Question

When the following reaction is balanced in basic solution using the lowest possible whole number coefficients, the coefficient of OH- will be??

CN-(aq) + MnO4-(aq) --> CNO-(aq) + MnO2(s)

Answer 1

Balance equations:

1.Write half reactions

CN-(aq) --> CNO-(aq)

MnO4-(aq) -->  MnO2(s)

2. Balance elements other than O and H.

CN-(aq) --> CNO-(aq)

MnO4-(aq) -->  MnO2(s)

3: Add H2O to balance oxygen

CN-(aq) + H2O --> CNO-(aq)

MnO4-(aq) -->  MnO2(s) + 2H2O

4: Balance hydrogen with protons.

CN-(aq) + H2O --> CNO-(aq) + 2H+

MnO4-(aq) + 4H+ -->  MnO2(s) + 2H2O

5: Balance the charge with e-

CN-(aq) + H2O --> CNO-(aq) + 2H+ + 2e-

MnO4-(aq) + 4H+ + 3e- -->  MnO2(s) + 2H2O

6: Scale the reactions so that they have an equal amount of electrons. In this case, it is already done.

3 ( CN-(aq) + H2O --> CNO-(aq) + 2H+ + 2e-)

2 ( MnO4-(aq) + 4H+ + 3e- -->  MnO2(s) + 2H2O)

Get:

3CN-(aq) + 3H2O --> 3CNO-(aq) + 6H+ + 6e-

2 MnO4-(aq) + 8H+ + 6e- --> 2MnO2(s) + 4H2O

7: Add the reactions and cancel the electrons.

3CN-(aq) + 3H2O + 2 MnO4-(aq) + 8H+ + 6e --> 3CNO-(aq) + 6H+ + 6e- + 2MnO2(s) + 4H2O

Cancel E-

3CN-(aq) + 3H2O + 2 MnO4-(aq) + 8H+--> 3CNO-(aq) + 6H+ + 2MnO2(s) + 4H2O

Cancel H2O

3CN-(aq) + 2 MnO4-(aq) + 8H+--> 3CNO-(aq) + 6H+ + 2MnO2(s) + H2O

Cancel H+

3CN-(aq) + 2 MnO4-(aq) + 2H+--> 3CNO-(aq) + 2MnO2(s) + H2O

8: Add OH- to balance H+.

3CN-(aq) + 2 MnO4-(aq) + 2H+ + 2OH- --> 3CNO-(aq) + 2MnO2(s) + H2O + 2OH-

9: Combine OH- ions and H+ ions that are present on the same side to form water.

3CN-(aq) + 2 MnO4-(aq) + 2H2O(l) --> 3CNO-(aq) + 2MnO2(s) + H2O + 2OH-

10: Cancel common terms.

3CN-(aq) + 2 MnO4-(aq) + H2O(l) --> 3CNO-(aq) + 2MnO2(s) + 2OH-

Final!

3 CN-(aq) + 2 MnO4-(aq) + H2O(l) --> 3CNO-(aq) + 2 MnO2(s) + 2 OH-(aq)

Therefore, the OH- stoichiometric coefficient is 2