To determine the standard cell notation for a galvanic cell made with aluminum (Al) and nickel (Ni), we need to follow the rules for writing cell notations:
1. Identify the anode and the cathode. In a galvanic cell, the anode is where oxidation (loss of electrons) occurs, and the cathode is where reduction (gain of electrons) occurs. Aluminum (Al) is more active than nickel (Ni), meaning Al will act as the anode, and Ni will act as the cathode.
2. Write the components of the anode on the left and the components of the cathode on the right, separated by a double vertical line ([tex]$||$[/tex]), which represents the salt bridge.
3. Within each half-cell, separate the solid and aqueous phases with a single vertical line ([tex]$|$[/tex]).
So, for an aluminum-nickel galvanic cell:
- Anode (oxidation): Aluminum (Al) metal will lose electrons to form aluminum ions ([tex]$Al^{3+}(aq)$[/tex]).
- Cathode (reduction): Nickel ions ([tex]$Ni^{2+}(aq)$[/tex]) will gain electrons to form nickel (Ni) metal.
Putting it all together:
- Anode side: [tex]$Al(s) \rightarrow Al^{3+}(aq) + 3e^-$[/tex]
- Cathode side: [tex]$Ni^{2+}(aq) + 2e^- \rightarrow Ni(s)$[/tex]
- Overall cell reaction: [tex]$2Al(s) + 3Ni^{2+}(aq) \rightarrow 2Al^{3+}(aq) + 3Ni(s)$[/tex]
The correct standard cell notation follows:
[tex]\[ Al(s) | Al^{3+}(aq) || Ni^{2+}(aq) | Ni(s) \][/tex]
So, the correct answer in this case is:
C. [tex]$Al (s) \left| Al ^{3+}(a q) \right| \left| Ni ^{2+}(a q) \right| Ni (s)$[/tex]
