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.
Concentration of KCl solution (mol/L) | Conductivity at 298.15 K (S cm-1) | Molar Conductivity at 298.15 K (S cm2 mol-1) |
---|---|---|
1.000 | 0.1113 | 111.3 |
0.100 | 0.0129 | 129.0 |
0.010 | 0.00141 | 141.0 |