To determine the standard cell potential for the given voltaic cell, we will follow these steps:
1. Identify the half-reactions and their standard reduction potentials (given):
- Cadmium (Cd) half-reaction:
[tex]\[
\mathrm{Cd}^{2+}(aq) + 2e^- \rightarrow \mathrm{Cd}(s) \quad [E^\circ_{\mathrm{Cd}^{2+}} = -0.40\, \text{V}]
\][/tex]
- Silver (Ag) half-reaction:
[tex]\[
\mathrm{Ag}^+(aq) + e^- \rightarrow \mathrm{Ag}(s) \quad [E^\circ_{\mathrm{Ag}^+} = +0.80\, \text{V}]
\][/tex]
2. Determine the oxidation and reduction half-reactions:
- The half-cell with the higher reduction potential will undergo reduction (cathode).
- The half-cell with the lower reduction potential will undergo oxidation (anode).
In this case:
- The silver (Ag) half-reaction has a higher standard reduction potential (+0.80 V). Therefore, it occurs at the cathode (reduction).
- The cadmium (Cd) half-reaction has a lower standard reduction potential (-0.40 V). Therefore, it occurs at the anode (oxidation).
3. Write the overall cell reaction:
At the anode (oxidation):
[tex]\[
\mathrm{Cd}(s) \rightarrow \mathrm{Cd}^{2+}(aq) + 2e^-
\][/tex]
At the cathode (reduction):
[tex]\[
\mathrm{Ag}^+(aq) + e^- \rightarrow \mathrm{Ag}(s)
\][/tex]
4. Calculate the standard cell potential (E°_cell):
The formula for the standard cell potential is:
[tex]\[
E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}
\][/tex]
5. Substitute the given values:
[tex]\[
E^\circ_{\text{cell}} = E^\circ_{\mathrm{Ag}^+} - E^\circ_{\mathrm{Cd}^{2+}}
\][/tex]
[tex]\[
E^\circ_{\text{cell}} = 0.80\, \text{V} - (-0.40\, \text{V})
\][/tex]
6. Perform the calculation:
[tex]\[
E^\circ_{\text{cell}} = 0.80\, \text{V} + 0.40\, \text{V} = 1.20\, \text{V}
\][/tex]
Therefore, the standard cell potential for this voltaic cell is [tex]\( +1.20\, \text{V} \)[/tex].
The correct answer is:
[tex]\[
\boxed{+1.20\,\text{V}}
\][/tex]
