Question

Please show work:

How many electrons are transferred in the following redox reaction when it is balanced with the smallest possible integer coefficients? MnO4–(aq) + NH4+(aq) → Mn(OH)2(s) + N2H4(aq) (basic solution) [Note that H2O(l) and OH–(aq) may have to be added where necessary to balance the equation.]

Answer 1

First, define the “ACIDIC” solution/conditions as H+ presence and

Basic solution implies OH- once it is balanced.

Also; note that ALL species must be balanced, as well as charges

Typical steps:

1) split half redox cells

MnO4–(aq) + NH4+(aq) → Mn(OH)2(s) + N2H4(aq)

MnO4- = Mn(OH)2

NH4+ = N2H4

2) balance atoms other than O,H

MnO4- = Mn(OH)2

2NH4+ = N2H4

3) balance O by adding H2O

MnO4- = Mn(OH)2 + 2H2O

2NH4+ = N2H4

4) balance H by adding H+

6H+ + MnO4- = Mn(OH)2 + 2H2O

2NH4+ = N2H4 + 4H+

5) balance charge by adding e-

5e- + 6H+ + MnO4- = Mn(OH)2 + 2H2O

2NH4+ = N2H4 + 4H+ + 2e-

6) balance e- by multiplying by the Greatest common divisor

10e- + 12H+ + 2MnO4- = 2Mn(OH)2 + 4H2O

10NH4+ = 5N2H4 + 20H+ + 10e-

7) Add both equations

10NH4+ + 10e- + 12H+ + 2MnO4- = 2Mn(OH)2 + 4H2O+5N2H4 + 20H+ + 10e-

8) simplify repeating elements, H+, H2O, and e- typically

10NH4+ + 2MnO4- = 2Mn(OH)2 + 4H2O + 5N2H4 + 8H+

9) add OH- if we need basic media, otherwise this is done.

8OH- + 10NH4+ + 2MnO4- = 2Mn(OH)2 + 4H2O + 5N2H4 + 8H+ + 8OH-

10) Simplify again

8OH- + 10NH4+ + 2MnO4- = 2Mn(OH)2 + 4H2O + 5N2H4 + 8H2O

8OH- + 10NH4+ + 2MnO4- = 2Mn(OH)2 + 12H2O + 5N2H4  

we cancelled 10 electrons, so 10 electrons are being transferred