$$ \begin{aligned} &\text { The e.m.f. of the cell } M \mid M^{n+}(0.02 M) \| \mathrm{H}^{+}(1 M) \mathrm{H}_{2(\mathrm{~g})}(1 \mathrm{~atm}) \mathrm{Pt} \text { at }\\\ &25^{\circ} \mathrm{C} \text { is } 0.81 \mathrm{~V} \text { . Calculate the valence of metal if } E^{\circ}{ }_{\ldots 1 / 2}^{n+}=0.76 \mathrm{~V} \end{aligned} $$

Short Answer

Expert verified
Calculate the natural logarithm of \frac{1}{0.02}, multiply it with \frac{8.314 \cdot 298}{96500}, and then divide the difference between the standard electrode potential and the cell potential by this product to get the valence \(n\).

Step by step solution

01

- Understand the Nernst Equation

The Nernst equation relates the cell potential to the standard electrode potential, temperature, and the activities (or concentrations) of the reactants and products. It is given by: \(E = E^\circ - \frac{RT}{nF} \ln\frac{[\text{Products}]}{[\text{Reactants}]}\) where - \(E\) is the cell potential, - \(E^\circ\) is the standard cell potential, - \(R\) is the universal gas constant, - \(T\) is the temperature in Kelvin, - \(n\) is the number of moles of electrons exchanged, - \(F\) is the Faraday's constant, - \([\text{Products}]\) and \([\text{Reactants}]\) represent the activities or concentrations of the products and reactants, respectively.
02

- Apply the Nernst Equation

For the cell described, the e.m.f. (cell potential) is known to be 0.81 V and the standard electrode potential is 0.76 V. Therefore, we can set up the equation: \(0.81 V = 0.76 V - \frac{(8.314 \cdot 298)}{n \cdot 96500} \ln\frac{1}{0.02}\) Here we plugged in \(R=8.314 \mathrm{\frac{J}{mol \cdot K}}\), \(T=298K\) (equivalent to 25°C), and \(F=96500 \mathrm{\frac{C}{mol}}\). We need to determine the unknown valence \(n\), which is the number of moles of electrons transferred.
03

- Solve for the Valence

Rearranging the equation to solve for \(n\): \(n = \frac{(8.314 \cdot 298)}{(0.76 V - 0.81 V) \cdot 96500} \ln\frac{1}{0.02}\) Solve the natural logarithm and then the fraction to find the value of \(n\).
04

- Calculate the Valence

Upon calculating the natural logarithm and division, we will obtain a value for \(n\) which represents the valence of the metal. Since \(n\) must be a whole number, round off to the nearest whole number if necessary.

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Key Concepts

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

Cell Potential Calculation
Cell potential calculation is central to understanding the workings of electrochemical cells. The overall voltage, or electromotive force (e.m.f.), of a cell determines its ability to drive an electric current through a circuit.
To calculate the cell potential, we must consider both the standard electrode potential and the actual conditions under which the cell operates. The Nernst equation allows us to calculate the cell potential under non-standard conditions by incorporating the concentrations of the reactants and products involved in the cell's electrochemical reaction.
When using the Nernst equation, as we've seen in the solution, we take the standard electrode potential and adjust it based on the ratio of product and reactant concentrations. This reflects the real-world scenario where concentrations often differ from the standard state. To ensure a solid understanding of the cell potential calculation, remember that the cell potential diminishes as the reaction proceeds and the concentration of the reactants decreases.
Electrochemistry
Electrochemistry is the branch of chemistry that studies the interrelation of electrical currents and chemical reactions. It involves the conversion between chemical energy and electrical energy, which is the basis for batteries, fuel cells, and electrolysis.
Key concepts that we encounter in electrochemistry include oxidation and reduction processes, which occur at the electrodes during the operation of an electrochemical cell. The flow of electrons from one substance to another is what generates an electric current outside the cell and a transfer of charge through ions within the cell. Electrochemistry is not only about generating electrical energy but also encompasses the use of electrical energy to drive chemical reactions that are not spontaneous, a process known as electrolysis.
In the context of our exercise, we dive into the practical application of electrochemistry through understanding cell potential and its calculation, which directly relates to the efficiency and functionality of electrochemical cells.
Standard Electrode Potential
The standard electrode potential (denoted as E°) is a measure of the propensity of a substance to lose or gain electrons in an electrochemical reaction, also known as a redox reaction.
It is a half-cell potential measured under standard conditions: a solute concentration of 1 M, a pressure of 1 atm for gases, and a specified temperature, usually 25°C (298 K). By convention, the standard electrode potential for hydrogen is set at 0 V, and other potentials are determined relative to it.
This concept is crucial because it allows chemists to predict the direction of redox reactions and to construct electrochemical cells with a desired e.m.f. Understanding the standard electrode potential is also instrumental in figuring out which metals can displace others from solutions, and in determining the spontaneity of redox reactions. It's also a valuable tool for troubleshooting exercises such as the one given, wherein we apply the standard electrode potential to find an unknown metal valence.

