Question

Use the half-reaction method to balance the following equation in basic solution: CN^- + MnO_4^- Right arrow CNO^- + MnO_2

Answer 1

consider the given reaction

CN- + MnO4- ---> CNO- + MnO2

first consider the reduction reaction

MnO4- ---> MnO2

first balance the Mn atoms

now balance the oxygen atoms using H20

MnO4- ---> MnO2 + 2H20

now balance H atoms using H+

MnO4- + 4H+ ---> MnO2 + 2H20

now balance charge with electrons

MnO4- + 4H+ + 3e- ---> MnO2 + 2H20

Now

consider the oxidation reaction

CN- ---> CN0-

using the rules given above

CN- + H20 ---> CN0-

CN- + H20 ---> CNO- + 2H+

CN- + H20 ---> CNO- + 2H+ + 2e-

now

equate the electrons on both equations

2MnO4- + 8H+ + 6e- ---> 2MnO2 + 4H20

3CN- + 3H20 ---> 3CNO- + 6H+ + 6e-

cancel out the common terms

we get

2 Mn04- + 3CN- + 2H+ --> 2MnO2 + 3 CNO- + H20

since the solution is basic

add OH- on both sides to convert H+ into water

2 Mn04- + 3CN- + 2H+ + 2OH- --> 2MnO2 + 3 CNO- + H20 + 2OH-

2 Mn04- + 3CN- + 2H20 --> 2MnO2 + 3 CNO- + H20 + 2OH-

2 Mn04- + 3CN- + H20 --> 2MnO2 + 3 CNO- + 2 OH-

so

the balanced reaction is

2 Mn04- + 3CN- + H20 --> 2MnO2 + 3 CNO- + 2 OH-