Question

In this activity, you will practice balancing overall redox reactions in acidic and basic solutions. To do this, you will apply the half-reaction method for balancing redox equations.

You may reference this Periodic Table and these steps to help complete the problems. Record all of your solutions.

Part A
NH4+ + MnO4- → NO3- + Mn2+ (acidic solution)

Part B
Cr(OH)3 + ClO3- → CrO42- + Cl- (basic solution)

Answer 1

• Part A : The given unbalanced redox reaction is,

`NH^+_4+MnO^-_4\rightarrow NO^-_3+Mn^(2+)`

Now we have to balance the given redox reaction in acidic medium by half reaction method.

1st half reaction :
`8H^++MnO^-_4+5e^-\rightarrow Mn^(2+)+4H_2O`

2nd half reaction :
`3H_2O+NH^+_4\rightarrow NO^-_3+10H^++8e^-`

For net balanced reaction, 1st half reaction is multiplying by 8 and 2nd half reaction is multiplying by 5

The net balanced redox reaction is,

`8MnO^-_4+14H^++5NH^+_4\rightarrow 8Mn^(2+)+17H_2O+5NO^-_3`

• Part B : The given unbalanced redox reaction is,

`Cr(OH)_3+ClO^-_3\rightarrow CrO^(2-)_4+Cl^-`

Now we have to balance the given redox reaction in basic medium by half reaction method.

1st half reaction :
`Cr(OH)_3+H_2O+5OH^-\rightarrow CrO^(2-)_4+5H_2O+3e^-`

2nd half reaction :
`ClO^-_3+6H_2O+6e^-\rightarrow Cl^-+3H_2O+6OH^-`

For net balanced reaction, 1st half reaction is multiplying by 2.

The net balanced redox reaction is,

`2Cr(OH)_3+ClO^-_3+4OH^-\rightarrow 2CrO^(2-)_4+Cl^-+5H_2O`