To determine the cell potential for the given electrochemical cell, let's analyze the provided standard electrode potentials step-by-step.
1. Identify the given standard electrode potentials:
- For the [tex]\( \text{Sn}^{4+}/\text{Sn}^{2+} \)[/tex] couple: [tex]\( E^\circ_{\text{Sn}} = +0.15 \, \text{V} \)[/tex]
- For the [tex]\( \text{Cr}^{3+}/\text{Cr} \)[/tex] couple: [tex]\( E^\circ_{\text{Cr}} = -0.73 \, \text{V} \)[/tex]
2. Determine which electrode potential will be the cathode and which will be the anode:
- The cathode is where the reduction takes place and has the more positive standard electrode potential.
- The anode is where oxidation takes place and has the more negative standard electrode potential.
3. Assign the roles:
- For the [tex]\( \text{Sn}^{4+}/\text{Sn}^{2+} \)[/tex] couple, [tex]\( E^\circ_{\text{Sn}} = +0.15 \, \text{V} \)[/tex], this is more positive, so it will be the cathode.
- For the [tex]\( \text{Cr}^{3+}/\text{Cr} \)[/tex] couple, [tex]\( E^\circ_{\text{Cr}} = -0.73 \, \text{V} \)[/tex], this is more negative, so it will be the anode.
4. Calculate the cell potential (E_cell):
- The cell potential is given by the difference between the cathode and anode potentials:
[tex]\[
E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}}
\][/tex]
- Substituting the given values:
[tex]\[
E_{\text{cell}} = (+0.15 \, \text{V}) - (-0.73 \, \text{V})
\][/tex]
[tex]\[
E_{\text{cell}} = 0.15 \, \text{V} + 0.73 \, \text{V}
\][/tex]
[tex]\[
E_{\text{cell}} = 0.88 \, \text{V}
\][/tex]
Therefore, the cell potential for the given electrochemical cell is:
[tex]\[
\boxed{+0.88 \, \text{V}}
\][/tex]
So, the correct answer is (A) [tex]\(+0.88 \, \text{V}\)[/tex].
