Case study (4 Marks)

Question 14 Marks

Standard electrode potentials are used for various processes :

It is used to measure relative strengths of various oxidants and reductants.
It is used to calculate standard cell potential.
It is used to predict possible reactions.

A set of half$-$reactions $($in acidic medium$)$ along with their standard reduction potential$, E^\circ\ ($in volt$)$ values are given below :
$\ce{I_2 + 2e^- \rightarrow 2I^- ; E^\circ = 0.54 V}$
$\ce{Cl_2 + 2e^- \rightarrow 2Cl^- ; E^\circ = 1.36 V}$
$\ce{Mn^{3+} +e^- \rightarrow Mn^{2+}; E^\circ = 1.50 V}$
$\ce{Fe^{3+} + e^- \rightarrow Fe^{2+}; E^\circ = 0.77 V}$
$\ce{O2 + 4H^+ + 4e^- \rightarrow 2H2O ; E^\circ = 1.23 V}$
The following questions are multiple choice questions. Choose the most appropriate answer :

Which of the following statements is correct?

$Cl^-$ is oxidised by $\ce{O_2}$.
$Fe^{2+}$ is oxidised by iodine.
$I^-$ is oxidised by chlorine.
$Mn^{2+}$ is oxidised by chlorine.

$Mn^{3+}$ is not stable in acidic medium, while $Fe^{3+}$ is stable because :

$\ce{O2}$ oxidises $Mn^{2+}$ to $Mn^{3+}$
$\ce{O2}$ oxidises both $Mn^{2+}$ to $Mn^{3+}$ and $Fe^{2+}$ to $Fe^{3+}$
$Fe^{3+}$ oxidises $\ce{H2O}$ to $\ce{O2}$
$Mn^{3+}$ oxidises $\ce{H2O}$ to $\ce{O2}$

The strongest reducing agent in the aqueous solution is :

$I^-$
$Cl^-$
$Mn^{2+}$
$Fe^{2+}$

The emf for the following reaction is :

$\text{I}_2+\text{KCl}\rightleftharpoons2\text{KI}+\text{Cl}_2$

$-0.82 V$
$+0.82 V$
$-0.73 V$
$+0.73 V$

Which of the following statements is correct for the following reaction?

$Fe^{3+} + Mn^{2+} \rightarrow Fe^{2+ }+ Mn^{3+}$

The emf of the cell is positive.
$Fe^{3+}$ oxidises $Mn^{2+}$.
The reaction does not occur.
All are correct.

Answer

$(c)\ I^-$ is oxidised by chlorine.

The half cell having the higher reduction potential will undergo reduction process.

$(d)\ Mn^{3+}$ oxidises $H_2O$ to $O_2$

Electrode potential of $Mn^{3+}$ is higher than $O_2.$

$(a)\ I^-$

Due to least electrode potential value.

$(a)\ -0.82 V$

Half reactions:
$I_2 2e^- \rightarrow 2I^-$
$2CI^- \rightarrow CI_2 + 2e^-$
$\text{Reduction}\text{ E}^\circ= 0.54\text{V}\\\text{Oxidation}\text{ E}^\circ=-1.36\text{V}\\\overline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\\\ \text{e.m.f}=-0.82\text{V}$

$(c)$ The reaction does not occur.

$Fe^{3+} + Mn^{2+} \rightarrow Mn^{3+} + Fe^{2+}$
Two half reactions:
$Fe^{3+ }+ e^- \rightarrow Fe^{2+}$
$Mn^{2+ }\rightarrow Mn^{3+} + e^-$
$\text{Reduction}\text{ E}^\circ= 0.77\text{V}\\\text{Oxidation}\text{ E}^\circ=-1.50\text{V}\\\overline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\\\ \text{e.m.f}=-0.73\text{V}$
Since, emf is negative, the reaction does not occur i.e.$, Fe^{3+}$ does not oxidise $Mn^{2+}.$

