Answer :
To determine the standard cell notation for a galvanic cell made with zinc (Zn) and gold (Au), let's follow these steps:
1. Identify the half-reactions:
- Oxidation (anode reaction): This is where the metal loses electrons (is oxidized).
- Zinc (Zn) is more reactive than gold (Au), meaning zinc will be the one that gets oxidized.
- The oxidation half-reaction for zinc is: \( \text{Zn(s)} \rightarrow \text{Zn}^{2+}(\text{aq}) + 2e^- \)
- Reduction (cathode reaction): This is where the ions gain electrons (are reduced).
- Gold ions will be reduced because zinc is oxidized.
- The reduction half-reaction for gold ions is: \( \text{Au}^+(\text{aq}) + e^- \rightarrow \text{Au(s)} \)
2. Construct the cell notation:
Standard cell notation is written as:
\( \text{Anode (s)} | \text{Anode ion (aq)} || \text{Cathode ion (aq)} | \text{Cathode (s)} \)
- From the oxidation half-reaction, the anode is zinc and the notation for the anode part is: \( \text{Zn(s)} | \text{Zn}^{2+}(\text{aq}) \)
- From the reduction half-reaction, the cathode is gold and the notation for the cathode part is: \( \text{Au}^+(\text{aq}) | \text{Au(s)} \)
3. Combine the parts:
Using the standard form, combine the anode and cathode parts of the cell notation:
[tex]\[ \text{Zn(s)} | \text{Zn}^{2+}(\text{aq}) || \text{Au}^+(\text{aq}) | \text{Au(s)} \][/tex]
By carefully combining these components, we find the correct standard cell notation.
Given the options:
A. \( \text{Au}^+(\text{aq}) | \text{Au(s)} || \text{Zn(s)} | \text{Zn}^{2+}(\text{aq}) \)
B. \( \text{Zn(s)} | \text{Zn}^{2+}(\text{aq}) || \text{Au}^+(\text{aq}) | \text{Au(s)} \)
C. \( \text{Zn}^{2+}(\text{aq}) | \text{Zn(s)} || \text{Au(s)} | \text{Au}^+(\text{aq}) \)
D. \( \text{Au(s)} | \text{Au}^+(\text{aq}) || \text{Zn}^{2+}(\text{aq}) | \text{Zn(s)} \)
The correct notation is:
B. \( \text{Zn(s)} | \text{Zn}^{2+}(\text{aq}) || \text{Au}^+(\text{aq}) | \text{Au(s)} \)
Therefore, the answer is B.
1. Identify the half-reactions:
- Oxidation (anode reaction): This is where the metal loses electrons (is oxidized).
- Zinc (Zn) is more reactive than gold (Au), meaning zinc will be the one that gets oxidized.
- The oxidation half-reaction for zinc is: \( \text{Zn(s)} \rightarrow \text{Zn}^{2+}(\text{aq}) + 2e^- \)
- Reduction (cathode reaction): This is where the ions gain electrons (are reduced).
- Gold ions will be reduced because zinc is oxidized.
- The reduction half-reaction for gold ions is: \( \text{Au}^+(\text{aq}) + e^- \rightarrow \text{Au(s)} \)
2. Construct the cell notation:
Standard cell notation is written as:
\( \text{Anode (s)} | \text{Anode ion (aq)} || \text{Cathode ion (aq)} | \text{Cathode (s)} \)
- From the oxidation half-reaction, the anode is zinc and the notation for the anode part is: \( \text{Zn(s)} | \text{Zn}^{2+}(\text{aq}) \)
- From the reduction half-reaction, the cathode is gold and the notation for the cathode part is: \( \text{Au}^+(\text{aq}) | \text{Au(s)} \)
3. Combine the parts:
Using the standard form, combine the anode and cathode parts of the cell notation:
[tex]\[ \text{Zn(s)} | \text{Zn}^{2+}(\text{aq}) || \text{Au}^+(\text{aq}) | \text{Au(s)} \][/tex]
By carefully combining these components, we find the correct standard cell notation.
Given the options:
A. \( \text{Au}^+(\text{aq}) | \text{Au(s)} || \text{Zn(s)} | \text{Zn}^{2+}(\text{aq}) \)
B. \( \text{Zn(s)} | \text{Zn}^{2+}(\text{aq}) || \text{Au}^+(\text{aq}) | \text{Au(s)} \)
C. \( \text{Zn}^{2+}(\text{aq}) | \text{Zn(s)} || \text{Au(s)} | \text{Au}^+(\text{aq}) \)
D. \( \text{Au(s)} | \text{Au}^+(\text{aq}) || \text{Zn}^{2+}(\text{aq}) | \text{Zn(s)} \)
The correct notation is:
B. \( \text{Zn(s)} | \text{Zn}^{2+}(\text{aq}) || \text{Au}^+(\text{aq}) | \text{Au(s)} \)
Therefore, the answer is B.
