Problem 140 - A Complete Resource Book in Chemistry for JEE Main

Question

The standard reduction potentials of \(\mathrm{Cu}^{2+} / \mathrm{Cu}\) and \(\mathrm{Cu}^{2+} /\) \(\mathrm{Cu}^{+}\)are \(0.337 \mathrm{~V}\) and \(0.153 \mathrm{~V}\) respectively. The standard electrode potential of \(\mathrm{Cu}^{+} /\)Cu half cell is (a) \(0.184 \mathrm{~V}\) (b) \(0.827 \mathrm{~V}\) (c) \(0.521 \mathrm{~V}\) (d) \(0.490 \mathrm{~V}\)

Step-by-Step Solution

Answer

The standard electrode potential for \(\mathrm{Cu}^{+} / \mathrm{Cu}\) is \(0.184 \mathrm{~V}\), option (a).

Step 1: Understand the Given Potentials

We are given the standard reduction potentials for the following reactions: \( \mathrm{Cu}^{2+} / \mathrm{Cu}\) is \(0.337 \mathrm{~V}\) and \(\mathrm{Cu}^{2+} / \mathrm{Cu}^{+}\) is \(0.153 \mathrm{~V}\). We need to find the standard electrode potential of \(\mathrm{Cu}^{+} / \mathrm{Cu}\).

Step 2: Set Up the Reaction Equations

We know the reduction equations: (1) \(\mathrm{Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu}\) with a potential of \(0.337 \mathrm{~V}\), and (2) \(\mathrm{Cu}^{2+} + e^- \rightarrow \mathrm{Cu}^{+}\) with a potential of \(0.153 \mathrm{~V}\).

Step 3: Find Potential for Cu+/Cu

To find the standard electrode potential of \(\mathrm{Cu}^{+} / \mathrm{Cu}\), we need the reverse reaction of (2) which is \(\mathrm{Cu}^{+} \rightarrow \mathrm{Cu} + e^-\), since \(\mathrm{Cu}^{2+} / \mathrm{Cu}^{+}\) is \(0.153 \mathrm{~V}\), and \(\mathrm{Cu}^{2+} / \mathrm{Cu}\) is \(0.337 \mathrm{~V}\). Subtract \(0.153 \mathrm{~V}\) from \(0.337 \mathrm{~V}\).

Step 4: Perform the Calculation

Calculate \(E^0_{\mathrm{Cu}^{+}/\mathrm{Cu}} = E^0_{\mathrm{Cu}^{2+}/\mathrm{Cu}} - E^0_{\mathrm{Cu}^{2+}/\mathrm{Cu}^{+}} = 0.337 \mathrm{~V} - 0.153 \mathrm{~V} = 0.184 \mathrm{~V}\).

Step 5: Identify the Correct Answer

The calculated potential of \(0.184 \mathrm{~V}\) matches option (a), so the standard electrode potential for \(\mathrm{Cu}^{+} / \mathrm{Cu}\) is \(0.184 \mathrm{~V}\).

Key Concepts

Standard Reduction Potential

The concept of standard reduction potential is key in understanding electrochemical reactions. It tells us how likely a species will gain electrons, i.e., be reduced. A higher reduction potential means a species is more willing to accept electrons. These potentials are always measured under standard conditions: 25°C, 1 M concentration, and 1 atm pressure.

Standard reduction potential is typically denoted by the symbol \(E^0\). It is crucial because it allows us to predict the direction of electron flow in an electrochemical cell. In simple terms, it helps us determine which half-reaction will occur as a reduction and which one as oxidation. For example, copper has different reduction potentials; for \(\mathrm{Cu}^{2+} / \mathrm{Cu}\), it is \(0.337 \mathrm{~V}\), and for \(\mathrm{Cu}^{2+} / \mathrm{Cu}^{+}\), it is \(0.153 \mathrm{~V}\).

• Higher \(E^0\): indicates better access to electrons.
• Measured in volts (V).
• Used under standard conditions of temperature, concentration, and pressure.

Electrode Potential Calculation

Calculating electrode potentials involves manipulating given standard reduction potentials to find unknown potentials of half-cells. The key idea is to understand the reactions in play and then apply a straightforward subtraction to compute the needed potential.

In the exercise, we're tasked with calculating the standard electrode potential for \(\mathrm{Cu}^{+} / \mathrm{Cu}\). This involves the standard reduction potentials for reactions of \(\mathrm{Cu}^{2+}\) with \(\mathrm{Cu}^{+}\) and \(\mathrm{Cu}\). We have:

• \(\mathrm{Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu}\) \(E^0 = 0.337 \mathrm{~V}\)
• \(\mathrm{Cu}^{2+} + e^- \rightarrow \mathrm{Cu}^{+}\) \(E^0 = 0.153 \mathrm{~V}\)

To find \(E^0_{\mathrm{Cu}^{+}/\mathrm{Cu}}\), reverse the second equation: \(\mathrm{Cu}^{+} \rightarrow \mathrm{Cu} + e^-\). The potential for this is the reverse of the reaction, subtracting the reduction potential of \(\mathrm{Cu}^{2+} / \mathrm{Cu}^{+}\) from that of \(\mathrm{Cu}^{2+} / \mathrm{Cu}\).

• Calculation: \(E^0_{\mathrm{Cu}^{+}/\mathrm{Cu}} = 0.337 - 0.153 = 0.184 \mathrm{~V}\)
• The result: \(0.184 \mathrm{~V}\), indicating this is the potential required for the \(\mathrm{Cu}^{+} / \mathrm{Cu}\) conversion.

Electrochemistry Concepts

Electrochemistry is the field that studies chemical reactions involving electric currents and potentials. It explains how electrons move between different elements and ions in solutions, sparking reactions through redox processes, where reduction and oxidation happen simultaneously.

A basic electrochemical setup includes two electrodes immersed in electrolyte solutions. One electrode will undergo oxidation (losing electrons), while the other undergoes reduction (gaining electrons). These reactions occur at the anode and cathode, respectively.

• Anode: oxidation occurs, electrons are lost.
• Cathode: reduction occurs, electrons are gained.
• Salt bridge: keeps electrical neutrality by allowing ion movement.

Understanding electrode potentials and their calculations helps us predict the behavior of these electrochemical cells. By knowing which way electrons will flow, we can harness electrical energy efficiently. This is critical in areas like battery design and industrial processes. The principles of electrochemistry power many of the devices we depend on daily, turning chemical energy into electrical power.