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Question 24 Marks

Nemst equation relates the reduction potential of an electrochemical reaction to the standard potential and activities of the chemical species undergoing oxidation and reduction. Let us consider the reaction $, \text{M}^{\text{n+}}_{(\text{aq})}\xrightarrow{\ \ \ \ \ \ \ \ }\text{nM}_\text{(s)}$ For this reaction, the electrode potential measured with respect to standard hydrogen electrode can be given as $\text{E}_{\Big(\frac{\text{M}^{\text{n+}}}{\text{M}}\Big)}=\text{E}^\circ_{\Big(\frac{\text{M}^\text{n+}}{\text{M}}\Big)}-\frac{\text{RT}}{\text{nF}}\text{ln}\frac{[\text{M}]}{[\text{M}^{\text{n}+}]}$ In these questions $(Q$. No. $i-iv),$ a statement of assertion followed by a statement ofreason is given. Choose the correct answer out of the following choices.

Assertion and reason both are correct statements and reason is correct explanation for assertion.
Assertion and reason both are correct statements but reason is not correct explanation for assertion.
Assertion is correct statement but reason is wrong statement.
Assertion is wrong statement but reason is correct statement.

Assertion : For concentration cell, $\text{Zn}_{(\text{s})}|\text{ Zn}^{2+}_{\text{(aq)}}||\text{ Zn}^{2+}_{(\text{aq})}|\text{ Zn}\\\ \ \ \ \ \ \ \ \ \ \ \ \text{C}_1\ \ \ \ \ \ \ \ \text{C}_2$

For spontaneous cell reaction $, C_1 < C_2$
Reason : For concentration cell, $\text{E}_\text{cell}=\frac{\text{RT}}{\text{nF}}\log\frac{\text{C}_2}{\text{C}_1}$
For spontaneous reaction, $\text{E}_\text{cell}=+\text{ve}\Rightarrow\text{C}_2>\text{C}_1$

Assertion : For the cell reaction, $\text{Zn}_{(\text{s})}+\text{Cu}^{2+}_{(\text{aq})}\xrightarrow{\ \ \ \ \ }\text{Zn}^{2+}_{(\text{aq})}+\text{Cu}_{(\text{s})}$ voltmeter gives zero reading at equilibrium.

Reason : At the equilibrium, there is no change in concentration of $\ce{Cu2+}$ and $\ce{Zn2+}$ ions.

Assertion : The Nernst equation gives the concentration dependence of emf of the cell.

Reason : In a cell, current flows from cathode to anode.

Assertion : Increase in the concentration of copper half cell in a cell, increases the emfofthe cell.

Reason : $\text{E}_\text{cell}=\text{E}^\circ_\text{cell}+\frac{0.059}{2}\log\frac{[\text{Cu}^{2+}]}{[\text{Zn}^{2+}]}$

Assertion : Electrode potential for the electrode $\frac{\text{Mn}^+}{\text{Mn}}$ with concentration is given by the expression under $\text{STP}$ conditions.

$\text{E}=\text{E}^\circ+\frac{0.059}{\text{n}}\log[\text{Mn}^{+}]$
Reason : $\text{STP}$ conditions require the temperature to be $273K$.

Answer

$(a)$ Assertion and reason both are correct statements and reason is correct explanation for assertion.

$\log\Big(\frac{\text{C}_1}{\text{C}_2}\Big)<0$ for spontaneity.
$\therefore\text{C}_1<\text{C}_2$

$(a)$ Assertion and reason both are correct statements and reason is correct explanation for assertion.

$(b)$ Assertion and reason both are correct statements but reason is not correct explanation for assertion.

$(a)$ Assertion and reason both are correct statements and reason is correct explanation for assertion.

$(d)$ Assertion is wrong statement but reason is correct statement.

Nernst equation is measured at $298K$. At $\text{STP}$ conditions, temperature to be $273K$.

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Question 34 Marks

The potential of each electrode is known as electrode potential. Standard electrode potential is the potential when concentration of each species taking part in electrode reaction is unity and the reaction is taking place at $298K$. By convention, the standard electrode potential of hydrogen $\text{(SHE)}$ is $0.0V$. The electrode potential value for each electrode process is a measure of relative tendency of the active species in the process to remain in the oxidised/ reduced form. The negative electrode potential means that the redox couple is stronger reducing agent than $\frac{\text{H}^+}{\text{H}_2}$ couple. A positive electrode potential means that the redox couple is a weaker reducing agent than the $\frac{\text{H}^+}{\text{H}_2}$ couple. Metals which have higher positive value of standard reduction potential form the oxides of greater thermal stability. In these questions $(Q$. No. $i-iv),$ a statement of assertion followed by a statement ofreason is given. Choose the correct answer out of the following choices.

