Problem 93 - A Complete Resource Book in Chemistry for JEE Main

Question

The emf of the cell \(\mathrm{Cu}(\mathrm{s})\left|\mathrm{Cu}^{2+}(\mathrm{aq}) \| \mathrm{Ag}^{+}(\mathrm{aq})\right| \mathrm{Ag}(\mathrm{s})\) \(E^{\circ} \mathrm{Cu} / \mathrm{Cu}^{2 \prime}=-0.34 \mathrm{~V} ; E^{\circ} \mathrm{Ag} / \mathrm{Ag}^{\prime}=-0.80 \mathrm{~V}\) (a) \(+0.46 \mathrm{~V}\) (b) \(-0.46 \mathrm{~V}\) (c) \(+1.14 \mathrm{~V}\) (d) \(-1.14 \mathrm{~V}\)

Step-by-Step Solution

Answer

The emf of the cell is +1.14 V, corresponding to option (c).

Step 1: Identify the Half-Reactions

The given cell consists of two half-reactions: the reduction of silver ions and the oxidation of copper metal. The half-reactions are:- Silver half-reaction: \( \mathrm{Ag}^{+} (\mathrm{aq}) + e^{-} \rightarrow \mathrm{Ag} (\mathrm{s}) \)- Copper half-reaction (reverse of reduction, hence oxidation): \( \mathrm{Cu} (\mathrm{s}) \rightarrow \mathrm{Cu}^{2+} (\mathrm{aq}) + 2e^{-} \)

Step 2: Use Standard Electrode Potentials

Given are the standard reduction potentials:- \( E^{\circ} \mathrm{Ag} / \mathrm{Ag}^{+} = +0.80 \mathrm{~V} \)- \( E^{\circ} \mathrm{Cu}^{2+} / \mathrm{Cu} = +0.34 \mathrm{~V} \)Note: We will reverse the sign of the copper half-reaction's potential since it is oxidizing. Then, \( E^{\circ} \mathrm{Cu} / \mathrm{Cu}^{2+} = -0.34 \mathrm{~V} \).

Step 3: Calculate the Cell Potential

The cell potential \( E^{\circ}_{\text{cell}} \) is determined by adding the reversed oxidation potential of copper and the reduction potential of silver:\[E^{\circ}_{\text{cell}} = E^{\circ} \mathrm{Ag} / \mathrm{Ag}^{+} - E^{\circ} \mathrm{Cu}^{2+} / \mathrm{Cu}\]Substitute the given potentials:\[E^{\circ}_{\text{cell}} = +0.80 \mathrm{~V} - (-0.34 \mathrm{~V}) = +0.80 \mathrm{~V} + 0.34 \mathrm{~V}\]Finally, we get:\[E^{\circ}_{\text{cell}} = +1.14 \mathrm{~V}\]

Step 4: Select the Correct Answer

Compare the calculated cell potential to the given options. The result, \(+1.14 \mathrm{~V}\), matches option (c).

Key Concepts

Electrode Potential

Electrode potential is a crucial concept in electrochemistry that describes the voltage or electrical potential difference between an electrode and its surrounding electrolytic solution. It's essentially the measure of the willingness of an electrode to lose or gain electrons. This is typically expressed in volts (V). The standard electrode potential is measured under standard conditions, which include a temperature of 25°C, and a pressure of 1 atm for gases, and all solutions at 1 M concentration.

Understanding electrode potential helps predict the direction of electron flow in electrochemical cells. If the electrode potential is negative, as in the case of copper with -0.34 V, it indicates a stronger tendency to lose electrons (oxidation). Conversely, a positive electrode potential, like silver's +0.80 V, suggests a tendency to gain electrons (reduction). This positive and negative potential difference is what drives the electron flow from one side of the cell to the other.

Half-Reactions

Half-reactions are integral to understanding electrochemical processes. They explain the two parts of an electrochemical reaction — oxidation and reduction. In our given cell, we have two specific half-reactions:

• The oxidation half-reaction: - Copper metal loses electrons to form copper ions, \( \mathrm{Cu} (\mathrm{s}) \rightarrow \mathrm{Cu}^{2+} (\mathrm{aq}) + 2e^{-} \).
• The reduction half-reaction: - Silver ions gain electrons to form silver metal, \( \mathrm{Ag}^{+} (\mathrm{aq}) + e^{-} \rightarrow \mathrm{Ag} (\mathrm{s}) \).

By separating these reactions, we can better understand which reactions are occurring at the anode (where oxidation occurs) and the cathode (where reduction occurs). The combination of these half-reactions gives us the full picture of electron exchange in the cell.

Cell Potential

Cell potential, also known as electromotive force (emf), is the measure of the electrical energy provided by an electrochemical cell. It is determined by the difference in electrode potentials between the cathode and anode.

The overall cell potential \( E^{\circ}_{\text{cell}} \) is calculated by combining the potentials from the two half-reactions. In our exercise:

• Silver's reduction potential is +0.80 V.
• Copper's oxidation potential is effectively the negative of its reduction potential, -0.34 V.

The cell potential can be calculated by adding these values: \[ E^{\circ}_{\text{cell}} = +0.80 \mathrm{~V} - (-0.34 \mathrm{~V}) = +0.80 \mathrm{~V} + 0.34 \mathrm{~V} = +1.14 \mathrm{~V}\]
This positive emf value indicates that the reaction is spontaneous under standard conditions.

Standard Reduction Potential

The standard reduction potential is a measure of the tendency of a chemical species to be reduced, and it helps in ranking elements according to their reactivity. It is usually measured relative to the standard hydrogen electrode, which is assigned a value of 0 volts.

For our focus in electrochemical cells, the standard reduction potential guides the understanding of the direction of electron flow.

• An element with a higher standard reduction potential, like silver (+0.80 V), is more likely to gain electrons compared to one with a lower potential.
• Copper, with a standard reduction potential of +0.34 V in its reduced form, indicates a lesser tendency to gain electrons.

By comparing the standard reduction potentials, we ensure that the cell's chemical reactions align correctly with electron flow, allowing us to calculate the correct cell potential. This comparison, thus, plays a vital role in designing and predicting the behavior of electrochemical cells.