To solve for the cell potential of a voltaic cell made from copper (Cu) and silver (Ag), we need to use the standard reduction potentials ([tex]\( E^\circ \)[/tex]) provided. Here are the given reactions:
- [tex]\( Cu^{2+} + 2e^- \rightarrow Cu \, ( E^\circ = 0.52 \, \text{V} ) \)[/tex]
- [tex]\( Ag^+ + e^- \rightarrow Ag \, ( E^\circ = 0.80 \, \text{V} ) \)[/tex]
When constructing a voltaic cell, the cathode is the electrode where reduction occurs, and the anode is the electrode where oxidation occurs. The cell potential can be determined by taking the difference between the standard reduction potentials of the cathode and the anode.
1. Identify the role of each electrode:
- [tex]\( Ag^+ \)[/tex] is reduced to [tex]\( Ag \)[/tex], so silver acts as the cathode.
- [tex]\( Cu^{2+} \)[/tex] is reduced to [tex]\( Cu \)[/tex], but in a voltaic cell it will be oxidized from [tex]\( Cu \)[/tex] to [tex]\( Cu^{2+} \)[/tex], so copper acts as the anode.
2. Write down the reduction potential for the cathode:
- [tex]\( E^\circ_{\text{cathode}}(Ag^+ + e^- \rightarrow Ag) = 0.80 \, \text{V} \)[/tex]
3. Write down the reduction potential for the anode. Since the anode undergoes oxidation, we consider the oxidation potential. The reduction potential is given, so we take the negative of the reduction potential to get the oxidation potential of copper:
- [tex]\( E^\circ_{\text{anode}}(Cu \rightarrow Cu^{2+} + 2e^-) = -0.52 \, \text{V} \)[/tex]
4. Calculate the cell potential:
The electrical potential of the cell is given by the difference between the cathode and anode potentials:
[tex]\[
E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}
\][/tex]
Substituting the values:
[tex]\[
E^\circ_{\text{cell}} = 0.80 \, \text{V} - 0.52 \, \text{V}
\][/tex]
5. Perform the subtraction:
[tex]\[
E^\circ_{\text{cell}} = 0.28 \, \text{V}
\][/tex]
Therefore, the cell potential for the voltaic cell made from these two metals is:
[tex]\[
\boxed{0.28 \, \text{V}}
\][/tex]
So, the correct answer is:
B) 0.28 V
