Question #cc55e

1 Answer
May 8, 2016

Answer:

#4Mn^(2+)+5BiO_3^(+)+H_2O->4MnO_4^(-)+5Bi^(3+)+2H^(+)#

Explanation:

To balance the following redox equation:

#Mn^(2+)+BiO_3^(+)->MnO_4^(-)+Bi^(3+)#

we can split this reaction into two half equations:
Oxidation: #Mn^(2+)->MnO_4^(-)#

Reduction: #BiO_3^(+)->Bi^(3+)#

I am going to use the following rules:

  1. Balance the elements other than oxygen and hydrogen
  2. Balance oxygen using water #H_2O#
  3. Balance Hydrogen using #H^+#
  4. Balance the charge using electrons #e^-#

Oxidation: #Mn^(2+)+4H_2O->MnO_4^(-)+8H^(+)+5e^(-)color(red)(xx4#

Reduction: #BiO_3^(+)+6H^(+)+4e^(-)->Bi^(3+)+3H_2Ocolor(red)(xx5#

Redox: #4Mn^(2+)+5BiO_3^(+)+H_2O->4MnO_4^(-)+5Bi^(3+)+2H^(+)#

Here is a video that explains in details the balancing process:
Balancing Redox Reactions | Acidic Medium.