Assertion and reason both are correct statements and reason is correct explanation for assertion.
Assertion and reason both are correct statements but reason is not correct explanation for assertion.
Assertion is correct statement but reason is wrong statement.
Assertion is wrong statement but reason is correct statement.

Assertion : An electrochemical cell can be set $-$ up only if the redox reaction is spontaneous.

Reason : A reaction is spontaneous if the free energy change is negative.

Assertion : The standard electrode potential of hydrogen is $0.0V.$

Reason : It is by convention.

Assertion : The more negative is the standard reduction potential, greater is its ability to displace $H_2 $ from acid.

Reason : Strength of reducing agent increases with the increase in negative value of the standard reduction potential.

Assertion : The negative value of standard reduction potential means that reduction takes place on this electrode with reference to hydrogen electrode.

Reason : The standard electrode potential of a half cell has a fixed value.

Assertion : The absolute value of electrode potential cannot be determined experimentally.

Reason : The electrode potential values are generally determined with respect to $\text{SHE}$.

Answer

$(b)$ Assertion and reason both are correct statements but reason is not correct explanation for assertion.
$(a)$ Assertion and reason both are correct statements and reason is correct explanation for assertion.
$(a)$ Assertion and reason both are correct statements and reason is correct explanation for assertion. More negative is the standard reduction potential, greater is its ability to displace hydrogen from acid.
$(d)$ Assertion is wrong statement but reason is correct statement. A negative value of standard reduction potential means that oxidation takes place on the electrode with reference to $\text{SHE}.$
$(a)$ Assertion and reason both are correct statements and reason is correct explanation for assertion.

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Question 44 Marks

Electrical work done in unit time is equal to electrical potential multiplied by total charge passed. ln order to obtain maximum work from a cell, the charge has to be passed reversibly. The reversible work done by a cell is equal to decrease in its Gibb's energy. Hence, Gibb's energy of reaction is given by $\Delta\text{G}=\text{nFE}_\text{cell}$ Hence, Eis the emfof the cell and $nF$ is the amount of energy. In these questions $(Q$. No. $i-Iv),$ a statement of assertion followed by a statement ofreason is given. Choose the correct answer out of the following choices.

Assertion and reason both are correct statements and reason is correct explanation for assertion.
Assertion and reason both are correct statements but reason is not correct explanation for assertion.
Assertion is correct statement but reason is wrong statement.
Assertion is wrong statement but reason is correct statement.

Assertion : $\Delta\text{G}^\circ=-\text{nFE}^\circ$

Reason : $E^\circ$ should be positive for a spontaneous reaction.

Assertion : An electrochemical cell can be set up only if the red ox reaction is spontaneous.

Reason : A reaction is spontaneous if free energy change is negative.

Assertion : For an electrochemical cell, $\Delta\text{G}<0$ and $\text{E}_\text{cell}>0.$

Reason : The given cell is non $-$ spontaneous.

Assertion : Current stops flowing when $E_\text{cell} = 0$.

Reason : Equilibrium of the cell reaction is attained.

Assertion :$ E_\text{cell}$ should have a positive value for the cell to function.

Reason : $E_\text{cell} = E_\text{cathode} - E_\text{anode}$

Answer

$(b)$ Assertion and reason both are correct statements but reason is not correct explanation for assertion.

$(b)$ Assertion and reason both are correct statements but reason is not correct explanation for assertion.

If redox reaction is spontaneous, $\Delta\text{G}$ is $-ve$ and hence $, E^\circ$ is positive.
$-\Delta\text{G}^\circ=\text{nFE}^\circ\text{cell}$

$(c)$ Assertion is correct statement but reason is wrong statement.

$\Delta\text{G}<0$ and $\text{E}_\text{cell}>0$ ; spontaneous.

