Everything Science Grade 12

4.5 CHAPTER 4. ELECTROCHEMICAL REACTIONS

Step 5 : Multiply each half reaction by a suitable number so that the

number of electrons released in the oxidationhalf reaction is made

equal to the number ofelectrons that are accepted in the reduction half

reaction.

We need to multiply thereduction half reactionby three so that

the number of electronson either side are balanced. This gives:

3S^2 −→ 3S + 6e−

Step 6 : Combine the two halfreactions to get a finalequation for the

overall reaction.

Cr 2 O^27 −+ 14H++ 3S^2 −→ 3S + 2Cr3++ 7H 2 O

Step 7 : Do a final check to make sure that the equation is balanced

Example 10: Balancing redox reactions in an alkaline medium

QUESTION

If ammonia solution isadded to a solution that contains cobalt(II) ions, a

complex ion is formed,called the hexaaminecobalt(II) ion (Co(NH 3 )2+ 6 ). In

a chemical reaction with hydrogen peroxide solution, hexaaminecobalt ions

are oxidised by hydrogen peroxide solution tothe hexaaminecobalt(III) ion

Co(NH 3 )3+ 6. Write a balanced equation for this reaction.

SOLUTION

Step 1 : Write down the oxidation half reaction

Co(NH 3 )2+ 6 → Co(NH 3 )3+ 6

Step 2 : Balance the number ofatoms on both sides ofthe equation.

The number of atoms are the same on both sides.

Step 3 : Once the atoms are balanced, check that the charges balance.

The charge on the left of the equation is +2, but the charge on

the right is +3. One electron must be added to the right hand side

to balance the charges in the equation.The halfreaction is now:

Co(NH 3 )2+ 6 → Co(NH 3 )3+ 6 + e−

Step 4 : Repeat the above steps, but this time using the reduction half

reaction.