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The standard reduction potentials for \(\mathrm{Zn}^{2+} / \mathrm{Zn}, \mathrm{Ni}^{2+} / \mathrm{Ni}\), and \(\mathrm{Fe}^{2+} / \mathrm{Fe}\) are \(-0.76,-0.23\) and \(-0.44 \mathrm{~V}\) respectively. The reaction \(\mathrm{X}+\mathrm{Y}^{2+} \rightarrow \mathrm{X}^{2+}+\mathrm{Y}\) will be spontaneous when (a) \(\mathrm{X}=\mathrm{Fe}, \mathrm{Y}=\mathrm{Zn}\) (b) \(\mathrm{X}=\mathrm{Ni}, \mathrm{Y}=\mathrm{Zn}\) (c) \(\mathrm{X}=\mathrm{Ni}, \mathrm{Y}=\mathrm{Fe}\) (d) \(\mathrm{X}=\mathrm{Zn}, \mathrm{Y}=\mathrm{Ni}\)

Short Answer

Expert verified
Option (d) \(\mathrm{X} = \mathrm{Zn}\), \(\mathrm{Y} = \mathrm{Ni}\) is spontaneous.

Step by step solution

01

Understanding the Problem

We are given standard reduction potentials and a reaction: \(\mathrm{X}+\mathrm{Y}^{2+} \rightarrow \mathrm{X}^{2+}+\mathrm{Y}\). We need to determine which combinations of \(\mathrm{X}\) and \(\mathrm{Y}\) will make the reaction spontaneous. A reaction is spontaneous if the calculated cell potential \(E^\circ_{\text{cell}}\) for this reaction is positive.
02

Standard Reduction Potentials

The standard reduction potentials given are: \(\mathrm{Zn}^{2+}/\mathrm{Zn} = -0.76 \mathrm{~V}\), \(\mathrm{Ni}^{2+}/\mathrm{Ni} = -0.23 \mathrm{~V}\), and \(\mathrm{Fe}^{2+}/\mathrm{Fe} = -0.44 \mathrm{~V}\). These values will be used to calculate \(E^\circ_{\text{cell}}\) for each combination.
03

Calculating Cell Potential for Each Option

For each option, the equation \(E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}\) is used. The reduction potential that is more positive will serve as the cathode (reduction reaction site), while the more negative will serve as the anode (oxidation reaction site).
04

Option (a) \(\mathrm{X} = \mathrm{Fe}\), \(\mathrm{Y} = \mathrm{Zn}\)

Here, Fe is oxidized and Zn is reduced. \[E^\circ_{\text{cell}} = E^\circ_{\text{Zn}^{2+}/\text{Zn}} - E^\circ_{\text{Fe}^{2+}/\text{Fe}} = (-0.76) - (-0.44) = -0.32 \mathrm{~V}\]. Since \(E^\circ_{\text{cell}}\) is negative, this reaction is not spontaneous.
05

Option (b) \(\mathrm{X} = \mathrm{Ni}\), \(\mathrm{Y} = \mathrm{Zn}\)

Here, Ni is oxidized and Zn is reduced. \[E^\circ_{\text{cell}} = E^\circ_{\text{Zn}^{2+}/\text{Zn}} - E^\circ_{\text{Ni}^{2+}/\text{Ni}} = -0.76 - (-0.23) = -0.53 \mathrm{~V}\]. Since \(E^\circ_{\text{cell}}\) is negative, this reaction is not spontaneous.
06

Option (c) \(\mathrm{X} = \mathrm{Ni}\), \(\mathrm{Y} = \mathrm{Fe}\)

Here, Ni is oxidized and Fe is reduced. \[E^\circ_{\text{cell}} = E^\circ_{\text{Fe}^{2+}/\text{Fe}} - E^\circ_{\text{Ni}^{2+}/\text{Ni}} = -0.44 - (-0.23) = -0.21 \mathrm{~V}\]. Since \(E^\circ_{\text{cell}}\) is negative, this reaction is not spontaneous.
07

Option (d) \(\mathrm{X} = \mathrm{Zn}\), \(\mathrm{Y} = \mathrm{Ni}\)

Here, Zn is oxidized and Ni is reduced. \[E^\circ_{\text{cell}} = E^\circ_{\text{Ni}^{2+}/\text{Ni}} - E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.23 - (-0.76) = +0.53 \mathrm{~V}\]. Since \(E^\circ_{\text{cell}}\) is positive, this reaction is spontaneous.

