A solution containing silver (I) ion (an oxidizing agent) is mixed with a solution containing iron (II) ion (a reducing agent).
a. What is the balanced equation?
b. If the contents of the reaction mixture described above are filtered, what substances, if any, would remain on the filter paper?
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For A, I can't think of any other product other than AgFe, which makes no sense.
Help? Thanks!
The oxidizing agent, Ag+, gained an electron and is reduced to Ag(s).
The reducing agent loses an electron and is oxidized from Fe^2+ to Fe^3+:
Ag^+(aq) + Fe^2+(aq) --> Ag(s) + Fe^3+(aq)
To balance the equation for this redox reaction, we first need to identify the oxidation states of each element involved.
In the given reaction, silver (I) ion (Ag⁺) acts as the oxidizing agent, meaning it is reduced. On the other hand, iron (II) ion (Fe²⁺) acts as the reducing agent, meaning it is oxidized.
Let's assign oxidation states:
Ag⁺ + Fe²⁺ ---> Ag⁰ + Fe³⁺
Now, let's balance the equation by ensuring that the number of atoms on each side of the equation is equal.
Ag⁺ + Fe²⁺ ---> Ag⁰ + Fe³⁺
Since there is only one Ag and one Fe atom on each side, the equation is balanced.
The balanced equation is:
2 Ag⁺ + Fe²⁺ ---> 2 Ag⁰ + Fe³⁺
Regarding part b of the question, if the reaction mixture is filtered, the substances that would remain on the filter paper depend on their solubility and physical properties.
- Silver (Ag⁰) is a solid and insoluble in water. It would, therefore, remain on the filter paper.
- Iron (III) ion (Fe³⁺) would likely remain in the solution as it is usually present as an ionic species dissolved in water.
So, if the reaction mixture is filtered, silver (Ag⁰) would be expected to remain on the filter paper, while the solution containing iron (III) ions would pass through the filter.