Problem 179 - A Complete Resource Book in Chemistry for JEE Main

Question

For the redox reaction \(\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(0.1 \mathrm{M}) \longrightarrow \mathrm{Zn}^{2+}(1 \mathrm{M})+\mathrm{Cu}(\mathrm{s})\) Taking place in a cell, \(E_{\text {call }}^{\circ}\) is \(1.10\) volt. \(E_{\text {cell }}\) for the cell will be \((2.303 \mathrm{RT} / F=0.0591)\) [2003] (a) \(2.14 \mathrm{~V}\) (b) \(1.80 \mathrm{~V}\) (c) \(1.07 \mathrm{~V}\) (d) \(0.82 \mathrm{~V}\)

Step-by-Step Solution

Answer

The cell potential \( E_{\text{cell}} \) is \( 1.07 \mathrm{~V} \), option (c).

Step 1: Write the Nernst Equation

The Nernst equation for the cell reaction is given by: \[ E = E^{\circ} - \frac{0.0591}{n} \log \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \] where \( E^{\circ} \) is the standard cell potential, \( n \) is the number of moles of electrons transferred in the reaction, and \( \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \) is the reaction quotient.

Step 2: Identify the Number of Electrons Transferred

For the given reaction, \( \mathrm{Zn}(\mathrm{s}) + \mathrm{Cu}^{2+}(0.1 \mathrm{M}) \rightarrow \mathrm{Zn}^{2+}(1 \mathrm{M}) + \mathrm{Cu}(\mathrm{s}) \), the number of electrons \( n \) transferred is 2, as zinc goes from 0 to +2 oxidation state and copper goes from +2 to 0.

Step 3: Substitute Values into Nernst Equation

Substitute \( E^{\circ} = 1.10 \; \text{V} \), \( n = 2 \), \( [\text{Zn}^{2+}] = 1 \; \text{M} \), and \( [\text{Cu}^{2+}] = 0.1 \; \text{M} \) into the Nernst Equation: \[ E = 1.10 - \frac{0.0591}{2} \log \left( \frac{1}{0.1} \right) \]

Step 4: Calculate the Logarithmic Term

Calculate the logarithmic term: \[ \log \left( \frac{1}{0.1} \right) = \log(10) = 1 \]

Step 5: Compute the Nernst Equation

Plug the logarithmic result into the equation to get \[ E = 1.10 - \frac{0.0591}{2} \times 1 \] Simplifying, \[ E = 1.10 - 0.02955 \]

Step 6: Final Calculation

Perform the final calculation: \[ E = 1.07045 \]

Step 7: Determine the Closest Option

The closest option to the calculated value is \(1.07 \mathrm{~V}\) (option c).

Key Concepts

Redox Reaction

Redox reactions, short for reduction-oxidation reactions, involve the transfer of electrons between chemical species. In the reaction \[\mathrm{Zn} (\mathrm{s}) + \mathrm{Cu}^{2+} (0.1\, \mathrm{M}) \rightarrow \mathrm{Zn}^{2+} (1\, \mathrm{M}) + \mathrm{Cu} (\mathrm{s})\], zinc is oxidized, meaning it loses electrons, while copper ions are reduced by gaining electrons. This electron transfer is vital for the cell's function.
In a redox reaction, oxidation and reduction always occur simultaneously. Oxidation is the loss of electrons, whereas reduction refers to the gain of electrons.

• The species that loses electrons is said to be oxidized.
• The species that gains electrons is called the reducing agent.
• Balancing redox reactions is essential for precisely determining the number of electrons transferred.

Understanding these principles allows us to use redox reactions in electrochemical cells like batteries, where chemical energy is converted to electrical energy.

Standard Cell Potential

Standard cell potential, denoted as \( E^{\circ} \), quantifies the voltage or electromotive force (emf) of a cell under standard conditions. Standard conditions mean that all reactants and products are at 1 M concentration, gases are at 1 atm pressure, and the measurements occur at 25°C.
For the given reaction, \( E^{\circ} = 1.10 \text{ V} \). This value tells us how much energy per charge is transferred during the reaction when the cell operates in its standard state. A positive \( E^{\circ} \) indicates a spontaneous reaction.

• It is determined using a reference, often the standard hydrogen electrode (SHE), which is assigned a potential of 0.00 volts.
• Different metals and ions exhibit unique standard potentials, defining their ability to be oxidized or reduced.
• Understanding \( E^{\circ} \) helps in predicting the feasibility and directionality of reactions.

Reaction Quotient

The reaction quotient, \( Q \), is a measure of the relative amounts of reactants and products present during a reaction at any point in time. For the given reaction, it's expressed as \( \frac{[\mathrm{Zn}^{2+}]}{[\mathrm{Cu}^{2+}]} \). In the Nernst Equation, this represents the ratio of concentrations of products over reactants, which affects the actual cell potential. When the reaction is not at equilibrium, \( Q \) differs from the equilibrium constant \( K \). Comparing \( Q \) with \( K \) indicates the direction the reaction needs to proceed:

• If \( Q < K \), the reaction will proceed forward to produce more products.
• If \( Q > K \), the reverse reaction is favored.
• If \( Q = K \), the reaction is at equilibrium.

It's crucial in electrochemical cells since the cell potential changes with \( Q \). As in this scenario, \( Q = 10 \) because the concentrations of products versus reactants influence the energy change in the cell.

Oxidation State

The oxidation state or oxidation number of an atom in a compound provides valuable information about its electron control. In our redox reaction, zinc transitions from a metallic (0 oxidation state) to a cationic form (\( +2 \) oxidation state), while copper moves from \( +2 \) ion to its elemental state.
Understanding oxidation states is essential as it allows you to:

• Determine which element is oxidized and which is reduced.
• Balance redox reactions, especially complex ones.
• Predict the number of electrons transferred in a reaction (\( n \)), essential for using the Nernst Equation.

By analyzing these changes, we can easily track electron movement and ensure a systematic approach to balancing equations. Oxidation states offer insight into both molecular composition and redox activity.

Electrode Potential

Electrode potential refers to the ability of an element in a redox reaction to either gain or lose electrons. It's measured as the voltage between an electrode and a standard reference (like SHE). For any metal, this value indicates its tendency to oxidize or reduce.
The cell potential \( E \), relevant in this context, illustrates the flow of electrons from anode (oxidation takes place) to cathode (reduction occurs):

• The higher the electrode potential, the stronger the tendency to gain electrons and undergo reduction.
• The cell voltage reflects electron flow, influenced by the concentration of ions as shown by the Nernst Equation.

In essence, electrode potential captures the driving force behind electron motion, critical for calculating cell voltage and understanding electrochemical reactions. This value aids in assessing the efficiency and energy output of electrochemical cells efficiently.