Write the nernst equation and e.m.f. of the following cells at \(298 \mathrm{~K}\) : (a) \(\mathrm{Mg}(\mathrm{s})\left|\mathrm{Mg}^{2+}(0.001 \mathrm{M}) \| \mathrm{Cu}^{2+}(0.0001 \mathrm{M})\right| \mathrm{Cu}(\mathrm{s})\) (b) \(\mathrm{Fe}(\mathrm{s})\left|\mathrm{Fe}^{2+}(0.001 M) \| \mathrm{H}^{\mathrm{i}}(1 \mathrm{M})\right| \mathrm{H}_{2}(\mathrm{~g})(1 \mathrm{bar}) \mid \mathrm{Pt}(\mathrm{s})\) (c) \(\mathrm{Sn}(\mathrm{s})\left|\mathrm{Sn}^{2+}(0.050 \mathrm{M}) \| \mathrm{H}^{+}(0.020 \mathrm{M})\right| \mathrm{H}_{2}(\mathrm{~g})(1 \mathrm{bar}) \mid \mathrm{Pt}(\mathrm{s})\) (d) \(\mathrm{Pt}(\mathrm{s})\left|\mathrm{Br}_{2}(\mathrm{l})\right| \mathrm{Br}^{-}(0.010 \mathrm{M}) \| \mathrm{H}^{+}(0.030 \mathrm{M}) \mid \mathrm{H}_{2}(\mathrm{~g})(\mathrm{I} \mathrm{bar}) ! \mathrm{Pt}(\mathrm{s})\) Given : \(E_{\text {OP }}^{\circ} \mathrm{Mg}=2.36 \mathrm{~V}, E_{\mathrm{OP}}^{\circ} \mathrm{Cu}=-0.34 \mathrm{~V}, E_{\mathrm{OP}}^{\mathrm{O}} \mathrm{Fe}=0.44 \mathrm{~V}\) \(E_{\mathrm{OP}}^{\circ} \mathrm{Sn}=0.14 \mathrm{~V}\) and \(E_{\mathrm{OP}}^{\circ} \mathrm{Br}_{2}=-1.09 \mathrm{~V}\) respectively.

Short Answer

Expert verified
To find the e.m.f. of each cell, one must determine the standard cell potential (E^0), calculate the reaction quotient (Q), and then apply the Nernst equation for each specific cell at 298 K, taking into account the number of moles of electrons transferred and the concentrations given for each cell.

Step by step solution

01

- Understand the Nernst Equation

The Nernst equation is used to calculate the cell potential (E) at non-standard conditions. It is expressed as: \[ E = E^0 - \frac{RT}{nF} \ln Q \] where: \(E^0\) is the standard cell potential, R is the ideal gas constant (8.314 J\(\cdot\)mol\(^{-1}\)\(\cdot\)K\(^{-1}\)), T is the temperature in kelvins (T = 298 K), n is the number of moles of electrons transferred in the redox reaction, F is the Faraday constant (96485 C/mol), and Q is the reaction quotient.
02

- Determine Standard Cell Potential (E^0)

For each cell, calculate the standard cell potential using the given standard reduction potentials. It is given by: \[ E^0_{cell} = E^0_{cathode} - E^0_{anode} \]
03

- Calculate the Reaction Quotient (Q)

Use the concentrations of the ionic species in the cell to calculate the reaction quotient, Q. It takes the form: \[ Q = \frac{\text{Products}}{\text{Reactants}} \] with each concentration raised to a power corresponding to its coefficient in the balanced chemical equation.
04

- Utilize the Nernst Equation for Each Cell

Use the determined E^0 and Q for each cell to calculate the cell potential (E) under the given conditions using the Nernst equation. Since T is given as 298 K, R, F, and T will be constants. The value for n will be determined based on the specific redox reaction occurring in each cell.
05

- Calculate the E.m.f. for the Given Cells

Substitute all values into the Nernst equation to find the e.m.f. of each cell. Perform the calculation for each cell separately.

