We typically think of Hydrogen as having an oxidation number of +1. However when it is bonded to a less electronegative element such as Na it is actually assigned an oxidation number of -1.

Recall that oxidation always occurs at the anode (in both the electrochemical and galvanic cells). loses two electrons in this case to become . The presence of is hinted by the ionic compound .

Since pH > pKA, the deprotonated form of the acid is predominant.

Larger halogen size leads to greater acidity because of weaker H-X interactions.

Let's break down the reaction into two separate reactions:

and

We can see that copper loses electrons, while silver gains electrons. Recall that oxidation is loss and reduction is gain, with regard to electrons. Copper is oxidized and silver is reduced.

However, this question asks for the oxidizing agent and reducing agent. Recall that the oxidizing agent is reduced, while the reducing agent is oxidized. Since copper is oxidized, it is the reducing agent. Similarly, since silver is reduced, it is the oxidizing agent.

The Ka is the acid dissociation constant, and thus it is what determines how strong the acid is. Stronger acids dissociate to a greater extent and produce lower pH values.

Hydrogen doesn't change. Cu2+ doesn't change (partnered with S2- then with SO42-). Sulfur goes from S2- and S6+(paired with 6 O2- with a 2– charge), showing an oxidation. Nitrogen goes from N5+ to N2+ meaning it was reduced.

The common factor between 2 e- and 5 e- is 10. Therefore the number of electrons involved is 10 e-.

The pH at half the equivalence point is equal to the pKa of the acid.

Larger halogen size leads to greater acidity because of weaker H-X interactions.