Question

Balancing, oxidation-reduction reactions by half-reaction method. Can someone explain how to do this???

Answer 1

Answer: H2 + 2Ag+ -> 2Ag + 2H+

Step-by-step explanation:

H2 + Ag+ -> Ag + H+

Oxidation half reaction: H2 -> H+

H2 -> 2H+ + 2e-

Reduction half reaction: Ag+ -> Ag

Ag+ + e -> Ag

Balance the numbers of electrons in each half equations

H2 -> 2H+ + 2e- x1

Ag+ + e- -> Ag x2

H2 -> 2H+ + 2e-

2Ag+ + 2e- -> 2Ag

Combine both half equations

H2 + 2Ag+ + 2e- -> 2Ag + 2H+ + 2e-

Canceling out the equal number of electrons on both sides

H2 + 2Ag+ -> 2Ag + 2H+