To solve the problem, we need to calculate the cell potential of the electrochemical cell formed by connecting the Sn$^{4+}$/Sn$^{2+}$ and Cr$^{3+}$/Cr couples in their standard states.
1. Understanding the Standard Electrode Potentials:
The standard electrode potential (E°) for the two couples is given as:
The cell potential (E°cell) is calculated by taking the difference between the electrode potentials of the two half-reactions. The couple with the more positive potential will undergo reduction, while the other will undergo oxidation.
2. Identifying the Anode and Cathode:
In this case: - Sn$^{4+}$ will be reduced to Sn$^{2+}$ because it has the more positive electrode potential (+0.15 V), so it will act as the cathode. - Cr will be oxidized to Cr$^{3+}$ because it has the more negative electrode potential (-0.74 V), so it will act as the anode.
3. Calculating the Cell Potential:
The cell potential (E°cell) is given by the equation: E°cell = E°(cathode) - E°(anode)
Substituting the values: E°cell = (+0.15 V) - (-0.74 V)
E°cell = 0.15 V + 0.74 V = 0.89 V
Final Answer:
The cell potential will be 0.89 V.
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