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The standard oxidation potential \(E^{\circ}\) for the half reactions are as \(\mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+}+2 \mathrm{e}^{-} ; E^{\circ}=+0.76 \mathrm{~V}\) \(\mathrm{Fe} \longrightarrow \mathrm{Fe}^{2+}+2 \mathrm{e} ; E^{\circ}=+0.41 \mathrm{~V}\) The emf for the cell reaction \(\mathrm{Fe}^{2+}+\mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{Fe}\) is (a) \(+1.17 \mathrm{~V}\) (b) \(-0.35 \mathrm{~V}\) (c) \(+0.35 \mathrm{~V}\) (d) \(0.117 \mathrm{~V}\)

Short Answer

Expert verified
None of the options directly match; a reassessment suggests understanding values for favorability was key.

Step by step solution

01

Identify Relevant Reactions

The cell consists of two half-reactions: one for zinc (Zn) and one for iron (Fe). Here, the relevant half-reactions are: 1. Anode (Zn is oxidized): \( \mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+} + 2 \mathrm{e}^- \)2. Cathode (Fe is reduced): \( \mathrm{Fe}^{2+} + 2 \mathrm{e}^- \longrightarrow \mathrm{Fe} \)
02

Determine Oxidation and Reduction Potentials

The given standard potentials are for oxidation:- \( \mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+} + 2 \mathrm{e}^-; E^{\circ} = +0.76 \mathrm{~V} \)- \( \mathrm{Fe} \longrightarrow \mathrm{Fe}^{2+} + 2 \mathrm{e}^-; E^{\circ} = +0.41 \mathrm{~V} \)For reduction, these potentials are reversed. For reductions:- \( \mathrm{Fe}^{2+} + 2 \mathrm{e}^- \longrightarrow \mathrm{Fe}; E^{\circ} = -0.41 \mathrm{~V} \)
03

Calculate Cell EMF

The cell EMF (electromotive force) is calculated using the formula:\[ E_{\text{cell}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ} \]Here:- Anode, where oxidation happens: \( E_{\text{anode}}^{\circ} = +0.76 \text{ V} \)- Cathode, where reduction happens: \( E_{\text{cathode}}^{\circ} = -0.41 \text{ V} \)Thus:\[ E_{\text{cell}}^{\circ} = -0.41 \text{ V} - (+0.76 \text{ V}) = -0.41 \text{ V} - 0.76 \text{ V} = -1.17 \text{ V} \]
04

Determine Correct Answer from Options

The calculated EMF, \(-1.17 \text{ V} \), is compared to the given options.- The expected correct answer should be opposite to \(-1.17 \text{ V}\) values, accounting for zinc and iron roles. However, since this scenario describes an unfavorable reaction; none of the answers match exactly.- A reassessment of roles indicates incorrect inference by leaving emf for positive redox direction possibly wanted; still, readjust for favorable cell gives correct positive prediction.

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Key Concepts

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

Electrochemistry
Electrochemistry is the branch of chemistry that deals with the study of chemical reactions that involve the transfer of electrons. These reactions are common in electrochemical cells, where chemical energy is converted into electrical energy, or vice versa. Understanding electrochemistry is crucial in many applications, ranging from batteries for storing energy to electroplating that enhances the properties of metals.

In the context of electrochemical cells, there are two main half-reactions: oxidation, where electrons are lost, and reduction, where electrons are gained. These half-reactions occur at different electrodes within the cell. The electrode where oxidation takes place is known as the anode, while the electrode for reduction is called the cathode.
  • Oxidation takes place at the anode and involves the release of electrons.
  • Reduction occurs at the cathode and involves the acceptance of electrons.
Understanding these basic components alongside the flow of electrons helps explain how electrochemical cells function and generate electric current. When defining an electrochemical cell, it is important to identify the redox couple involved and understand which species will undergo oxidation or reduction, as this determines the movements of electrons and the direction of the current within the cell.
Standard Oxidation Potentials
Standard oxidation potentials refer to the inherent ability of a chemical species to lose electrons when compared to a standard hydrogen electrode under standard conditions. These potentials are measured in volts (V) and provide a quantifiable measurement of a species' tendency to be oxidized. The more positive the potential, the greater the ability of the species to undergo oxidation.

In the standard tables of electrode potentials, the oxidation potential indicates how easily a species can lose electrons. For example, zinc's oxidation potential is given as +0.76 V, indicating its relative ease of losing electrons compared to iron, which has a lower oxidation potential of +0.41 V.
  • The standard state assumes a pressure of 1 atm, temperature around 25°C, and 1 M concentration for solutions.
  • Standard oxidation potentials are important in predicting the direction in which an electrochemical reaction will proceed.
To convert these values into reduction potentials, which describe the tendency to gain electrons, the sign of the oxidation potential is reversed. Thus, a high positive oxidation potential becomes a negative reduction potential when the perspective of the reactants such as electrons gained is considered. Accurately understanding these potentials is vital in computing the cell potential and determining the feasibility and directionality of electrochemical reactions.
Electromotive Force (EMF)
Electromotive force, abbreviated as EMF, is a critical concept in electrochemistry that represents the voltage developed by any source of electrical energy such as a battery or galvanic cell. The EMF of an electrochemical cell is calculated using the difference between the reduction potentials of the cathode and anode.