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Most popular questions from this chapter

Given the standard electrode potentials; \(\mathrm{K}^{+} / \mathrm{K}=-2.93 \mathrm{~V}, \quad \mathrm{Ag}^{+} / \mathrm{Ag}=0.80 \mathrm{~V}, \quad \mathrm{Hg}^{2+} / \mathrm{Hg}=0.79 \mathrm{~V}\) \(\mathrm{Mg}^{2+} / \mathrm{Mg}=-2.37 \mathrm{~V}, \quad \mathrm{Cr}^{3+} / \mathrm{Cr}=-0.74 \mathrm{~V}\) Arrange these metals in their increasing order of reducing power.

For the cell : $$ \text { Zn }\left|\begin{array}{c} \mathrm{Zn}_{\text {aq }}^{2+} \\ 1 M \end{array}\right|\left|\begin{array}{c|c} \mathrm{Cu}_{a q}^{2+} \\ 2 M \end{array}\right| \mathrm{Cu} $$ Calculate the values for ; (a) cell reaction, (b) \(E_{\text {cell }}^{\circ}\). (c) \(E_{\text {cell }}\) (d) the minimum concentration of \(\mathrm{Cu}^{2+}\) at which cell reaction is spontaneous if \(\mathrm{Zn}^{2+}\) is \(1 M\) (e) does the displacement of \(\mathrm{Cu}^{2+}\) goes almost to completion. Given : \(E_{R P_{\mathrm{Cu}^{2+} / \mathrm{Cu}}}^{\circ}=+0.35 \mathrm{~V}\) $$ E_{R P_{\mathrm{Zn}^{2+} / Z n}}^{\circ}=-0.76 \mathrm{~V} $$

The e.m.f. of a cell corresponding to the reaction : $$ \begin{aligned} \mathrm{Zn}(s)+2 \mathrm{H}^{+}(\mathrm{aq}) \longrightarrow \mathrm{Zn}^{2+}+& \mathrm{H}_{2}(\mathrm{~g}) \\ (0.1 \mathrm{M}) &(1 \mathrm{~atm}) \end{aligned} $$ is \(0.28 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\) and \(E_{\mathrm{Zn} / \mathrm{Zn}^{2+}}^{\circ}=0.76 \mathrm{~V}\) (i) Write half cell reactions. (ii) Calculate \(\mathrm{pH}\) of the solution at \(\mathrm{H}\) electrode.

Two metals \(A\) and \(B\) have \(E_{\mathrm{RP}}^{\circ}=+0.76 \mathrm{~V}\) and \(-0.80 \mathrm{~V}\) respectively. Which will liberate \(\mathrm{H}_{2}\) from \(\mathrm{H}_{2} \mathrm{SO}_{4} ?\)

A metal is known to form fluoride \(M \mathrm{~F}_{2}\). When 10 ampere electricity is passed throwgh i multen salt for \(330 \mathrm{sec}, 1.95 \mathrm{~g}\) metal is deposited. Find out the atomic weight of metal. What will be the quantity of charge required to deposit the same mass of \(\mathrm{Cu}\) from \(\mathrm{CuSO}_{4}\) (aq.) (At. wt. of \(\mathrm{Cu}=63.6\) )

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