Problem 104 - A Complete Resource Book in Chemistry for JEE Main

Question

Two electrochemical cells \(\mathrm{Zn}\left|\mathrm{Zn}^{2+} \| \mathrm{Cu}^{2+}\right| \mathrm{Cu}\) and \(\mathrm{Fe}\left|\mathrm{Fe}^{2+} \| \mathrm{Cu}^{2+}\right| \mathrm{Cu}\) are con- nected in series. What will be the net emf of the cell at \(25^{\circ} \mathrm{C} ?\) Given: \(\mathrm{Zn}^{2+} \mid \mathrm{Zn}=-0.73 \mathrm{~V}\), \(\mathrm{Cu}^{2+} \mid \mathrm{Cu}=+0.34 \mathrm{~V}\) \(\mathrm{Fe}^{2+} \mid \mathrm{Fe}=-0.41 \mathrm{~V}\) (a) \(+1.85\) (b) \(-1.85 \mathrm{~V}\) (c) \(+0.83 \mathrm{~V}\) (d) \(-0.83 \mathrm{~V}\)

Step-by-Step Solution

Answer

The net EMF is approximately +1.82 V. Closest to option (a) +1.85 V.

Step 1: Identify cell reactions and potentials

For the cell \( \mathrm{Zn}|\mathrm{Zn}^{2+} \| \mathrm{Cu}^{2+}| \mathrm{Cu} \), the standard electrode potentials are: \( \mathrm{Zn}^{2+} /\mathrm{Zn} = -0.73 \mathrm{~V} \) and \( \mathrm{Cu}^{2+} /\mathrm{Cu} = +0.34 \mathrm{~V} \). For the cell \( \mathrm{Fe}|\mathrm{Fe}^{2+} \| \mathrm{Cu}^{2+}| \mathrm{Cu} \), the standard electrode potentials are: \( \mathrm{Fe}^{2+} /\mathrm{Fe} = -0.41 \mathrm{~V} \) and \( \mathrm{Cu}^{2+} /\mathrm{Cu} = +0.34 \mathrm{~V} \).

Step 2: Calculate EMF of the first cell

The standard EMF of the first cell \( \mathrm{Zn}|\mathrm{Zn}^{2+} \| \mathrm{Cu}^{2+}| \mathrm{Cu} \) is calculated as: \( E_{\text{cell 1}} = E_{\text{cathode}} - E_{\text{anode}} = 0.34 \mathrm{~V} - (-0.73 \mathrm{~V}) = 1.07 \mathrm{~V} \).

Step 3: Calculate EMF of the second cell

The standard EMF of the second cell \( \mathrm{Fe}|\mathrm{Fe}^{2+} \| \mathrm{Cu}^{2+}| \mathrm{Cu} \) is calculated as: \( E_{\text{cell 2}} = E_{\text{cathode}} - E_{\text{anode}} = 0.34 \mathrm{~V} - (-0.41 \mathrm{~V}) = 0.75 \mathrm{~V} \).

Step 4: Calculate the net EMF of the series

When cells are connected in series, the net EMF is the sum of the individual EMFs. Therefore, the net EMF is: \( E_{\text{net}} = E_{\text{cell 1}} + E_{\text{cell 2}} = 1.07 \mathrm{~V} + 0.75 \mathrm{~V} = 1.82 \mathrm{~V} \).

Step 5: Compare with answer choices

The calculated net EMF is \( +1.82 \mathrm{~V} \). Compare this value to the given options: (a) \(+1.85 \mathrm{~V}\), (b) \(-1.85 \mathrm{~V}\), (c) \(+0.83 \mathrm{~V}\), (d) \(-0.83 \mathrm{~V}\). None of the options exactly match \(+1.82\), indicating a potential miscalculation or typo in the options. The closest logical choice would be (a) \(+1.85 \mathrm{~V}\).

Key Concepts

Electrode Potentials

In electrochemical cells, the concept of electrode potentials is essential for understanding the direction and spontaneity of the reaction. An electrode potential, sometimes called a standard electrode potential, represents the voltage developed at an electrode when it is in equilibrium with its ions in a solution. This is a key parameter in determining how readily an element can donate or accept electrons.

Each half-cell in an electrochemical system like the Zn|Cu or Fe|Cu cells has its own standard electrode potentials. These values can be found in reference tables. For instance, the potential for the \( \mathrm{Zn}^{2+} / \mathrm{Zn} \) system is \(-0.73 \mathrm{~V} \), indicating it has a tendency to donate electrons and undergo oxidation. Meanwhile, \( \mathrm{Cu}^{2+} / \mathrm{Cu} \) is \(+0.34 \mathrm{~V} \), suggesting it is more likely to accept electrons and be reduced. These values help us identify which component of the cell will serve as the anode (oxidation occurs) and which as the cathode (reduction occurs).

• Potential values: Specify the ability of an element to act as a reducing or oxidizing agent.
• More negative potential: More likely for the substance to donate electrons (oxidation).
• More positive potential: More likely for the substance to accept electrons (reduction).

Understanding these potentials lets us calculate the cell's overall electromotive force (EMF), crucial for device applications like batteries.

Cell EMF Calculation

Calculating the EMF of a cell gives insight into the potential energy change that can be harnessed from the cell's redox reactions. For any galvanic or voltaic cell, the standard EMF is calculated using the formula: \[ E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} \]In this formula, \(E_{\text{cathode}}\) represents the reduction potential of the cathode, and \(E_{\text{anode}}\) denotes the reduction potential observed at the anode. For the zinc-copper cell, it would be: \[ E_{\text{cell}} = 0.34 \mathrm{~V} - (-0.73 \mathrm{~V}) = 1.07 \mathrm{~V} \]

• The higher positive value indicates that the reaction will proceed spontaneously under standard conditions.
• It reflects the energy possible from each mole of electrons transferred.
• Simple calculations help dictate which materials to use in practical applications.

Once you become familiar with electrode potentials and EMF concepts, understanding and designing efficient energy-converting systems becomes straightforward.

Series Connection of Cells

Connecting multiple electrochemical cells in series can increase the overall voltage output of a system, which is beneficial for powering larger devices.
When cells are connected in a series, the total EMF is the algebraic sum of the EMFs of the individual cells. For the zinc-copper and iron-copper cells considered, their EMFs are \(1.07 \mathrm{~V} \) and \(0.75 \mathrm{~V} \) respectively. Adding these gives a net EMF of: \[ E_{\text{net}} = 1.07 \mathrm{~V} + 0.75 \mathrm{~V} = 1.82 \mathrm{~V} \]

• Series connections augment the system's potential by stacking voltage contributions from each cell.
• Useful in devices requiring higher voltage outputs, like flashlights.
• Mixed cell types can be connected if compatible, but expect varying individual contributions.

This setup's advantage is amplified total energy output, though care must be exercised to avoid imbalances, which can lead to inefficiencies or reduced operational lifetime.