To determine the half-reactions in a galvanic cell with Zn and Mg electrodes, we need to identify the oxidation and reduction processes occurring at each electrode.
1. Identify the anode and cathode:
- In a galvanic cell, the anode is where oxidation takes place (loss of electrons).
- The cathode is where reduction occurs (gain of electrons).
2. Determine the half-reactions:
- Zinc (Zn) and magnesium (Mg) are metals, and typically in their standard states, they will be oxidized to their ion forms in the anode.
- Mg has a higher reduction potential than Zn, meaning Mg will more likely undergo reduction in a galvanic cell.
3. Write the half-reactions:
- Anode (oxidation): This should involve zinc (Zn) since it will be oxidized. The half-reaction will be:
[tex]\[
Zn (s) \rightarrow Zn^{2+} (aq) + 2 e^-
\][/tex]
- Cathode (reduction): This should involve magnesium (Mg) ions being reduced. The half-reaction will be:
[tex]\[
Mg^{2+} (aq) + 2 e^- \rightarrow Mg (s)
\][/tex]
4. Match with the given choices:
- A. [tex]\( Zn^{2+} (aq) + 2 e^- \rightarrow Zn (s) \)[/tex] and [tex]\( Mg^{2+} (aq) + 2 e^- \rightarrow Mg (s) \)[/tex]
- B. [tex]\( Zn (s) \rightarrow Zn^{2+} (aq) + 2 e^- \)[/tex] and [tex]\( Mg^{2+} (aq) + 2 e^- \rightarrow Mg (s) \)[/tex]
- C. [tex]\( Zn^{2+} (aq) + 2 e^- \rightarrow Zn (s) \)[/tex] and [tex]\( Mg (s) \rightarrow Mg^{2+} (aq) + 2 e^- \)[/tex]
- D. [tex]\( Zn (s) \rightarrow Zn^{2+} (aq) + 2 e^- \)[/tex] and [tex]\( Mg (s) \rightarrow Mg^{2+} (aq) + 2 e^- \)[/tex]
From the analysis, the correct half-reactions for a galvanic cell with Zn and Mg electrodes are:
- Anode (oxidation): [tex]\( Zn (s) \rightarrow Zn^{2+} (aq) + 2 e^- \)[/tex]
- Cathode (reduction): [tex]\( Mg^{2+} (aq) + 2 e^- \rightarrow Mg (s) \)[/tex]
Thus, the correct answer is B:
[tex]\[
Zn (s) \rightarrow Zn^{2+} (aq) + 2 e^- \quad \text{and} \quad Mg^{2+} (aq) + 2 e^- \rightarrow Mg (s)
\][/tex]
