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The standard potentials of \(\mathrm{OCl} / \mathrm{Cl}^{-}\)and \(\mathrm{Cl} / 1 / 2 \mathrm{Cl}_{2}\) are \(0.94 \mathrm{~V}\) and \(1.36 \mathrm{~V}\) respectively. The \(\mathrm{E}^{\circ}\) value of \(\mathrm{OCl} / 1 / 2 \mathrm{Cl}_{2}\) will be a. \(-0.42 \mathrm{~V}\) b. \(2.3 \mathrm{~V}\) c. \(0.42 \mathrm{~V}\) d. \(0 . \mathrm{V}\)

Short Answer

Expert verified
The E° value of OCl / 1/2 Cl_2 is -0.42 V. (Answer a)

Step by step solution

01

Understanding Standard Potentials

The standard potential (E°) is a measure of the tendency of a chemical species to gain electrons and be reduced. It is given in volts and typically compared to the standard hydrogen electrode (SHE) as the reference point.
02

Given Potentials

We are given two redox reactions with their standard potentials: 1. f( ext{OCl} / ext{Cl}^- ) = 0.94 ext{ V} 2. f( ext{Cl}^- / rac{1}{2} ext{Cl}_2 ) = 1.36 ext{ V}.
03

Identifying Net Reaction

To find the f value of f( ext{OCl} / rac{1}{2} ext{Cl}_2 ) we need to calculate the standard potential for the overall reaction: f ext{OCl} ightarrow rac{1}{2} ext{Cl}_2.
04

Calculating the Target Potential

The f value for the target reaction can be determined by first reversing the f( ext{Cl}^- / rac{1}{2} ext{Cl}_2 ) reaction to rac{1}{2} ext{Cl}_2 ightarrow ext{Cl}^- which changes the potential to -1.36 V. Then add the two potentials: 0.94 ext{ V} + (-1.36 ext{ V}) = -0.42 ext{ V}.
05

Verifying the Calculation

The calculation confirms that f( ext{OCl} / ext{Cl}^- ) + [ - f( rac{1}{2} ext{Cl}_2 / ext{Cl}^- ) ] = f( ext{OCl} / rac{1}{2} ext{Cl}_2 ) which means: 0.94 ext{ V} - 1.36 ext{ V} = -0.42 ext{ V}. Therefore, the answer is correct.

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Key Concepts

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

Redox Reactions
Redox reactions are fundamental chemical processes where oxidation and reduction occur simultaneously. In simple terms, oxidation involves the loss of electrons, while reduction means the gain of electrons. Redox stands for "reduction-oxidation," highlighting their interplay.
Redox reactions are vital for various chemical processes, including:
  • Combustion
  • Photosynthesis
  • Metabolism in living organisms
In the context of electrochemical cells, a redox reaction is split into two half-reactions: one for oxidation and one for reduction. Each half-reaction has a standard potential, usually measured in volts. This potential allows us to predict the direction in which the reaction will proceed under standard conditions.
The stronger the oxidizing or reducing agent, the more spontaneous the reaction, which is crucial for processes like energy generation in batteries.
Electrochemical Cells
Electrochemical cells are devices that convert chemical energy into electrical energy or vice versa, using redox reactions. These cells consist of two electrodes: the anode and the cathode.
In a galvanic cell, for example:
  • The anode undergoes oxidation, while the cathode undergoes reduction.
  • Electrons flow from the anode to the cathode through an external circuit, generating electrical energy.
The cell's voltage is determined by the difference in the standard potentials of the two electrodes. This is known as the electromotive force (EMF) of the cell.
A positive EMF indicates a spontaneous reaction, a key factor in designing batteries and other power storage devices. Standard electrode potentials, like the ones given in the exercise, are essential for calculating this EMF, helping us understand the feasibility and efficiency of the cell's operation.
Standard Hydrogen Electrode
The standard hydrogen electrode (SHE) is a reference electrode used to measure electrode potentials. It is depicted by the half-reaction: \[ \text{H}_2(g) + 2\text{e}^- \rightarrow 2\text{H}^+(aq) \]This system is assigned a standard electrode potential of 0 volts, serving as a baseline for comparing other electrodes.
Key features of the SHE include:
  • A solution of hydrogen ions at a 1 M concentration.
  • Hydrogen gas at 1 atm pressure.
  • A platinum electrode as the interface.
It plays a critical role in determining the relative potentials of other electrodes, like in the original exercise. The potential difference between the SHE and another electrode will tell us how likely a reaction is to occur and in which direction it will proceed, supporting the design and understanding of electrochemical cells.

