Is the reaction involving a proton transfer a redox reaction?

Question

$$ \begin{align} \ce{H2S (aq) + NH3 (aq) &→ NH4+ (aq) + HS- (aq)}\tag{1}\\ \ce{Pb(s) + 2 FeCl3 (aq) &→ 2FeCl2 (aq) + PbCl2 (aq)}\tag{2} \end{align} $$

Which of the following statements correctly describes both of the equations above?

A. Both equations represent redox reactions.
B. Equation 1 represents a redox reaction but equation 2 does not.
C. Equation 2 represents a redox reaction but equation 1 does not.
D. Neither equation represents a redox reaction.

The answer key says that the correct answer is C. I picked A. But in equation 1, $\ce{H2S}$ donates a hydrogen to the $\ce{NH3}$ which would mean it's a redox as well. Both equations had donations and acceptances so I really don't understand why equation 1 is not a redox reaction.

Answer 1

Donation of a proton is not an indication of a redox reaction, rather an acid-base reaction (which may or may not be also involving a redox process). The only way to show the reduction/oxidation takes place is to demonstrate that the oxidation numbers do change:

$$ \begin{align} \ce{\overset{+1}{H}_2\overset{-2}{S} (aq) + \overset{-3}{N}\overset{+1}{H}_3 (aq) &→ \overset{-3}{N}\overset{+1}{H}_4+(aq) + \overset{+1}{H}\overset{-2}{S}^-(aq)}\tag{1}\\ \ce{\overset{0}{Pb}(s) + 2 \overset{+3}{Fe}\overset{-1}{Cl}_3 (aq) &→ 2 \overset{+2}{Fe}\overset{-1}{Cl}_2 (aq) + \overset{+2}{Pb}\overset{-1}{Cl}_2 (aq)}\tag{2} \end{align} $$

As you can see, oxidation numbers only change in reaction (2), and are preserved in reaction (1).

Answer 2

It is C indeed.

In a REDOX reaction must exist an oxidizing agent and a reducing agent. This means that at least one atom must be reduced while another must be oxidized. In other words, the acceptance and donations must be of electrons (not of atoms), from one specie to another. In your case in the reaction B is very simple, you have:

$$\ce{Pb (s) -> Pb^2+ (aq) + 2 e-}$$

This is your oxidation reaction and therefore the $\ce{Pb}$ is your reducing agent.

While:

$$\ce{Fe^3+ (aq) + e- -> Fe^2+ (aq)}$$

The $\ce{Fe}$ is reduced from $\ce{Fe^3+ (aq)}$ to $\ce{Fe^2+ (aq)}$.

So the net reaction is:

$$\ce{Pb (s) + 2 Fe^3+ (aq) -> 2 Fe^2+ (aq) + Pb^2+ (aq)}$$

For the first reaction, there is no variation in the oxidation number of any atom, this means that there is no net electron transfer and thus no REDOX process.