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.

Short Answer

Expert verified
To find the standard cell potentials, subtract the anode potential from the cathode potential for both reactions, then calculate \(\Delta_r G^o\) using \(\Delta_r G^o = -nFE_cell^o\) and the equilibrium constant using \(K = \exp(-\Delta_r G^o / RT)\).

Step by step solution

01

- Identify the Anode and Cathode for Reaction (a)

First, identify the oxidation and reduction half-reactions for reaction (a). Chromium is oxidized from Cr (solid) to Cr^3+ (aqueous), and cadmium is reduced from Cd^2+ (aqueous) to Cd (solid). Oxidation takes place at the anode, and reduction takes place at the cathode.
02

- Calculate the Standard Cell Potential for Reaction (a)

Use the standard reduction potentials given to calculate the standard cell potential (E_cell^o). The standard cell potential is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode: E_cell^o = E_cathode^o - E_anode^o. For reaction (a), E_cathode^o = E_Cr^o and E_anode^o = E_Cd^o.
03

- Apply the Formula

Calculate E_cell^o for reaction (a) by substituting the given values: E_cell^o_a = E_Cr^o - E_Cd^o = 0.74 V - 0.40 V.
04

- Identify the Anode and Cathode for Reaction (b)

Similarly, identify the oxidation and reduction half-reactions for reaction (b). Iron is oxidized from Fe^2+ (aqueous) to Fe^3+ (aqueous), and silver is reduced from Ag^+ (aqueous) to Ag (solid).
05

- Calculate the Standard Cell Potential for Reaction (b)

Determine E_cell^o for reaction (b) using the given standard reduction potentials and the formula: E_cell^o = E_cathode^o - E_anode^o. For reaction (b), E_cathode^o = E_Ag^o and E_anode^o = E_Fe^o.
06

- Apply the Formula

Calculate E_cell^o for reaction (b) by substituting the given values: E_cell^o_b = E_Ag^o - E_Fe^o = -0.80 V - (-0.77 V).
07

- Calculate the Standard Gibbs Free Energy Change

Use the formula \(\Delta_r G^o = -nFE_cell^o\) to calculate the standard Gibbs free energy change (\(\Delta_r G^o\)) for both reactions. First, determine the number of moles of electrons transferred (n) in each half-reaction, and then apply the formula using Faraday's constant (F = 96,485 C/mol).
08

- Calculate the Equilibrium Constant

Use the relationship between Gibbs free energy change and the equilibrium constant (K): \(\Delta_r G^o = -RT \ln K\), where R is the gas constant and T is the temperature (in Kelvin). Rearrange to solve for K: \(K = \exp(-\Delta_r G^o / RT)\), and calculate K for both reactions using the appropriate values for \(\Delta_r G^o\).

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Key Concepts

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

Galvanic Cell
The galvanic cell, also known as a voltaic cell, is an essential concept in electrochemistry that turns chemical energy into electrical energy through spontaneous redox reactions. It consists of two different metals connected by a salt bridge, with each metal in its respective electrolyte solution. One metal acts as an anode, losing electrons through oxidation, while the other acts as a cathode, gaining electrons through reduction.

Each half-cell has its standard electrode potential, and the difference between these potentials gives the standard cell potential of the galvanic cell. This potential is a measure of the electrical driving force of the cell. In a typical textbook exercise, students may be asked to calculate these potentials using given standard reduction potentials and then determine the direction of electron flow from anode to cathode.
Standard Gibbs Free Energy Change
The standard Gibbs free energy change (\(\triangle_r G^\text{o}\)) is a pivotal concept connecting electrochemistry and thermodynamics. It represents the maximum amount of work that can be done by a chemical reaction at standard conditions (1 bar pressure for gases, 1 M concentration for solutes, and 298.15 K temperature). A negative value indicates a spontaneous reaction.

For electrochemical cells, the relationship between the standard cell potential and standard Gibbs free energy change is defined by the formula:
\[ \triangle_r G^\text{o} = -nFE^\text{o}_\text{cell} \]
where \(n\) is the number of moles of electrons transferred in the reaction, \(F\) is Faraday's constant (approximately 96,485 C/mol), and \(E^\text{o}_\text{cell}\) is the standard cell potential. Students will often have to calculate this value after determining the standard cell potential to predict the spontaneity of a reaction.
Equilibrium Constant
The equilibrium constant (\(K\)) is a fundamental quantity in chemistry that expresses the ratio of the concentration of the products to the reactants at equilibrium. It provides insight into the extent of a chemical reaction and its dynamic state at a given temperature.

In electrochemistry, there is a direct relationship between the standard Gibbs free energy change and the equilibrium constant, described by:
\[ \triangle_r G^\text{o} = - RT \text{ln} K \]
Here, \(R\) is the universal gas constant and \(T\) represents the temperature in Kelvin. By rearranging this formula, students can calculate the equilibrium constant for a reaction occurring in a galvanic cell:
\[ K = \text{exp}\bigg(-\frac{\triangle_r G^\text{o}}{RT}\bigg) \]
An understanding of \(K\) helps predict favorability under standard conditions and is crucial when dealing with galvanic cells in thermodynamic calculations.
Electrochemistry
Electrochemistry is the study of chemical processes that cause electrons to move, creating an electric current. It bridges the gap between electrical energy and chemical reactions, including the important principles of galvanic cells, electrode potentials, and the relationship between electric potential, free energy, and chemical equilibrium.

Within this field, standard cell potentials provide essential information such as reaction spontaneity and battery voltage. Gibbs free energy change informs us about the feasibility and work potential of reactions. The equilibrium constant quantifies the position of equilibrium and reaction quotients. Fundamental understanding of these concepts allows students to not only solve textbook exercises but also gain deeper insights into the practical applications of electrochemistry in energy storage, corrosion, and metallurgy.

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Most popular questions from this chapter

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.

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

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.

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.

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

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