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.


Electricity is passed through an acidic solution of Cu$^{2+}$ till all the Cu$^{2+}$ was exhausted, leading to the deposition of 300 mg of Cu metal. However, a current of 600 mA was continued to pass through the same solution for another 28 minutes by keeping the total volume of the solution fixed at 200 mL. The total volume of oxygen evolved at STP during the entire process is ___ mL. (Nearest integer)
Given:
$\mathrm{Cu^{2+} + 2e^- \rightarrow Cu(s)}$
$\mathrm{O_2 + 4H^+ + 4e^- \rightarrow 2H_2O}$
Faraday constant = 96500 C mol$^{-1}$
Molar volume at STP = 22.4 L