$(a)$ Assertion and reason both are correct statements and reason is correct explanation for assertion.

$(b)$ Assertion and reason both are correct statements but reason is not correct explanation for assertion.

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Question 54 Marks

All chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules are present in a few gram of any chemical compound varying with their atomic / molecular masses. To handle such large number conveniently, the mole concept was introduced. All electrochemical cell reactions are also based on mole concept. For example, a $4.0$ molar aqueous solution of $\ce{NaCl}$ is prepared and $500\ mL$ of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrode. The amount of products formed can be calculated by using mole concept.
The following questions are multiple choice questions. Choose the most appropriate answer :

The total number of moles of chlorine gas evolved is :

$0.5$
$1.0$
$1.5$
$1.9$

If cathode is a $Hg$ electrode, then the maximum weight of amalgam formed from this solution is :

$300g$
$446g$
$396g$
$296g$

The total charge $($coulomb$)$ required for complete electrolysis is :

$186000$
$24125$
$48296$
$193000$

In the electrolysis, the number of moles of electrons involved are :

In electrolysis of aqueous $\text{NaCl}$ solution when $Pt$ electrode is taken, then which gas is liberated at cathode?

$\ce{H2}$ gas
$\ce{Cl2}$ gas
$\ce{O2}$ gas
None of these.

Answer

$(b)\ 1.0$

$\text{n}_\text{NaCl}=\frac{4\times500}{1000}=2\ \text{mol}$
$\therefore\text{n}_{\text{Cl}_2}=1\ \text{mol}$

$(b)\ 446g$

$\text{n}_\text{Na}$ deposited $= 2\ mol$
$\therefore\text{n}_{\text{Na}-\text{Hg}}$ formed $= 2\ mol$
$\therefore$ Mass of amalgam formed $= 2 \times 223 = 446g$

$(d)\ 193000$

$2\text{Na}^+ + 2\text{e}^-\rightarrow2\text{Na}\\\ \ \ \ \ \ \ \ \ \ \ \ \ (\text{2F})$
Total charge required $= 2F = 2 \times 96500 = 193000C$

$(a)\ 2$

$(a)\ H_2$ gas

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Question 64 Marks

Two types of conductors are generally used, metallic and electrolytic. Free electrons are the current carrier in metallic and in electrolytic conductors, free ions. Specific conductance or conductivity of an electrolytic solution is given by $\text{K}=\text{C}\times\frac{\text{l}}{\text{A}}$ where, $\text{C}\times\frac{1}{\text{R}}$ and $\frac{\text{l}}{\text{A}}=\text{G}^\star$ (cell constant) Molar conductance $(\wedge_\text{m})$ and equivalent conductance $(\wedge_\text{e})$ of an electrolyte solution are calculated as $\wedge_\text{m}=\frac{\text{K}\times1000}{\text{M}}$ or $\wedge_\text{e}=\frac{\text{K}\times1000}{\text{N}}$ where, M = molarity of solution and N is normality of solution. Molar conductance of strong electrolyte depends on the concentration. $\wedge_\text{m}=\wedge^\circ_{\text{m}^-}\text{b}\sqrt{\text{C}}$ $\wedge^\circ_\text{m}=$ molar conductance at infinite dilution, b = constant, C = cone.of solution In these questions (Q. No. i-iv), a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.

Assertion and reason both are correct statements and reason is correct explanation for assertion.
Assertion and reason both are correct statements but reason is not correct explanation for assertion.
Assertion is correct statement but reason is wrong statement.
Assertion is wrong statement but reason is correct statement.

Assertion: The molar conductivity of strong electrolyte decreases with increase in concentration.

Reason: At high concentration, migration ofions is slow.

Assertion: The molar conductance of weak electrolyte at infinite dilution is equal to the sum of molar conductance of cations and anions.

Reason: Kohlrausch's law is applicable for strong electrolytes.

Assertion: Equivalent conductance of all electrolytes increases with increasing concentration.

Reason: More number ofions are available per gram equivalent at higher concentration.

Assertion: Specific conductance decreases with dilution whereas equivalent conductance increases.

Reason: On dilution, number of ions per millilitre decreases but total number ofions increases considerably.

