Problem 17
Question
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) \(\mathrm{D}, \mathrm{A}, \mathrm{B}, \mathrm{C}\)
Step-by-Step Solution
Verified Answer
The correct sequence is (c) C, B, D, A.
1Step 1: 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.
2Step 2: 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.
3Step 3: 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).
4Step 4: 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).
5Step 5: 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.
Key Concepts
Electrode Potential SeriesElectrochemical DepositionRedox Reactions
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.
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’s deposition depends on enough potential being applied to overcome its reduction potential.
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.
Other exercises in this chapter
Problem 15
Consider the following cell reaction: [2011] \(2 \mathrm{Fe}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow 2 \mathrm{Fe}^{2+
View solution Problem 16
The rusting of iron takes place as follows \(2 \mathrm{H}^{+}+2 \mathrm{e}^{-}+1 / 2 \mathrm{O}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O}(1) ; E^{\circ}=+1.
View solution Problem 18
Given \(\mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+}(\mathrm{aq}) ; \mathrm{E}^{\circ}=+0.77 \mathrm{~V}\) \(\mathrm{Al}^{3+}(\math
View solution Problem 18
In the electrolytic cell, flow of electrons is from (a) Cathode to anode in solution (b) Cathode to anode through external supply (c) Cathode to anode through i
View solution