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Key Concepts

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

Standard Reduction Potential
In electrochemistry, the standard reduction potential is a fundamental concept used to gauge the tendency of a chemical species to gain electrons, or become reduced. Each species involved in a redox reaction has its unique standard reduction potential value, symbolized as \(E^\circ\).
These values are measured in volts and are referenced to a standard hydrogen electrode which is assigned a potential of 0 volts. The more positive the \(E^\circ\), the greater the substance's affinity for electrons. Conversely, a more negative \(E^\circ\) indicates a lower affinity and hence a stronger tendency to be oxidized.
  • Examples: Given potentials for reactions involving Zn, Ni, and Fe are -0.76 V, -0.23 V, and -0.44 V respectively. These numbers reflect the ability of Zn, Ni, and Fe to be reduced when reacting with other species.
  • Application: Standard reduction potentials are crucial for calculating whether an electrochemical reaction is spontaneous, which is significant in designing batteries and understanding corrosion processes.
Understanding which species can act as an oxidizing or reducing agent is foundational in predicting and controlling chemical reactions.
Spontaneity in Electrochemical Cells
For a reaction within an electrochemical cell to occur spontaneously, it must be energetically favorable. This is determined by calculating the cell potential \(E^\circ_{\text{cell}}\), which combines the potentials of the two half-reactions occurring.
The half-reaction with a higher or more positive standard reduction potential will undergo reduction, while the one with a lower potential will be oxidized. This yields:
  • \(E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}\)
For spontaneity in electrochemical cells, the \(E^\circ_{\text{cell}}\) must be positive:
  • A positive potential indicates that the energy released is capable of driving the reaction forward.
  • A negative potential signifies the reaction requires an input of energy, thus is non-spontaneous under standard conditions.
In our exercise, calculating \(E^\circ_{\text{cell}}\) for different combinations of metals illustrates this principle: only the Zn-Ni combination was spontaneous, reflecting a positive \(E^\circ_{\text{cell}} = +0.53 V\).
Electrochemical Cell Calculations
Electrochemical cell calculations involve determining the cell potential, which provides insights into the feasibility of a reaction. Calculating \(E^\circ_{\text{cell}}\) involves these steps:
  • Identify the cathode and anode based on the standard reduction potential values. The species with the higher \(E^\circ\) undergoes reduction.
  • Apply the formula \(E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}\).
By systematically substituting values in our exercise, we derived:
  • Option (d) \(\text{Zn-Ni}:\ E^\circ_{\text{cell}} = +0.53 \text{ V}\)
  • All other options resulted in negative \(E^\circ_{\text{cell}}\) values, indicating non-spontaneity.
Through these calculations, we also learn the practical implications:
  • Determining the likely direction of electron flow allows for better understanding of battery function and electric energy generation.

These insights allow scientists and engineers to devise efficient electrochemical systems.

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Most popular questions from this chapter

Standard reduction electrode potentials of three metals A, B and \(C\) are \(+0.5 \mathrm{~V},-3.0 \mathrm{~V}\) and \(-1.2 \mathrm{~V}\) respectively. The reducing power of these metals are (a) \(\mathrm{B}>\mathrm{C}>\mathrm{A}\) (b) \(A>B>C\) (c) \(C>B>A\) (d) \(\mathrm{A}>\mathrm{C}>\mathrm{B}\)

Equal quantities of electricity are passed through three voltameters containing \(\mathrm{FeSO}_{4}, \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}\), and \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\). Consider the following statements in this regard (1) the amount of iron deposited in \(\mathrm{FeSO}_{4}\) and \(\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) is equal (2) the amount of iron deposited in \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\) is two thirds of the amount of iron deposited in \(\mathrm{FeSO}_{4}\) (3) the amount of iron deposited in \(\mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3} \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\) is equal Of these statements (a) 1 and 2 are correct (b) 2 and 3 are correct (c) 1 and 3 are correct (d) 1,2 and 3 are correct

Corrosion of iron is essentially an electrochemical phenomenon where the cell reactions are (a) \(\mathrm{Fe}\) is oxidized to \(\mathrm{Fe}^{2+}\) and dissolved oxygen in water is reduced to \(\mathrm{OH}^{-}\) (b) \(\mathrm{Fe}\) is oxidized to \(\mathrm{Fe}^{3+}\) and \(\mathrm{H}_{2} \mathrm{O}\) is reduced to \(\mathrm{O}_{2}^{2-}\) (c) \(\mathrm{Fe}\) is oxidized to \(\mathrm{Fe}^{2+}\) and \(\mathrm{H}_{2} \mathrm{O}\) is reduced to \(\mathrm{O}_{2}^{-}\) (d) \(\mathrm{Fe}\) is oxidized to \(\mathrm{Fe}^{2+}\) and \(\mathrm{H}_{2} \mathrm{O}\) is reduced to \(\mathrm{O}_{2}\)

One litre of \(1 \mathrm{M} \mathrm{CuSO}_{4}\) solution is electrolysed. After passing 2F of electricity, molarity of \(\mathrm{CuSO}_{4}\) solution will be (a) \(\mathrm{M} / 2\) (b) \(\mathrm{M} / 4\) (c) \(\mathrm{M}\) (d) 0

For the electrolysis of \(\mathrm{CuSO}_{4}\) solution which is/are correct? (a) Cathode reaction: \(\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}\) using \(\mathrm{Cu}\) electrode (b) Anode reaction: \(\mathrm{Cu} \rightarrow \mathrm{Cu}^{2}++2 \mathrm{e}^{-}\)using \(\mathrm{Cu}\) electrode (c) Cathode reaction: \(2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2}\) using Pt electrode (d) Anode reaction: \(\mathrm{Cu} \rightarrow \mathrm{Cu}^{2+}+2 \mathrm{e}\) using \(\mathrm{Pt}\) electrode

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