To determine the coefficient of silver in the final, balanced equation for the given redox reaction, we need to balance the half-reactions for both oxidation and reduction.
First, let's write down the oxidation and reduction half-reactions:
1. The oxidation half-reaction:
[tex]\[ \text{Al (s)} \rightarrow \text{Al}^{3+} (aq) + 3e^{-} \][/tex]
2. The reduction half-reaction:
[tex]\[ \text{Ag}^{+} (aq) + e^{-} \rightarrow \text{Ag (s)} \][/tex]
To balance the overall redox reaction, we need to ensure that the electrons lost in the oxidation half-reaction are equal to the electrons gained in the reduction half-reaction.
The oxidation half-reaction involves the loss of 3 electrons by aluminum:
[tex]\[ \text{Al (s)} \rightarrow \text{Al}^{3+} (aq) + 3e^{-} \][/tex]
The reduction half-reaction involves the gain of 1 electron by each silver ion:
[tex]\[ \text{Ag}^{+} (aq) + e^{-} \rightarrow \text{Ag (s)} \][/tex]
To balance the electrons, we need to multiply the reduction half-reaction by 3 so that the number of electrons gained equals the number of electrons lost:
[tex]\[ 3[\text{Ag}^{+} (aq) + e^{-} \rightarrow \text{Ag (s)}] \][/tex]
This can be simplified to:
[tex]\[ 3\text{Ag}^{+} (aq) + 3e^{-} \rightarrow 3\text{Ag (s)} \][/tex]
Now, we can combine the oxidation and reduction half-reactions:
[tex]\[ \text{Al (s)} \rightarrow \text{Al}^{3+} (aq) + 3e^{-} \][/tex]
[tex]\[ 3\text{Ag}^{+} (aq) + 3e^{-} \rightarrow 3\text{Ag (s)} \][/tex]
By adding the two half-reactions together, the electrons cancel out:
[tex]\[ \text{Al (s)} + 3\text{Ag}^{+} (aq) \rightarrow \text{Al}^{3+} (aq) + 3\text{Ag (s)} \][/tex]
Therefore, the balanced equation is:
[tex]\[ \text{Al (s)} + 3\text{Ag}^{+} (aq) \rightarrow \text{Al}^{3+} (aq) + 3\text{Ag (s)} \][/tex]
The coefficient of silver (Ag) in the balanced equation is 3.
