Problem 72 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The standard reduction potential of \(\mathrm{Cu}^{++} / \mathrm{Cu}\) and \(\mathrm{Ag}^{+} / \mathrm{Ag}\) electrodes are \(0.337\) and \(0.799\) volt respectively. Construct a galvanic cell using these electrodes so that its standard e.m.f. is positive. For what concentration of \(\mathrm{Ag}^{+}\)will the e.m.f. of the cell, at \(25^{\circ} \mathrm{C}\), be zero if the concentration of \(\mathrm{Cu}^{++}\)is \(0.01 \mathrm{M}\) ?

Step-by-Step Solution

Answer

For the e.m.f. to be zero, \([\mathrm{Ag}^{+}]\) should be approximately \(10^{-7.32}\) M.

Step 1: Identify Half-Cell Reactions

The half-cell reaction for the silver electrode is \( \mathrm{Ag}^{+} + e^- \rightarrow \mathrm{Ag} \), with a reduction potential of 0.799 V. The half-cell reaction for the copper electrode is \( \mathrm{Cu}^{++} + 2e^- \rightarrow \mathrm{Cu} \), with a reduction potential of 0.337 V.

Step 2: Determine Cell Reaction

To create a galvanic cell with a positive standard e.m.f., we choose the silver reaction as the reduction half-reaction (cathode) and the copper reaction as the oxidation half-reaction (anode). The overall cell reaction becomes: \( 2\mathrm{Ag}^{+} + \mathrm{Cu} \rightarrow 2\mathrm{Ag} + \mathrm{Cu}^{++} \).

Step 3: Calculate Standard e.m.f.

The standard e.m.f. of the cell \( E^\circ_{\text{cell}} \) can be calculated using the formula \( E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \). Therefore, \( E^\circ_{\text{cell}} = 0.799 - 0.337 = 0.462 \) V.

Step 4: Apply Nernst Equation for Zero e.m.f.

The Nernst equation at \( 25^{\circ} \text{C} \) is given by: \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n}\log\left(\frac{[\mathrm{Cu}^{++}]}{[\mathrm{Ag}^{+}]^2}\right) \] We want \( E_{\text{cell}} = 0 \), so: \[ 0 = 0.462 - \frac{0.0591}{2}\log\left(\frac{0.01}{[\mathrm{Ag}^{+}]^2}\right) \].

Step 5: Solve for \\([\mathrm{Ag}^{+}]\\)

Rearrange the equation: \[ \frac{0.0591}{2}\log\left(\frac{0.01}{[\mathrm{Ag}^{+}]^2}\right) = 0.462 \]. Solving for \( \log\left(\frac{0.01}{[\mathrm{Ag}^{+}]^2}\right) \) gives \( \log\left(\frac{0.01}{[\mathrm{Ag}^{+}]^2}\right) = 15.64 \). Therefore, \( \frac{0.01}{[\mathrm{Ag}^{+}]^2} = 10^{15.64} \).

Step 6: Calculate \\([\mathrm{Ag}^{+}]\\) Concentration

Calculating \( [\mathrm{Ag}^{+}] \), we find \( [\mathrm{Ag}^{+}]^2 = \frac{0.01}{10^{15.64}} \). Taking the square root, \([\mathrm{Ag}^{+}] \approx 10^{-7.32} \) M.

Key Concepts

Standard Reduction Potential

In electrochemistry, the standard reduction potential is crucial for understanding how different chemical species gain electrons in a reduction reaction. This potential measures a half-cell's ability to attract electrons and is expressed in volts.

• Each half-cell or half-reaction has a standard reduction potential, which serves as a baseline to compare against others. Known potentials make it easier to predict how new reactions will behave.
• The value of the standard reduction potential indicates a species' tendency to be reduced: the more positive the value, the greater the propensity for reduction.

To measure these potentials, a standard hydrogen electrode (SHE) is used as a reference point, defined to have a potential of 0 volts. For example, the half-cell reaction for silver ions, \(\mathrm{Ag}^{+} + e^- \rightarrow \mathrm{Ag} \)\, has a reduction potential of 0.799 V, signifying that silver ions readily accept electrons. In comparison, the copper half-cell reaction, \(\mathrm{Cu}^{++} + 2e^- \rightarrow \mathrm{Cu}\), shows a potential of 0.337 V.
These differences in potential help create voltages in electrochemical cells, forming the core mechanics behind galvanic cells.

Galvanic Cell

The galvanic cell is a vital component in electrochemistry, commonly used to convert chemical energy into electrical energy. It comprises two different electrodes immersed in electrolyte solutions, where spontaneous redox reactions occur.

• In a galvanic cell, the anode undergoes oxidation, losing electrons, while the cathode undergoes reduction, gaining electrons.
• The electron flow from anode to cathode through an external circuit generates electric current, which we harness as electricity.

When constructing a galvanic cell, it is crucial to select electrode reactions that result in a positive standard e.m.f (electromotive force).
For example, combining the half-cell reactions of silver, with a higher standard reduction potential, with copper, results in the overall reaction: \(2\mathrm{Ag}^{+} + \mathrm{Cu} \rightarrow 2\mathrm{Ag} + \mathrm{Cu}^{++}\). This configuration ensures a positive cell potential, allowing for efficient energy generation.

Nernst Equation

One of the most enriching equations in electrochemistry is the Nernst Equation. It predicts the cell potential under non-standard conditions by relating it to ion concentrations and temperature.

• The standard form of the Nernst equation at room temperature (\(25^{\circ} \text{C}\)) is:\[E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n}\log\left(\frac{[\text{Products}]}{[\text{Reactants}]}\right)\]
• Here, \(n\) refers to the number of electrons transferred in the reaction.

For the discussed galvanic cell, if we desire the e.m.f. to be zero, we have zero potential driving the reaction forward means equilibrium is achieved.
From the original step-by-step solution, we use the Nernst Equation to calculate for \([\mathrm{Ag}^{+}]\) concentration:
By setting \(E_{\text{cell}} = 0\) and calculating backward, we can determine the concentration needed for equilibrium, which corresponds to when our energy output is zero.
This demonstrates how real-world conditions differ from standard laboratory ones. The Nernst equation provides a highly useful tool for understanding and predicting such variations.