When the following redox equation is balanced in acidic solution what is the coefficient in front of H+(aq)? Ag(s) + NO3-(aq) → Ag+(aq) + NO2(g)

Question

When the following redox equation is balanced in acidic solution what is the coefficient in front of H⁺_(aq)?

Ag_(s)  + NO₃^-_(aq)  → Ag⁺_(aq)  + NO_2(g)

Answer 1

Hello Holly

Separate the entire chemical reaction into 2 half reactions.

Reaction #1

Ag (s) --> Ag+ (aq)

There are no oxygens involved with this reaction, therefore, you only need to balance the charges. Charge on the left side is zero, but +1 on the right. Add one electron to the right to cancel the +1 charge and get a net charge of zero.

Ag (s) --> Ag+ (aq) + e-

Reaction #2

NO3- (aq) --> NO2 (g)

Balance the oxygens: 3 on the reactant side and 2 on the product side. Add one water molecule to the product side to give 3 oxygens as well.

NO3- (aq) --> NO2 (g) + H2O (l)

Now balance the hydrogens by adding H+ on the left side since it's in an acidic solution. 2 H's on the right, so add 2H+ to the left.

2H+(aq) + NO3- (aq) --> NO2 (g) + H2O (l)

Balance the charges next. +1 on the left, zero on the right. Add one electron to the left for the charges to equal zero.

e- + 2H+(aq) + NO3- (aq) --> NO2 (g) + H2O (l)

Add both half reactions.

Ag (s) --> Ag+ (aq) + e-

e- + 2H+(aq) + NO3- (aq) --> NO2 (g) + H2O (l)

Electrons must be on opposite sides of the equation and they must have the same coefficient in order to cancel. In both half reactions, we have only one electron so both are balanced.

Net reaction is:

Ag (s) + 2H+(aq) + NO3- (aq) --> Ag+ (aq) + NO2 (g) + H2O (l)