The EMF is often denoted by the Greek letter "\(E^{\circ}_{\text{cell}}\)" when referring to standard conditions. It is vital in determining the voltage that can be expected from a cell before any current is drawn from it. The positive value of EMF indicates a spontaneous reaction, meaning the cell can do work on its own. A negative EMF suggests a non-spontaneous reaction.
  • The formula to calculate the cell EMF is: \[E_{\text{cell}}^{\circ} = E_{\text{cathode}}^{\circ} - E_{\text{anode}}^{\circ}\]
  • In a practical sense, EMF helps battery-powered devices function as expected by providing a steady flow of electrons through a circuit.
By comprehending EMF, students are better equipped to understand how energy conversions occur within cells and how this principle is applied in designing efficient and reliable electrochemical systems. A strong grasp of EMF and how it is affected by cell components, like electrode materials and electrolyte concentration, is crucial for optimizing electrochemical devices.

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Most popular questions from this chapter

Match the following $$ \begin{array}{ll} \text { Column-I } & \text { Column-II } \\ \hline \begin{array}{l} \text { (a) } 50 \text { \% solution of } \mathrm{H}_{2} \mathrm{SO}_{4} \\ \text { using Pt electrodes } \end{array} & \text { (p) } \mathrm{H}_{2} \text { is evolved at } \\ \text { cathode } \\ \text { (b) } \begin{array}{l} \text { Dilute solution } \mathrm{NaCl} \\ \text { using Pt electrodes } \end{array} & \text { (q) } \mathrm{O}_{2} \text { is evolved at } \\ \text { (c) } \begin{array}{l} \text { Dilute solution of } \mathrm{H}_{2} \mathrm{SO}_{4} \\ \text { using Cu electrodes } \end{array} & \text { (r) } \mathrm{Cl}_{2} \text { is evolved at } \\ \text { (d) } \begin{array}{c} \text { Concentrated solution of } \\ \text { LiCl using Pt electrodes. } \end{array} & \text { (s) } \mathrm{H}_{2} \mathrm{~S}_{2} \mathrm{O}_{8} \text { is } \\ \text { formed at anode } \\ & \text { (t) non-spontaneous } \\ \text { process } \end{array} $$

The standard reduction potentials of \(\mathrm{Zn}^{2+} \mid \mathrm{Zn}\) and \(\mathrm{Cu}^{2+}\) \(\mathrm{Cu}\) are \(-0.76 \mathrm{~V}\) and \(+0.34 \mathrm{~V}\) respectively. What is the cell emf (in V) of the following cell? (RT \(/ F=0.059\) ) \(\mathrm{Zn}\left|\mathrm{Zn}^{2+}(0.05 \mathrm{M}) \| \mathrm{Cu}^{2+}(0.005 \mathrm{M})\right| \mathrm{Cu}\) (a) \(1.1295\) (b) \(1.0705\) (c) \(1.1\) (d) \(1.041\)

A current of \(15 \mathrm{amp}\) is employed to plate Nickel in a \(\mathrm{NiSO}_{4}\) bath. Both \(\mathrm{Ni}\) and \(\mathrm{H}_{2}\) are formed at the cathode. If \(9.9 \mathrm{~g}\) of \(\mathrm{Ni}\) are deposited with the simultaneous liberation of \(2.51\) litres of \(\mathrm{H}_{2}\) measured at STP, what is the current efficiency for the deposition of Ni? (Atomic weight of \(\mathrm{Ni}=58.7\) ) (a) \(60 \%\) (b) \(70 \%\) (c) \(80 \%\) (d) \(56 \%\)

Given \(\mathrm{E}^{\circ} \mathrm{Cr}^{3+} / \mathrm{Cr}=-0.72 \mathrm{~V}, \mathrm{E}^{\circ} \mathrm{Fe}^{2+} / \mathrm{Fe}=-0.42 \mathrm{~V}\). The potential for the cell \(\mathrm{Cr}\left|\mathrm{Cr}^{3+}(0.1 \mathrm{M}) \| \mathrm{Fe}^{2+}(0.01 \mathrm{M})\right| \mathrm{Fe}\) is (a) \(0.26 \mathrm{~V}\) (b) \(0.399 \mathrm{~V}\) (c) \(-0.339 \mathrm{~V}\) (d) \(-0.26 \mathrm{~V}\)

\(\mathrm{aHNO}_{3}+\mathrm{bH}_{2} \mathrm{~S} \longrightarrow \mathrm{cNO}+\mathrm{d} \mathrm{H}_{2} \mathrm{O}+3 \mathrm{~S} .\) What are the values of \(\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d} ?\) (a) \(2,3,2,2\) (b) \(2,3,2,4\) (c) \(3,3,2,4\) (d) \(4,3,1,2\)

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