/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 For the following cell, \(\mathr... [FREE SOLUTION] | 91Ó°ÊÓ

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For the following cell, \(\mathrm{Zn}(\mathrm{s})\left|\mathrm{ZnSO}_{4}(\mathrm{aq}) \| \mathrm{CuSO}_{4}(\mathrm{aq})\right| \mathrm{Cu}(\mathrm{s})\) when the concentration of \(\mathrm{Zn}^{2+}\) is 10 times the concentration of \(\mathrm{Cu}^{2+}\), the expression for \(\Delta G\) (in \(\mathrm{J} \mathrm{mol}^{-1}\) ) is \([F\) is Faraday constant; \(R\) is gas constant; \(T\) is temperature; \(E^{0}\) (cell) \(\left.=1.1 \mathrm{~V}\right]\) [Adv. 2017] (a) \(1.1 \mathrm{~F}\) (b) \(2.303 R T-2.2 F\) (c) \(2.303 R T+1.1 F\) (d) \(-2.2 F\)

Short Answer

Expert verified
The correct expression is option (b): \(2.303RT - 2.2F\).

Step by step solution

01

Identify the Nernst Equation

The Nernst Equation is used to calculate the potential of an electrochemical cell under non-standard conditions. It is given by: \[ E = E^0 - \frac{RT}{nF} \ln Q \]where \( E \) is the cell potential under non-standard conditions, \( E^0 \) is the standard cell potential (1.1 V in this case), \( R \) is the gas constant, \( T \) is the temperature in Kelvin, \( n \) is the number of moles of electrons exchanged, \( F \) is the Faraday constant, and \( Q \) is the reaction quotient.
02

Calculate the Reaction Quotient Q

For the given cell, the reaction is: \[ \text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s) \]The reaction quotient \( Q \) is given by:\[ Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \]Since \([\text{Zn}^{2+}]\) is 10 times \([\text{Cu}^{2+}]\), \( Q = 10 \).
03

Determine the Expression for ΔG

The change in Gibbs free energy \( \Delta G \) is related to the cell potential by the equation:\[ \Delta G = -nFE \]Substituting for \( E \) from the Nernst equation:\[ \Delta G = -nF \left( E^0 - \frac{RT}{nF} \ln Q \right) \]\[ \Delta G = -nFE^0 + RT \ln Q\]
04

Substitute Known Values and Simplify

Given that \( n = 2 \) because 2 moles of electrons are exchanged, \( E^0 = 1.1 \, \text{V} \), and \( Q = 10 \):\[ \Delta G = -2F(1.1) + 2.303RT \ln(10) \]Simplifying gives:\[ \Delta G = -2.2F + 2.303RT \]
05

Identify the Correct Answer Choice

Look at the given options to find which expresses the derived equation for \( \Delta G \). The equation \( \Delta G = 2.303RT - 2.2F \) matches option (b), which is the correct expression for \( \Delta G \) under the given conditions.

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Key Concepts

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

Nernst Equation
The Nernst Equation is a powerful tool used in electrochemistry to determine the potential of an electrochemical cell under non-standard conditions. It provides insights into how cell potential changes with varying concentrations of ions. The equation is expressed as:- \[ E = E^0 - \frac{RT}{nF} \ln Q \]- Here:
  • \( E \) is the cell potential under actual conditions.
  • \( E^0 \) refers to the standard cell potential.
  • \( R \) is the universal gas constant, \( 8.314 \) J/(molâ‹…K).
  • \( T \) is the temperature in Kelvin.
  • \( n \) denotes the number of moles of electrons involved in the reaction.
  • \( F \) is Faraday's constant (approximately 96485 C/mol).
  • \( Q \) is the reaction quotient, representing the ratio of product and reactant concentrations.
Breaking it down, the equation shows us how a difference in ion concentration can influence the voltage of the cell. For example, in our reaction, where the concentration of \( \mathrm{Zn}^{2+} \) is 10 times that of \( \mathrm{Cu}^{2+} \), the Nernst equation helps calculate the resulting cell potential, factoring in this concentration gradient.
Reaction Quotient
The reaction quotient, \( Q \), is a crucial aspect in understanding the Nernst Equation's utility. It is a measure of the ratio of concentrations of products to reactants at any point in a reaction that is not necessarily at equilibrium. The expression for \( Q \) changes depending on the balanced chemical equation. For our zinc-copper cell system, the reaction is:- \[ \text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s) \]- From this, \( Q \) is determined by:- \[ Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \]- Because the zinc ion concentration is provided as 10 times the copper ion concentration, this gives us \( Q = 10 \).
Reacting systems constantly aim to reach equilibrium. As such, the reaction quotient provides a snapshot of the reaction's status relative to equilibrium:
  • If \( Q < K \) (equilibrium constant), the reaction proceeds forward.
  • If \( Q = K \), the system is at equilibrium.
  • If \( Q > K \), the reaction shifts backward.
In conjunction with the Nernst Equation, \( Q \) becomes a vital component in predicting cell potentials under varying conditions.
Gibbs Free Energy
Gibbs Free Energy (\( \Delta G \)) is a term that combines enthalpy and entropy to determine the spontaneity of a reaction. It tells us whether a process can occur under constant pressure and temperature. When it comes to electrochemical cells, \( \Delta G \) can be directly linked to the cell potential by:- \[ \Delta G = -nFE \]- Where:
  • \( \Delta G \) is the change in Gibbs Free Energy.
  • \( n \) refers to the number of moles of electrons exchanged in the reaction.
  • \( F \) stands for Faraday's constant.
  • \( E \) is the cell potential.
By substituting the Nernst equation into this expression, we can understand how changes in ion concentration and temperature affect \( \Delta G \):- \[ \Delta G = -nF \left( E^0 - \frac{RT}{nF} \ln Q \right) \]- For our problem:- \[ \Delta G = -2F(1.1) + 2.303RT \ln(10) \]- This shows how increasing the concentration of \( \mathrm{Zn}^{2+} \) impacts the Gibbs Free Energy, ultimately affecting the reaction's spontaneity. When \( \Delta G \) is negative, the reaction is spontaneous, moving towards equilibrium naturally; a positive \( \Delta G \) indicates non-spontaneity under the given conditions.