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Key Concepts

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

Electrochemistry
Electrochemistry is a fascinating branch of chemistry that explores the relationship between electrical energy and chemical reactions. Central to this field is the study of redox reactions, where oxidation (loss of electrons) and reduction (gain of electrons) occur simultaneously. These reactions can be harnessed in electrochemical cells, which convert chemical energy into electrical energy or vice versa. Two main types of electrochemical cells are galvanic cells, which generate electrical energy from spontaneous chemical reactions, and electrolytic cells, which use electrical energy to drive non-spontaneous chemical reactions.
Key components of electrochemical cells include electrodes (where the redox reactions occur) and electrolytes (substances that conduct ions and complete the circuit). By analyzing these reactions, scientists and engineers can develop batteries, fuel cells, and various forms of corrosion protection, as well as processes for refining metals and electroplating.
Cell Potential
The cell potential, often denoted as 'E', is a measure of the driving force behind an electrochemical reaction and indicates how much voltage a cell can produce. It's pivotal because it helps predict whether a reaction will proceed spontaneously. The direction of electron flow in an electrochemical cell is from the anode (where oxidation takes place) to the cathode (where reduction occurs). If the cell potential is positive, the reaction is spontaneous; if negative, the reaction is non-spontaneous and requires an external power source to proceed.
Understanding cell potential is crucial when designing batteries and other energy-conversion devices. It enables scientists and engineers to estimate how much electrical energy can be obtained from a given chemical reaction or conversely, how much energy needs to be invested to achieve a chemical transformation in the case of electrolytic cells.
Standard Reduction Potential
Standard reduction potential (often represented as \(E^0\)) is a fundamental aspect of electrochemistry that provides the innate tendency of a substance to gain electrons and thus be reduced. It is measured under standard conditions, which include a solute concentration of 1 M, a pressure of 1 bar for gases, and a temperature of 298 K. Each half-reaction has its associated standard reduction potential.
The more positive the value of \(E^0\), the greater the species' tendency to be reduced. When comparing two half-reactions, the one with the higher reduction potential will act as the cathode, and the other as the anode in an electrochemical cell. The standard cell potential (\(E^0_{cell}\)) can then be calculated by taking the difference between the cathode's and anode's standard reduction potentials. This value helps us to predict the voltage that a cell can produce under standard conditions and to compare the relative strengths of different oxidizing and reducing agents.
Reaction Quotient
The reaction quotient (denoted as Q) is a mathematical expression that describes the ratio of concentrations of products to reactants at a given moment during a chemical reaction, each raised to the power of their stoichiometric coefficients. While the equilibrium constant (K) describes the ratio at equilibrium, Q can be used to determine the reaction's direction at any point.
In the context of electrochemistry and the Nernst equation, the reaction quotient is instrumental in calculating the cell potential at non-standard conditions. Accurately determining the reaction quotient is essential for predicting how the cell potential will shift when the concentrations of reactants or products change. This is especially useful for understanding how batteries will perform under varying conditions, for example, as they become discharged and the concentrations of the involved species evolve.

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Most popular questions from this chapter

$$ \text { Write equations for the electrolysis of } \mathrm{CaH}_{2} \text { in fused state. } $$

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

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.

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.

Calculate the standard cell potentials of galvanic cell in which the following reactions take place : (Given \(E_{\text {OP }}^{\circ} \mathrm{Cr}, \mathrm{Cd}, \mathrm{Fe}^{2+}, \mathrm{Ag}\) are \(0.74 \mathrm{~V}, 0.40 \mathrm{~V},-0.77 \mathrm{~V}\) and \(-0.80 \mathrm{~V}\) respectively) (a) \(2 \mathrm{Cr}_{(\mathrm{s})}+3 \mathrm{Cd}^{2+}\) (aq.) \(\longrightarrow 2 \mathrm{Cr}^{3+}{ }_{\text {(aq. })}^{3+} 3 \mathrm{Cd}\) (b) \(\mathrm{Fe}^{2+}{ }_{\text {(aq.) }}+\mathrm{Ag}_{(\mathrm{aq})}^{+} \longrightarrow \mathrm{Fe}^{3+}{ }_{\text {(aq.) }}+\mathrm{Ag}_{(\mathrm{s})}\) Calculate the \(\Delta_{\mathrm{r}} G^{\circ}\) and equilibrium constant of the reactions.

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