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Sagot :
Certainly! To determine the coefficient of silver in the balanced redox reaction, we'll need to follow these steps:
1. Identify the oxidation half-reaction and reduction half-reaction:
- Oxidation half-reaction (aluminum loses electrons):
[tex]\[ Al(s) \longrightarrow Al^{3+}(aq) + 3e^- \][/tex]
- Reduction half-reaction (silver ion gains electrons):
[tex]\[ Ag^+(aq) + e^- \longrightarrow Ag(s) \][/tex]
2. Balance the electrons lost in oxidation with the electrons gained in reduction:
- To have the same number of electrons in both half-reactions, the reduction half-reaction needs to be multiplied by the appropriate factor to balance the electrons. Since one [tex]\( Ag^+ \)[/tex] ion gains one electron in the reduction half-reaction, and [tex]\( Al \)[/tex] loses three electrons in the oxidation half-reaction, the reduction half-reaction should be multiplied by 3 to balance the total number of electrons.
[tex]\[ 3(Ag^+(aq) + e^- \longrightarrow Ag(s)) \][/tex]
Which simplifies to:
[tex]\[ 3Ag^+(aq) + 3e^- \longrightarrow 3Ag(s) \][/tex]
3. Combine the half-reactions to balance the overall equation:
- Now, add the balanced oxidation and reduction half-reactions together:
[tex]\[ Al(s) \longrightarrow Al^{3+}(aq) + 3e^- \][/tex]
[tex]\( + \)[/tex]
[tex]\[ 3Ag^+(aq) + 3e^- \longrightarrow 3Ag(s) \][/tex]
This will cancel out the electrons from both sides, leading to:
[tex]\[ Al(s) + 3Ag^+(aq) \longrightarrow Al^{3+}(aq) + 3Ag(s) \][/tex]
4. Determine the coefficient of silver (Ag) in the balanced equation:
- From the balanced equation above:
[tex]\[ Al(s) + 3Ag^+(aq) \longrightarrow Al^{3+}(aq) + 3Ag(s) \][/tex]
The coefficient of silver (Ag) is 3.
So, the coefficient of silver in the final balanced equation is [tex]\(\boxed{3}\)[/tex].
1. Identify the oxidation half-reaction and reduction half-reaction:
- Oxidation half-reaction (aluminum loses electrons):
[tex]\[ Al(s) \longrightarrow Al^{3+}(aq) + 3e^- \][/tex]
- Reduction half-reaction (silver ion gains electrons):
[tex]\[ Ag^+(aq) + e^- \longrightarrow Ag(s) \][/tex]
2. Balance the electrons lost in oxidation with the electrons gained in reduction:
- To have the same number of electrons in both half-reactions, the reduction half-reaction needs to be multiplied by the appropriate factor to balance the electrons. Since one [tex]\( Ag^+ \)[/tex] ion gains one electron in the reduction half-reaction, and [tex]\( Al \)[/tex] loses three electrons in the oxidation half-reaction, the reduction half-reaction should be multiplied by 3 to balance the total number of electrons.
[tex]\[ 3(Ag^+(aq) + e^- \longrightarrow Ag(s)) \][/tex]
Which simplifies to:
[tex]\[ 3Ag^+(aq) + 3e^- \longrightarrow 3Ag(s) \][/tex]
3. Combine the half-reactions to balance the overall equation:
- Now, add the balanced oxidation and reduction half-reactions together:
[tex]\[ Al(s) \longrightarrow Al^{3+}(aq) + 3e^- \][/tex]
[tex]\( + \)[/tex]
[tex]\[ 3Ag^+(aq) + 3e^- \longrightarrow 3Ag(s) \][/tex]
This will cancel out the electrons from both sides, leading to:
[tex]\[ Al(s) + 3Ag^+(aq) \longrightarrow Al^{3+}(aq) + 3Ag(s) \][/tex]
4. Determine the coefficient of silver (Ag) in the balanced equation:
- From the balanced equation above:
[tex]\[ Al(s) + 3Ag^+(aq) \longrightarrow Al^{3+}(aq) + 3Ag(s) \][/tex]
The coefficient of silver (Ag) is 3.
So, the coefficient of silver in the final balanced equation is [tex]\(\boxed{3}\)[/tex].
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