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Most popular questions from this chapter

A gas \(X\) at 1 atm is bubbled through a solution containing a mixture of \(1 \mathrm{M} Y^{-}\)and \(1 \mathrm{M} Z^{-}\)at \(25^{\circ} \mathrm{C}\). If the reduction potential of \(Z>Y>X\) then, [1999 - 2 Marks] (a) \(Y\) will oxidize \(X\) and \(\operatorname{not} Z\) (b) \(Y\) will oxidize \(Z\) and \(\operatorname{not} X\) (c) \(Y\) will oxidize both \(X\) and \(Z\) (d) \(Y\) will reduce both \(X\) and \(Z\)

For the cell \(\mathrm{Zn}(\mathrm{s}) \mid \mathrm{Zn}^{2+}(\mathrm{aq}) \| \mathrm{M}^{\mathrm{x}+}\) (aq) \(\mid \mathrm{M}(\mathrm{s})\), different half cells and their standard electrode potentials are given below: (aq)/ (aq)/ \(\mathrm{Au}^{3+}(\mathrm{aq}) / \mathrm{Ag}^{+}(\mathrm{aq}) / \mathrm{Fe}^{3+}(\mathrm{aq}) / \mathrm{Fe}^{2+}(\mathrm{a}\) \(\mathrm{Au}(\mathrm{s}) \quad \mathrm{Ag}(\mathrm{s}) \quad \mathrm{Fe}^{2+}(\mathrm{aq}) \mathrm{Fe}(\mathrm{s})\) \(\frac{\mathrm{M}^{\mathrm{x}+}(\mathrm{aq}) /}{\mathrm{M}(\mathrm{s})}\) (aq) 1 \(\mathrm{E}_{\mathrm{M}^{\mathrm{x}+} / \mathrm{M}}^{\circ} /(\mathrm{V}) \overline{1.40}\) 1 0 \(\begin{array}{lll}0.80 & 0.77 & -0.44\end{array}\) If \(\mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}=-0.76 \mathrm{~V}\), which cathode will give a maximum value of \(\mathrm{E}_{\mathrm{cell}}^{\circ}\) per electron transferred? [Main Jan. \(11,2019(\mathrm{I})]\) (a) \(\mathrm{Ag}^{+} / \mathrm{Ag}\) (b) \(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\) (c) \(\mathrm{Au}^{3+} / \mathrm{Au}\) (d) \(\mathrm{Fe}^{2+} / \mathrm{Fe}\)

The Edison storage cells is represented as \(\mathrm{Fe}(\mathrm{s})|\mathrm{FeO}(\mathrm{s})| \mathrm{KOH}(\mathrm{aq})\left|\mathrm{Ni}_{2} \mathrm{O}_{3}(\mathrm{~s})\right| \mathrm{Ni}(\mathrm{s})\) The half-cell reactions are : \(\mathrm{Ni}_{2} \mathrm{O}_{3}(\mathrm{~s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\rightleftharpoons 2 \mathrm{e}^{-} 2 \mathrm{NiO}_{(s)}+2 \mathrm{OH}^{-}\) \(E^{\circ}=+0.40 \mathrm{~V}\) \(\mathrm{FeO}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(1)+2 \mathrm{e}^{-} \rightleftharpoons \mathrm{Fe}(\mathrm{s})+2 \mathrm{OH}^{-} ; E^{0}=-0.87 \mathrm{~V}\) (i) What is the cell reaction? (ii) What is the cell e.m.f ? How does it depend on the concentration of \(\mathrm{KOH}\) ? (iii) What is the maximum amount of electrical energy that can be obtained from one mole of \(\mathrm{Ni}_{2} \mathrm{O}_{3} ?\)

The rusting of iron takes place as follows \(2 \mathrm{H}^{+}+2 \mathrm{e}^{-}+1 / 2 \mathrm{O}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O}(1) ; E^{\circ}=+1.23 \mathrm{~V}\) \(\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(\mathrm{s}) ; E^{\circ}=-0.44 \mathrm{~V}\) Calculate \(\Delta G^{\circ}\) for the net process (a) \(-322 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(-161 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(-152 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(-76 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

The standard reduction potential for the half-cell \(\mathrm{NO}_{3}^{-}(\mathrm{aq})+2 \mathrm{H}^{+}(\mathrm{aq})+\mathrm{e} \rightarrow \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}\) is \(0.78 \mathrm{~V}\) (i) Calculate the reduction potential in \(8 \mathrm{M} \mathrm{H}^{+}\) (ii) What will be the reduction potential of the half-cell in a neutral solution? Assume all the other species to be at unit concentration.

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