A solution containing silver (I) ion (an oxidizing agent) is mixed with a solution containing iron (II) ion (a reducing agent).

If silver (I) is the oxidizing agent, then it will gain an electron and become Ag (s)

If Fe(II) is the reducing agent, then it loses an electron, and becomes Fe(III) ion.

a. Ag+ + Fe2+ = Ag + Fe3+

a. The balanced equation for the reaction between silver (I) ion and iron (II) ion can be written as follows:

2 Ag⁺(aq) + Fe²⁺(aq) → 2 Ag(s) + Fe³⁺(aq)

In this reaction, silver (I) ion (Ag⁺) is reduced to elemental silver (Ag), while iron (II) ion (Fe²⁺) is oxidized to iron (III) ion (Fe³⁺).

b. If the reaction mixture is filtered, the solid product formed (silver, Ag) will remain on the filter paper. Iron (III) ion (Fe³⁺) will stay in the solution and pass through the filter, as it is in the form of dissolved ions.

To determine the balanced equation for the reaction, let's break down the process step by step:

a. Balancing the equation:
The reaction involves the oxidation of silver(I) ion (Ag+) and the reduction of iron(II) ion (Fe2+). In order to balance the equation, we need to make sure that the number of atoms of each element is the same on both sides.

First, let's write the oxidation half-reaction:
Ag+ (aq) → Ag(s)

Next, let's write the reduction half-reaction:
Fe2+ (aq) → Fe(s)

Now, we can combine these two half-reactions to form the balanced equation:
2 Ag+ (aq) + Fe2+ (aq) → 2 Ag (s) + Fe(s)

b. Filtering the reaction mixture:
When the reaction mixture is filtered, the solid products formed (silver and iron) will remain on the filter paper, while the remaining substances (such as any unreacted solution or excess ions) will pass through the filter paper as a filtrate.

Therefore, the substances that would remain on the filter paper would be the solid silver (Ag) and solid iron (Fe).

To summarize:
a. The balanced equation for the reaction is:
2 Ag+ (aq) + Fe2+ (aq) → 2 Ag (s) + Fe(s)

b. The substances that would remain on the filter paper would be silver (Ag) and iron (Fe).