Question

The electrode potentials for \(\mathrm{Cu}_{\text {(aq) }}^{2 *}+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}_{\text {(aq) }}^{*} \quad\) and \(\mathrm{Cu}_{\text {(aq) }}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}_{(\mathrm{s})}\) are \(+0.15 \mathrm{~V}\) and \(+0.50 \mathrm{~V}\) respectively. The value of \(E_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}\) will be : (1) \(0.500 \mathrm{~V}\) (2) \(0.325 \mathrm{~V}\) (3) \(0.650 \mathrm{~V}\) (4) \(0.150 \mathrm{~V}\)

Short answer

The value of \(E_{\text {Cu}^{2+} /\text{ Cu}}^{\beta }= 0.325 \text{ V} \). The correct answer is (2) 0.325 V.

Step-by-step solution

- Write down the given electrode potentials

Write the given electrode potentials:1. \(\text{Cu}^{2+} + e^{-} \rightarrow \text{Cu}^+\), with \(E^0 = +0.15V\)2. \(\text{Cu}^+ + e^{-} \rightarrow \text{Cu} \), with \(E^0 = +0.50V \)

- Understand the combined reaction

The combined reaction for \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \) should be calculated by summing these individual potentials.

- Apply the formula

Use the relationship that the equilibrium potential for the full reaction can be found by adding individual half-cell potentials: \(\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}\) We sum the potentials: \[ E_{\text{Cu}^{2+}/\text{Cu}}^0= (E_{\text{Cu}^{2+}/\text{Cu}^+}^0 +E_{\text{Cu}^+/\text{Cu}}^0 ) / 2\]So,\[ E_{\text{Cu}^{2+}/\text{Cu}}^0 = \frac{0.15V + 0.50V}{2} \]

- Calculate the value

Perform the calculation: \[ E_{\text{Cu}^{2+}/\text{Cu}}^0 = \frac{0.15 + 0.50}{2} \] = \( 0.325V \)

- Identify the correct answer

From the calculation, \( E_{\text {Cu}^{2+} /\text{ Cu}}^{\beta }= 0.325 \text{V} \). So, the correct answer is (2) \( 0.325 \text {V}. \)

Key concepts

electrochemistry

Electrochemistry is the branch of chemistry that studies the relationship between electricity and chemical reactions. It looks at how electrons move in redox reactions, either producing electricity or being driven by an external electric current. This field is crucial for understanding batteries, fuel cells, and electrolysis.
In an electrochemical cell, two electrodes are in contact with an electrolyte. Electrodes are the points where electricity enters or leaves. The electrolyte is a substance containing free ions that conduct electric current.
There are two main types of electrochemical cells:

• Galvanic cells: Generate electric current from chemical reactions.
• Electrolytic cells: Use electric current to drive chemical reactions.

Understanding electrochemistry helps us make practical applications, such as designing better batteries for electronic devices and electric vehicles.

redox reactions

Redox reactions are chemical reactions that involve the transfer of electrons between substances. The term 'redox' is a combination of two processes: reduction and oxidation.
Oxidation is the loss of electrons, while reduction is the gain of electrons. Here are a few key points to understand:

• An element that loses electrons is oxidized.
• An element that gains electrons is reduced.
• Oxidizing agents gain electrons and are reduced.
• Reducing agents lose electrons and are oxidized.

In the exercise, we deal with the following reactions:
1. \(\text{Cu}^{2+} + e^{-} \rightarrow \text{Cu}^+\), which involves the reduction of \(\text{Cu}^{2+}\) to \(\text{Cu}^+\).
2. \(\text{Cu}^+ + e^{-} \rightarrow \text{Cu}\), which involves the further reduction of \(\text{Cu}^+\) to solid copper (Cu).
Understanding redox reactions allows us to grasp the fundamental mechanics of processes like corrosion, respiration, and combustion.

standard electrode potential

Standard electrode potential (\(E^0\)) is the measure of the driving force behind an electrochemical reaction under standard conditions \(1 M concentration, 1 atm pressure, and 25°C (298 K)\). The potential indicates how easily a species gains or loses electrons compared to the reference hydrogen electrode, which is assigned a potential of 0 volts.
Standard electrode potential is crucial for predicting the direction of electron flow in electrochemical cells. If the \(E^0\) value is positive, it means the reaction prefers to gain electrons and be reduced. Conversely, a negative \(E^0\) indicates a tendency to lose electrons and be oxidized.
In the provided exercise, we used standard electrode potentials to calculate the \(E_{\text{Cu}^{2+}/\text{Cu}}^0\) as follows:

• First, record the given potentials: \(E^0 = +0.15 V\) for \( \text{Cu}^{2+} + e^{-} \rightarrow \text{Cu}^+\) and \(E^0 = +0.50 V\) for \( \text{Cu}^+ + e^{-} \rightarrow \text{Cu} \).
  
• Combine these values to find the potential for \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \).

This helps predict the favorability of redox reactions and the cell potential, which is central to electrochemistry applications such as battery design and metal plating.