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Chapter 17: Problem 17

The standard electrode potentials \(\left(\mathrm{E}^{\circ} \mathrm{M}^{+} / \mathrm{M}\right)\) of four metals \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) and \(\mathrm{D}\) are \(-1.2 \mathrm{~V}, 0.6 \mathrm{~V}, 0.85 \mathrm{~V}\) and \(-0.76 \mathrm{~V}\), respectively. The sequence of deposition of metals on applying potential is: [Main Online April 9, 2014] (a) \(A, C, B, D\) (b) \(\mathrm{B}, \mathrm{D}, \mathrm{C}, \mathrm{A}\) (c) \(\mathrm{C}, \mathrm{B}, \mathrm{D}, \mathrm{A}\) (d) \(\mathrm{D}, \mathrm{A}, \mathrm{B}, \mathrm{C}\)

Short Answer

Expert verified
The correct sequence is (c) C, B, D, A.

Step by step solution

01

Understand the Problem

We are given standard electrode potentials for four metals (A, B, C, and D) and need to determine their deposition sequence when a potential is applied.
02

Recall the Concept of Electrode Potential

The electrode potential indicates the tendency of a species to gain electrons (be reduced). Metals with higher electrode potentials are more likely to be reduced and deposited first.
03

Arrange Electrode Potentials in Descending Order

The given electrode potentials are: A = -1.2 V, B = 0.6 V, C = 0.85 V, D = -0.76 V. Arrange these in descending order: C (0.85 V), B (0.6 V), D (-0.76 V), A (-1.2 V).
04

Determine the Deposition Sequence

Since metals with the highest potentials are deposited first, the metal C ( 0.85 V) will deposit first, followed by B ( 0.6 V), then D ( -0.76 V), and finally A ( -1.2 V).
05

Match with the Given Options

The correct sequence from Step 4 is C, B, D, A. Compare this sequence with the given options to find the matching one.

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Key Concepts

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

Electrode Potential Series
The electrode potential series is a valuable tool used to predict the order of metal deposition during electrochemical reactions. It involves organizing metals based on their standard electrode potentials, which reflect a metal's aptitude to be reduced by gaining electrons. When examining the potential values:
  • A low or negative potential suggests a weaker tendency to be reduced, while a positive potential implies a stronger reduction tendency.
By arranging metals such as A, B, C, and D in descending order of their electrode potentials, one can determine how and when they will deposit. Practically, metals with the highest electrode potentials get deposited first as they have the strongest desire to gain electrons.
Electrochemical Deposition
Electrochemical deposition is a fascinating process where metals are deposited from a solution onto a surface. This happens in applications like electroplating, where a metal coat is placed on an object to improve its appearance or resistance against corrosion. The sequence of deposition is crucial and is closely tied to the electrode potential series. Metals with higher electrode potentials are initiated to deposit first because they are more inclined to undergo reduction, in other words, to gain electrons.
  • For metals like C (0.85 V), B (0.6 V), D (-0.76 V), and A (-1.2 V), the higher potential of C makes it deposit first, followed by B, D, and finally A.
  • Each metal鈥檚 deposition depends on enough potential being applied to overcome its reduction potential.
By understanding these principles, the deposition sequence under applied potential can be accurately predicted.
Redox Reactions
Redox reactions, or reduction-oxidation reactions, are chemical processes where the oxidation state of atoms goes through a change. These reactions involve electron transfer between substances. One part of the substance gets oxidized while another gets reduced. In the context of electrochemical deposition, metals C and B are more likely to be reduced due to their higher electrode potentials.
  • Reduction means gaining electrons, while oxidation means losing them.
  • When potential is applied, metals will undergo reduction based on their place in the electrode potential series.
  • Subsequently, electron exchanges occur, and metals are deposited in the specific sequence predicted by their electrode potentials.
Understanding redox reactions assists in predicting the order of electrochemical processes and is vital in industrial applications like metal plating and energy storage technologies.

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Most popular questions from this chapter

While \(\mathrm{Fe}^{3+}\) is stable, \(\mathrm{Mn}^{3+}\) is not stable in acid solution because (a) \(\mathrm{O}_{2}\) oxideses \(\mathrm{Mn}^{2+}\) to \(\mathrm{Mn}^{3+}\) (b) \(\mathrm{O}_{2}\) oxideses both \(\mathrm{Mn}^{2+}\) to \(\mathrm{Mn}^{3+}\) and \(\mathrm{Fe}^{2+}\) to \(\mathrm{Fe}^{3+}\) (c) \(\mathrm{Fe}^{3+}\) oxideses \(\mathrm{H}_{2} \mathrm{O}\) to \(\mathrm{O}_{2}\) (d) \(\mathrm{Mn}^{3+}\) oxideses \(\mathrm{H}_{2} \mathrm{O}\) to \(\mathrm{O}_{2}\)

In a galvanic cell, the salt bridge (a) Does not participate chemically in the cell reaction (b) Stops the diffusion of ions from one electrode to another (c) Is necessary for the occurrence of the cell reaction (d) Ensures mixing of the two electrolytic solutions

The following electrochemical cell has been set up. \(\mathrm{Pt}(1)\left|\mathrm{Fe}^{3+}, \mathrm{Fe}^{2+}(a=1)\right| \mathrm{Ce}^{4+}, \mathrm{Ce}^{3+}(a=1) \mid \mathrm{Pt}(2)\) \(E^{\circ}\left(\mathrm{Fe}^{3+}, \mathrm{Fe}^{2+}\right)=0.77 \mathrm{~V}: E^{\circ}\left(\mathrm{Ce}^{4+} / \mathrm{Ce}^{3+}\right)=1.61 \mathrm{~V}\) If an ammeter is connected between the two platinum electrodes, predict the direction of flow of current. Will the current increase or decrease with time?

Identify the correct statement : [Main Online April 10, 2016] (a) Corrosion of iron can be minimized by forming a contact with another metal with a higher reduction potential (b) Iron corrodes in oxygen free water (c) Corrosion of iron can be minimized by forming an impermeable barrier at its surface (d) Iron corrodes more rapidly in salt water because its electrochemical potential is higher

Given below are the half-cell reactions: $$ \begin{aligned} &\mathrm{Mn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn} ; \mathrm{E}^{\circ}=-1.18 \mathrm{~V} \\ &2\left(\mathrm{Mn}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}\right) ; \mathrm{E}^{\circ}=+1.51 \mathrm{~V} \end{aligned} $$ The \(\mathrm{E}^{\circ}\) for \(3 \mathrm{Mn}^{2+} \rightarrow \mathrm{Mn}+2 \mathrm{Mn}^{3+}\) will be: (a) \(-2.69 \mathrm{~V} ;\) the reaction will not occur (b) \(-2.69 \mathrm{~V}\); the reaction will occur (c) \(-0.33 \mathrm{~V}\); the reaction will not occur (d) \(-0.33 \mathrm{~V}\); the reaction will occur
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