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