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

Answer :

• 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,

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