Assertion: The ratio of specific conductivity to the observed conductance does not depend upon the concentration of the solution taken in the conductivity cell.

Reason: Specific conductivity decreases with dilution whereas observed conductance increases with dilution.

Answer

(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.

Explanation:
$\wedge^\infty_{\text{AB}}=\lambda^\infty_{\text{A}^{+}}+\lambda^\infty_{\text{B}^-}$
Kohlrausch's law is applicable for weak electrolytes.

(d) Assertion is wrong statement but reason is correct statement.

Explanation:
At higher concentration, mobility of ions decreases. Hence, conductance decreases.

Explanation:
Total number of ions will increase slightly on dilution (not considerably).

(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.

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Question 74 Marks

Molar conductivity of ions are given as product of charge on ions to their ionic mobilities and Faradays constant.
$\lambda_\text{A}\text{n}+=\text{n}\mu_\text{A}\text{n}+\text{F} ($here $\mu$ is the ionic mobility of $A^{n+})$
For electrolytes say $A_xB_y,$ molar conductivity is given by
$\lambda_{\text{m}(\text{A}_\text{x}\text{B}_\text{y})}=\text{x}_\text{n}\mu_{\text{A}^\text{n}}+\text{F}+\text{y}_\text{m}\lambda_{\text{A}^\text{m}}-\text{F}$

Ions	Ionic mobility
$K^+$	$7.616 \times 10^{-4}$
$Ca^{2+}$	$12.33 \times 10^{-4}$
$Br^-$	$8.09 \times 10^{-4}$
$\text{SO}_{4}^{2-}$	$16.58 \times 10^{-4}$

The following questions are multiple choice questions. Choose the most appropriate answer:

At infinite dilution, the equivalent conductance of $\ce{CaSO_4}$ is:

$256 \times 10^{-4}$
$279$
$23.7$
$2.0 \times 10^{-8}$

If the degre
e of dissociation of $\ce{CaSO_4}$ solution is $10\%$ then equivalent conductance of $\ce{CaSO_4}$ is:

$3.59$
$36.9$
$27.9$
$30.6$

The correct order of equivalent conductance at infinite dilution of $\ce{LiCl, NaCl, KCl}$ is:

$\ce{LiCl = NaCl = KCl}$
$\ce{LiCl > NaCl > KCl}$
$\ce{KCl > LiCl > NaCl}$
$\ce{KCl > NaCl > LiCl}$

What is the unit of equivalent conductivity?

$ohm^{-1} \ cm^2 eq^{-1}$
$ohm \ cm^2 eq-1$
$ohm^{-1} \ cm eq^{-1}$
$ohm \ cm^{-2}eq^{-2}$

If the molar conductance value of $Ca^{2+}$ and $Cl^-$ at infinite dilution are $118.88 \times 10^{-4}\ m^2\ mho\ mol^{-1}$ and $77.33 \times 10^{-4}\ m^2\ mho\ mol^{-1}$ respectively then the molar conductance of $CaCl_2 ($ in $m^2 mho\ mol^{-1})$ will be:

$120.18 \times 10^{-4}$
$135 \times 10^{-4}$
$273.54 \times 10^{-4}$
$192.1 \times 10^{-4}$

Answer

$(b)\ 279$

Equivalent conductance of $\ce{CaSO_4:}$
$\alpha^\infty_{\text{CaSO}_4}=\lambda^\infty_{\text{Ca}^{2+}}+\lambda^\infty_{\text{SO}_4^{2-}}$
$\lambda^\infty_{\text{Ca}^{2+}}=(\mu_{\text{Ca}^{2+}})\text{F};$
$\lambda^\infty_{\text{SO}^{2-}_4}=(\mu_{\text{SO}^{2-}_4})\text{F}$
$\mu_{\text{Ca}^{2+}}$ and $\mu_{\text{SO}_4^{2-}} -$ are ionic mobilities.
$\alpha^\infty_{\text{CaSO}_4}=\text{F}(12.33+16.58)\times10^{-4}$
$=96500\times10^{-4}\times28.91=279$

