯3) Make the number of electrons equal:6e¯ + 6H+ + ClO3¯ ---> Cl¯ + 3H2O6H2O + 3SO2 ---> 3SO42¯ + 12H+ + 6e¯ <--- multiplied through by a factor of 34) Add the two half-reactions for the final answer:ClO3¯ + 3H2O + 3SO2 ---> 3SO42¯ + Cl¯ + 6H+Note that items duplicated on each side were cancelled out. The duplicates are 6e¯, 3H2O, and 6H+