Problem 114 - A Complete Resource Book in Chemistry for JEE Main

Question

The standard reduction potential for \(\mathrm{Fe}^{2+} / \mathrm{Fe}\) and \(\mathrm{Sn}^{2+} /\) Sn electrodes are \(-0.44\) and \(-0.14\) volts respectively. For the cell reaction \(\mathrm{Fe}^{2+}+\mathrm{Sn} \longrightarrow \mathrm{Fe}+\mathrm{Sn}^{2+}\) The standard emf is (a) \(+0.30 \mathrm{~V}\) (b) \(-0.58 \mathrm{~V}\) (c) \(+0.58 \mathrm{~V}\) (d) \(-0.300 \mathrm{~V}\)

Step-by-Step Solution

Answer

The standard emf is \(-0.30 \text{ V}\) which is option (d).

Step 1: Understand the Given Data and Reactions

The problem provides the standard reduction potential for two half-reactions: \( \mathrm{Fe}^{2+} / \mathrm{Fe} \) is \(-0.44\, \text{V}\) and \( \mathrm{Sn}^{2+} / \mathrm{Sn} \) is \(-0.14\, \text{V}\). We need to determine the standard electromotive force (emf) for the reaction: \( \mathrm{Fe}^{2+} + \mathrm{Sn} \rightarrow \mathrm{Fe} + \mathrm{Sn}^{2+} \).

Step 2: Write the Half-Reaction Equations

From the given reaction, identify the oxidation and reduction half-reactions.- \( \mathrm{Sn} \rightarrow \mathrm{Sn}^{2+} + 2e^- \) is the oxidation half-reaction.- \( \mathrm{Fe}^{2+} + 2e^- \rightarrow \mathrm{Fe} \) is the reduction half-reaction.

Step 3: Use the Formula for Standard EMF Calculation

The standard emf of the cell \(E^\circ_{\text{cell}}\) is calculated using the formula:\[E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}\]Here, \( \mathrm{Fe}^{2+} / \mathrm{Fe} \) is the cathode (\text{reduction}) and \( \mathrm{Sn}/\mathrm{Sn}^{2+} \) is the anode (\text{oxidation}).

Step 4: Substitute the Values into the Formula

Substitute the standard reduction potentials into the formula:\[ E^\circ_{\text{cell}} = (-0.44\, \text{V}) - (-0.14\, \text{V}) \]

Step 5: Calculate the Standard EMF

Simplify the equation by solving the subtraction:\[ E^\circ_{\text{cell}} = -0.44 \text{ V} + 0.14 \text{ V} = -0.30 \text{ V} \]

Step 6: Choose the Correct Answer from Options

The calculated standard emf is \(-0.30 \text{ V}\), which corresponds to option (d).

Key Concepts

Standard Electrode Potential

In the world of electrochemistry, understanding standard electrode potential is crucial. This potential indicates the capability of an electrode to gain electrons (also known as reduction) when compared to the standard hydrogen electrode, which has a potential of zero volts
.Think of it as a measurement of the tendency of a species to be reduced. Each half-cell has its own standard reduction potential, and it helps predict the direction of electron flow in electrochemical cells
.

• If an electrode has a more positive potential, it's more likely to undergo reduction.
• If it's more negative, it tends to oxidize.

In our problem, the standard reduction potentials are given for two half-reactions:

• The potential for the transition from \( \mathrm{Fe}^{2+} / \mathrm{Fe} \) is \(-0.44\, \text{V}\).
• The transition for \( \mathrm{Sn}^{2+} / \mathrm{Sn} \) stands at \(-0.14\, \text{V}\).

These values assist us in calculating the cell's overall potential, which predicts the feasibility and direction of the spontaneous reaction.

Half-Cell Reactions

Half-cell reactions are a fundamental component of electrochemical cells. Each electrochemical cell comprises two half-cells where oxidation and reduction occur separately
.In this exercise, we are handling two half-cells:

• The oxidation half-reaction is \( \mathrm{Sn} \rightarrow \mathrm{Sn}^{2+} + 2\mathrm{e}^- \). Here, tin is losing electrons, increasing its oxidation state.
• The reduction half-reaction is \( \mathrm{Fe}^{2+} + 2\mathrm{e}^- \rightarrow \mathrm{Fe} \). Here, iron gains electrons and gets reduced.

These reactions show how electrons are moved from one half-cell to another, which is the essence of generating electricity in an electrochemical cell. Recognizing which species undergoes oxidation and which undergoes reduction is key to understanding how a cell operates energetically and electronically.

Electromotive Force (EMF)

The electromotive force (EMF) of a cell is the driving force that pushes electrons through the external circuit. It represents the energy difference per charge between two electrodes and can be thought of as the 'battery' power of the electrochemical cell
.To compute the EMF for a cell, use this essential formula:\[E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}\]In essence:

• The cathode is the electrode at which reduction occurs; here, it's the \( \mathrm{Fe}^{2+} / \mathrm{Fe} \) system.
• The anode is where oxidation happens; in this problem, it's the \( \mathrm{Sn}/\mathrm{Sn}^{2+} \) system.

By substituting the given potentials into the formula:\[ E^\circ_{\text{cell}} = (-0.44\, \text{V}) - (-0.14\, \text{V}) = -0.30 \text{ V}\]This calculation shows the system's standard EMF, emphasizing how the cell functions energetically. Understanding EMF is vital in predicting whether reactions will occur spontaneously and in designing batteries and galvanic cells.