The given problem involves a galvanic cell with a specified cell notation: Tl$\vert^{Tl^+}_{_{(0.001M)}}\vert^{Cu^{2+}}_{_{(0.01M)}}$Cu. We are tasked with determining how the emf (electromotive force) of this cell could be increased. The correct answer is increasing the concentration of Cu$^{2+}$ ions.
Let's analyze why this choice is correct using the Nernst Equation, which describes how the emf of a cell changes with concentration. The general form of the Nernst Equation for a cell reaction is:
\(E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF} \ln Q\)
Where:
For the cell reaction:
\(\text{Tl}^+ + \text{Cu} \rightarrow \text{Tl} + \text{Cu}^{2+}\)
The reaction quotient \(Q\) can be expressed as:
\(Q = \frac{[\text{Tl}^+]}{[\text{Cu}^{2+}]}\)
To increase the emf of the cell (\(E_{\text{cell}}\)), we need to reduce the value of \(Q\). This can be done by increasing the concentration of the denominator (Cu$^{2+}$ ions) since \(Q\) is inversely proportional to the concentration of Cu$^{2+}$.
Now, let's rule out the other options:
Thus, the correct choice is increasing the concentration of Cu$^{2+}$ ions to increase the cell's emf.
The emf of the cell is given by the Nernst equation:
\[ E = E^\circ - \frac{0.0591}{n} \log \frac{\text{[Products]}}{\text{[Reactants]}}. \]
For the cell:
\[ \text{Tl|Tl}^+ \, || \, \text{Cu}^{2+}| \text{Cu}, \]
the cell reaction is:
\[ \text{Tl} + \text{Cu}^{2+} \rightarrow \text{Tl}^+ + \text{Cu}. \]
Effect of concentration: Increasing the concentration of Cu$^{2+}$ ions (product-side reactant) decreases the denominator in the reaction quotient, shifting the reaction to the right and increasing the emf.
Increasing the concentration of Tl$^+$ ions (reactant-side product) increases the numerator in the reaction quotient, decreasing the emf.
Decreasing concentrations of both ions would reduce the driving force of the reaction, lowering the emf.
Thus, the emf can be increased by: [Increasing the concentration of Cu$^{2+}$ ions (Option 4).]


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
The equivalent resistance between the points \(A\) and \(B\) in the given circuit is \[ \frac{x}{5}\,\Omega. \] Find the value of \(x\). 
Method used for separation of mixture of products (B and C) obtained in the following reaction is: 
In the following \(p\text{–}V\) diagram, the equation of state along the curved path is given by \[ (V-2)^2 = 4ap, \] where \(a\) is a constant. The total work done in the closed path is: 
Let \( ABC \) be a triangle. Consider four points \( p_1, p_2, p_3, p_4 \) on the side \( AB \), five points \( p_5, p_6, p_7, p_8, p_9 \) on the side \( BC \), and four points \( p_{10}, p_{11}, p_{12}, p_{13} \) on the side \( AC \). None of these points is a vertex of the triangle \( ABC \). Then the total number of pentagons that can be formed by taking all the vertices from the points \( p_1, p_2, \ldots, p_{13} \) is ___________.