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.
The molar conductance of an infinitely dilute solution of ammonium chloride was found to be 185 S cm$^{-1}$ mol$^{-1}$ and the ionic conductance of hydroxyl and chloride ions are 170 and 70 S cm$^{-1}$ mol$^{-1}$, respectively. If molar conductance of 0.02 M solution of ammonium hydroxide is 85.5 S cm$^{-1}$ mol$^{-1}$, its degree of dissociation is given by x $\times$ 10$^{-1}$. The value of x is ______. (Nearest integer)
Consider the following half cell reaction $ \text{Cr}_2\text{O}_7^{2-} (\text{aq}) + 6\text{e}^- + 14\text{H}^+ (\text{aq}) \longrightarrow 2\text{Cr}^{3+} (\text{aq}) + 7\text{H}_2\text{O}(1) $
The reaction was conducted with the ratio of $\frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}]} = 10^{-6}$
The pH value at which the EMF of the half cell will become zero is ____ (nearest integer value)
[Given : standard half cell reduction potential $\text{E}^\circ_{\text{Cr}_2\text{O}_7^{2-}, \text{H}^+/\text{Cr}^{3+}} = 1.33\text{V}, \quad \frac{2.303\text{RT}}{\text{F}} = 0.059\text{V}$
| 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 |

| S. No. | Particulars | Amount (in ₹ crore) |
|---|---|---|
| (i) | Operating Surplus | 3,740 |
| (ii) | Increase in unsold stock | 600 |
| (iii) | Sales | 10,625 |
| (iv) | Purchase of raw materials | 2,625 |
| (v) | Consumption of fixed capital | 500 |
| (vi) | Subsidies | 400 |
| (vii) | Indirect taxes | 1,200 |