91Ó°ÊÓ

Questions 32-36 refer to the following. Two half-cells are set up as follows: Half-Cell A: Strip of \(\mathrm{Cu}(s)\) in \(\mathrm{CuNO}_{3}(a q)\) Half-Cell B: Strip of \(\mathrm{Zn}(s)\) in \(\mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2}\) (aq) When the cells are connected according to the diagram below, the following reaction occurs: GRAPH CAN'T COPY $$2 \mathrm{Cu}^{+}(a q)+\mathrm{Zn}(s) \rightarrow 2 \mathrm{Cu}(s)+\mathrm{Zn}^{2+}(a q) E^{\circ}=+1.28 \mathrm{V}$$ If the \(\mathrm{Cu}^{+}+e^{-} \rightarrow \mathrm{Cu}(s)\) half-reaction has a standard reduction potential of \(+0.52 \mathrm{V},\) what is the standard reduction potential for the \(\mathrm{Zn}^{2+}+2 e^{-} \rightarrow \mathrm{Zn}(s)\) half-reaction? (A) \(+0.76 \mathrm{V}\) (B) \(-0.76 \mathrm{V}\) (C) \(+0.24 \mathrm{V}\) (D) \(-0.24 \mathrm{V}\)

Step by step solution

01

Identify the half-reactions

The two half-reactions are: \(Cu^{+}+e^{-}\rightarrow Cu(s)\) and \(Zn^{2+}+2e^{-}\rightarrow Zn(s)\). In the provided overall reaction, Copper is getting reduced (gains electrons) and Zinc is getting oxidized (loses electrons). This corresponds to the half-reactions except Zinc is undergoing oxidation, not reduction. Therefore, when calculating the potential, we need to be aware of the change.

02

Write down the known reduction potentials

We know that the standard reduction potential of the \(Cu^{+}+e^{-}\rightarrow Cu(s)\) reaction is \(+0.52V\). And the overall cell potential given is \(+1.28V\).

03

Calculate the standard reduction potential for the Zn half-reaction

The standard cell potential is the sum of the standard reduction potentials of the two half-reactions when the reactions are added together. The reduction potential of the Zinc reaction is the unknown we need to find. We use the equation: \(E_{cell} = E_{red, Cu} + E_{red, Zn}\). Substituting the known values: \(1.28V = 0.52V + E_{red, Zn}\). Solving for \(E_{red, Zn}\), we find \(E_{red, Zn} = 1.28V - 0.52V = 0.76V\). However, this is the value if Zinc was being reduced. In the overall reaction, it is being oxidised. Therefore, the reduction potential of the Zinc reaction is \(-0.76V\).

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Key Concepts

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

Standard reduction potential

Standard reduction potential is a measure of the tendency of a chemical species to acquire electrons and thereby be reduced. It is expressed in volts and usually measured under standard conditions, which are 25°C, 1 M concentration for solutions, and 1 atm pressure for gases.

In electrochemistry, a more positive standard reduction potential indicates a greater likelihood of the species accepting electrons and undergoing reduction. Conversely, a more negative potential suggests a species is less likely to be reduced.

• These potentials are listed in tables as reference points, making it easier to predict the direction of electron flow in electrochemical cells.
• They enable calculations like finding the cell potential, which is crucial for assessing the feasibility and spontaneity of a reaction.

Understanding standard reduction potentials helps predict the reactions that occur in voltaic cells, crucial for harnessing energy in batteries.

Half-cell reactions

Half-cell reactions represent one part of the two-part process in an electrochemical cell. Each half-cell involves a redox reaction: one for oxidation and one for reduction.

Consider the two half-cells from the example:

• Half-Cell A: Copper undergoes reduction: \(Cu^+ + e^- \rightarrow Cu(s) \)
• Half-Cell B: Zinc undergoes oxidation: \(Zn(s) \rightarrow Zn^{2+} + 2e^- \)

By writing these reactions separately, we can clearly see the electron exchange, making it easier to calculate the overall cell potential. In this context, the terms "oxidation" and "reduction" refer to the processes of losing and gaining electrons, respectively.

Each half-cell reaction is based on the respective reduction or oxidation potential provided in standard tables, making it critical for accurate cell design and calculations.

Voltage calculation

Calculating the voltage of an electrochemical cell involves adding the reduction potentials of each half-reaction. The overall cell potential \(E_{cell}\) is given by:
\[ E_{cell} = E_{red, \, cathode} + E_{red, \, anode} \]

In the case of the example, we want to find the contribution of the zinc half-reaction when copper's potential is known. The process is straightforward:

• Start with the known cell potential: \(1.28 \text{ V}\).
• The copper reduction potential is given as \(+0.52 \text{ V}\).
• We can rearrange terms to solve for zinc: \(E_{red, \, Zn} = 1.28 \text{ V} - 0.52 \text{ V} = 0.76 \text{ V}\).

This calculated \(0.76\text{ V}\) is for reduction. Since zinc is oxidized in the actual process, we flip the sign, resulting in a zinc reduction potential of \(-0.76\text{ V}\), demonstrating how the voltage reflects the electron flow direction.