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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 \(\mathrm{Cu}^{+} / \mathrm{Cu}\) half cell is a. \(0.184 \mathrm{~V}\) b. \(0.827 \mathrm{~V}\) c. \(0.521 \mathrm{~V}\) d. \(0.490 \mathrm{~V}\)

Short Answer

Expert verified
The standard electrode potential for \( \mathrm{Cu}^+ / \mathrm{Cu} \) is \(0.184 \text{ V}\). The correct answer is a.

Step by step solution

01

Identify the Given Information

We are given two standard reduction potentials: \( E^\circ(\mathrm{Cu}^{2+} / \mathrm{Cu}) = 0.337 \text{ V} \) and \( E^\circ(\mathrm{Cu}^{2+} / \mathrm{Cu}^+) = 0.153 \text{ V} \). We need to find the standard reduction potential for \( \mathrm{Cu}^+ / \mathrm{Cu} \).
02

Apply the Potential Difference Formula

The potential difference in the following two reduction reactions can help us find the unknown: \[\mathrm{Cu}^{2+} + e^- \rightarrow \mathrm{Cu}^+\] with standard potential \( E^\circ = 0.153 \text{ V} \), and \[\mathrm{Cu}^+ + e^- \rightarrow \mathrm{Cu}\] with unknown \( E^\circ \). These values must add up to the reaction \[\mathrm{Cu}^{2+} + 2e^- \rightarrow \mathrm{Cu}\] with given standard potential \( E^\circ = 0.337 \text{ V} \).
03

Subtract Potentials to Find the Unknown Value

To find \( E^\circ(\mathrm{Cu}^+ / \mathrm{Cu}) \), we subtract the potential of \( \mathrm{Cu}^{2+} / \mathrm{Cu}^+ \) from \( \mathrm{Cu}^{2+} / \mathrm{Cu} \): \[ E^\circ(\mathrm{Cu}^+ / \mathrm{Cu}) = E^\circ(\mathrm{Cu}^{2+} / \mathrm{Cu}) - E^\circ(\mathrm{Cu}^{2+} / \mathrm{Cu}^+) \]\[ E^\circ(\mathrm{Cu}^+ / \mathrm{Cu}) = 0.337 \text{ V} - 0.153 \text{ V} = 0.184 \text{ V} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Redox Reactions
Redox reactions, short for reduction-oxidation reactions, are fascinating processes where the transfer of electrons occurs between chemical species. In these reactions, one substance loses electrons, known as oxidation, and another gains electrons, known as reduction.
A good way to remember this process is through the acronym "OIL RIG": Oxidation Is Loss, Reduction Is Gain.
  • **Oxidation:** Loss of electrons, increase in oxidation state.
  • **Reduction:** Gain of electrons, decrease in oxidation state.
In the context of electrode potentials, these reactions often involve the conversion of different oxidation states of elements, such as copper existing in forms like \(\text{Cu}^{2+}\), \(\text{Cu}^{+}\), and \(\text{Cu}\). As electrons are transferred in redox reactions, it leads to the flow of electric current, which can be harnessed in electrochemical cells.
Standard Reduction Potential
Standard reduction potential, indicated as \(E^\circ\), represents the tendency of a chemical species to be reduced, under standard conditions. It's measured in volts (V) and provides a quantifiable way to predict the direction of electron flow in a redox reaction.

Having a higher standard reduction potential means a greater likelihood to gain electrons and be reduced. For instance, if you compare \(E^\circ(\text{Cu}^{2+} / \text{Cu}) = 0.337 \, \text{V}\) with \(E^\circ(\text{Cu}^{2+} / \text{Cu}^+) = 0.153 \, \text{V}\), it indicates \(\text{Cu}^{2+}\) is more eager to receive electrons than \(\text{Cu}^+\).
  • **Positive value:** Indicates a strong tendency to be reduced.
  • **Negative value:** Indicates a weaker tendency.
Calculating the differences in standard reduction potentials helps in determining unknown potentials, as seen with the half-cell reaction for \(\text{Cu}^+ / \text{Cu}\). This provides crucial insights for predicting how electrons move in a cell.
Half Cell Reactions
Half cell reactions are part of redox reactions where either oxidation or reduction takes place. Each half-cell is a component of an electrochemical cell, consisting of an electrode immersed in an electrolyte.
For copper, these reactions include:
  • **Reduction half-cell:** $$\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$$
  • **Oxidation half-cell:** $$\text{Cu} \rightarrow \text{Cu}^{2+} + 2e^-$$
In electrode potentials, each half-cell contributes a specific standard reduction potential. The combination of potentials from two half-cells gives the overall potential of the electrochemical cell.

Calculations of half-cell reactions help in figuring out unknown values, as in the example where knowing potentials for two separate reactions allowed the determination of \(E^\circ(\text{Cu}^+ / \text{Cu})\). When combined, these help in forming a complete cell equation and predicting cell behavior.

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Ammonia is always is added in this reaction. Which of the following must be incorrect? a. \(\mathrm{NH}_{3}\) combines with \(\mathrm{Ag}^{+}\)to form a complex b. \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\)is a stronger oxidizing reagent than \(\mathrm{Ag}^{+}\). c. In absence of \(\mathrm{NH}_{3}\) silver salt of gluconic acid is formed d. \(\mathrm{NH}_{3}\) has affected the standard reduction potential of glucose/gluconic acid electrode.

The cell reaction for a lead storage battery is $$ \begin{array}{r} \mathrm{Pb}(\mathrm{s})+\mathrm{PbO}_{2}(\mathrm{~s})+2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{HSO}_{4}^{-}(\mathrm{aq}) \\ \rightarrow 2 \mathrm{~Pb}\left(\mathrm{SO}_{4}\right)(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ \mathrm{E}^{\circ}=+1.92 \mathrm{~V} \end{array} $$ To provide a potential of about \(12 \mathrm{~V}\), one could a. Connect six cells in series. b. Adjust the \(\mathrm{pH}\) to 12 . c. Adjust the \(\mathrm{pH}\) to 1 . d. Greatly increase the surface area of the \(\mathrm{Pb}\) (s) and \(\mathrm{PbO}_{2}(\mathrm{~s})\).

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