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Most popular questions from this chapter

The \(\mathrm{emf}, \mathrm{E}\), is related to the change in Gibbs free energy, \(\Delta \mathrm{G}: \Delta \mathrm{G}=-\mathrm{nFE}\), where is the number of electrons transferred during the redox process and \(F\) is a unit called the Faraday. The faraday is the amount of charge on \(1 \mathrm{~mol}\) of electrons: \(1 \mathrm{~F}=96,500 \mathrm{C} / \mathrm{mol}\). Because \(\mathrm{E}\) is related to \(\Delta \mathrm{G}\), the sign of \(\mathrm{E}\) indicates whether a redox process is spontaneous: \(\mathrm{E}>0\) indicates a spontaneous process, and \(\mathrm{E}<0\) indicates a non-spontaneous one. \(\mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Ag}(\mathrm{s}) \mathrm{E}^{\circ}=+0.800 \mathrm{~V}\) \(\mathrm{AgBr}(\mathrm{s})+\mathrm{e}^{-} \rightarrow \mathrm{Ag}(\mathrm{s})+\mathrm{Br}^{-}(\mathrm{aq})\) \(\mathrm{E}^{\mathrm{o}}=+0.071 \mathrm{~V}\) \(\mathrm{Br}_{2}(\mathrm{l})+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-}(\mathrm{aq}) \mathrm{E}^{\circ}=+1.066 \mathrm{~V}\) Use some of the above data to calculate Ksp at \(25^{\circ} \mathrm{C}\) for \(\mathrm{AgBr}\). a. \(6.7 \times 10^{-12}\) b. \(2.8 \times 10^{-14}\) c. \(3.7 \times 10^{-19}\) d. \(4.9 \times 10^{-13}\)

In the following question two statements (Assertion) A and Reason (R) are given. Mark a. if \(\mathrm{A}\) and \(\mathrm{R}\) both are correct and \(\mathrm{R}\) is the correct explanation of \(\mathrm{A}\); b. if \(A\) and \(R\) both are correct but \(R\) is not the correct explanation of A; c. \(\mathrm{A}\) is true but \(\mathrm{R}\) is false; d. A is false but \(\mathrm{R}\) is true, e. A and \(R\) both are false. (A): When acidified zinc sulphate solution is electrolyzed between zinc electrodes. It is zinc that is deposited at the cathode and hydrogen evolution does not take place (R): The electrode potential of zinc is more negative than hydrogen as the overvoltage for hydrogen evolution on zinc is quite large.

The standard reduction potentials of \(\mathrm{Zn}\) and Ag in water are \(\mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \leftrightarrow \mathrm{Zn}\); \(\left(\mathrm{E}^{\circ}=-0.76 \mathrm{~V}\right)\) and \(\mathrm{Ag}^{+}+\mathrm{e}^{-} \leftrightarrow \mathrm{Ag}\) \(\left(\mathrm{E}^{\circ}=+0.80 \mathrm{~V}\right)\) at \(298 \mathrm{~K}\). Which of the following reaction is/are not feasible? a. \(\mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Zn}(\mathrm{s})+\mathrm{Ag}(\mathrm{s})\) b. \(\mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s}) \rightarrow 2 \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{Zn}(\mathrm{s})\) c. \(\mathrm{Zn}(\mathrm{s})+\mathrm{Ag}(\mathrm{s}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Ag}^{+}(\mathrm{aq})\) d. \(\mathrm{Zn}(\mathrm{s})+2 \mathrm{Ag}^{+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{Ag}\) (s)

The passage of electricity through certain electrolyte results in the liberation of \(\mathrm{H}_{2}\) gas at the cathode. The electrolyte could be: a. \(\mathrm{CuCl}_{2}(\mathrm{aq})\) b. \(\mathrm{NaCl}(\mathrm{aq})\) c. \(\mathrm{CaCl}_{2}(\mathrm{aq})\) d. \(\mathrm{AgNO}_{3}\) (aq)

The \(\mathrm{emf}, \mathrm{E}\), is related to the change in Gibbs free energy, \(\Delta \mathrm{G}: \Delta \mathrm{G}=-\mathrm{nFE}\), where is the number of electrons transferred during the redox process and \(F\) is a unit called the Faraday. The faraday is the amount of charge on \(1 \mathrm{~mol}\) of electrons: \(1 \mathrm{~F}=96,500 \mathrm{C} / \mathrm{mol}\). Because \(\mathrm{E}\) is related to \(\Delta \mathrm{G}\), the sign of \(\mathrm{E}\) indicates whether a redox process is spontaneous: \(\mathrm{E}>0\) indicates a spontaneous process, and \(\mathrm{E}<0\) indicates a non-spontaneous one. The standard electrode potential for the following reaction is \(+1.33 \mathrm{~V}\). What is the potential at \(\mathrm{pH}=2.0 ?\) $$ \begin{array}{r} \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq}, 1 \mathrm{M})+14 \mathrm{H}^{+}(\mathrm{aq})+6 \mathrm{e}^{-} \rightarrow \\ 2 \mathrm{Cr}^{3+}(\mathrm{aq}, 1 \mathrm{M})+7 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \end{array} $$ a. \(+2.184 \mathrm{~V}\) b. \(+0.5022 \mathrm{~V}\) c. \(+1.008 \mathrm{~V}\) d. \(+1.0542 \mathrm{~V}\)

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