$(c)\ 27.9$

$\alpha=\frac{\alpha_\text{C}}{\alpha^\infty}$
$\Rightarrow0.1=\frac{\alpha_{\text{C}}}{279}$
$\Rightarrow\alpha_\text{C}=27.9$

$(c)\ \ce{KCl > NaCl > LiCl}$

The ions formed are $Li^+, Na^+$ and $K^+,$ the hydration is maximum in case of $Li^+$ because of which its mobility is least and has least conductance.
Therefore, the correct order is $\ce{KCl > NaCl > LiCl.}$

$(a)\ ohm^{-1} \ cm^2 eq^{-1}$

$(c)\ 273.54 \times 10^{-4}$

$\alpha^\circ_{(\text{m}\text{CaCl}_2)}=\lambda^\circ_{\text{Ca}^{2+}}+2\lambda^\circ_{\text{Cl}^-}$
$= (118.88 \times 10^{-4}) + 2(77.33 \times 10^{-4})$
$= 273.54 \times 10^{-4 }m^{2 }mho\ mol^{-1}$

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Question 84 Marks

The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal $M$ is,
$M_{(s)} | M^+($aq.; $0.05$ molar$) \| M^+($aq; $1$ molar $) | M_{(s)}$
The following questions are multiple choice questions. Choose the most appropriate answer:

For the above cell,

$\text{E}_\text{cell}<0;\Delta\text{G}>0$
$\text{E}_\text{cell}>0;\Delta\text{G}<0$
$\text{E}_\text{cell}<0;\Delta\text{G}^\circ>0$
$\text{E}_\text{cell}>0;\Delta\text{G}^\circ<0$

If the $0.05$ molar solution of $M^+$ is replaced by a $0.0025$ molar $M^+$ solution, then the magnitude of the cell potential would be:

$130mV$
1$85mV$
$154mV$
$600mV$

The value of equilibrium constant for a feasible cell reaction is:

$ < 1$
$= 1$
$ > 1$
Zero

What is the emf of the cell when the cell reaction attains equilibrium?

$1$
$0$
$ > 1$
$ < 1$

The potential of an electrode change with change in:

Concentration ofions in solution.
Position of electrodes.
Voltage of the cell.
All of these.

Answer

$(b)\ \text{E}_\text{cell}>0;\Delta\text{G}<0$

$\ \ \text{M}\ \xrightarrow{\ \ \ \ \ \ \ }\text{M}^++\ \ \text{e}^- \ 1\text{M})\ \ \ \ (0.05\text{M})$
For concentration cell, $\text{E}_\text{cell}=\frac{0.059}{1}\log\frac{0.05}{1}$
$\text{E}_\text{cell}=\frac{0.059}{1}\log(5\times10^{-2})$
$\text{E}_\text{cell}=\frac{0.059}{1}[(-2)+\log5]-0.059(-2+0.698)$
$= -0.059(-1.302) = 0.0768 $
$\Delta\text{G}=-\text{nFE}_\text{cell}$
If $E_\text{cell}$ is positive, $\Delta\text{G}$ is negative.

$(c)\ 154mV$

$\frac{\text{E}_1}{\text{E}_2}=\frac{\log0.05}{\log0.0025}$
$\frac{\text{E}_1}{\text{E}_2}=\frac{\log5\times10^{-2}}{\log25\times10^{-4}}$
$\text{E}_1=0.0768$
$\frac{0.0168}{\text{E}_2}=\frac{-1.3}{-2.6}=\frac{1}{2}$ or $\text{E}_2=154\text{mV}$

$(c) > 1$

$\text{K}=\text{antilog}\Big(\frac{\text{nE}^\circ}{0.0591}\Big)$
For feasible cell $, E^\circ$ is positive, hence from the above equation $, K > 1$ for a feasible cell reaction.

$(b)\ 0$

$(a)$ Concentration ofions in solution.

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Question 94 Marks

The electrochemical cell shown below is concentration cell. $M|M^{2+}\ ($saturated solution of a sparingly soluble salt$, MX_2) || M^{2+} (0.001\ mol\ dm^{-3})\ |M$ The emfof the cell depends on the difference in concentrations of $M^{2+}$ ions at the two electrodes. The emf of the cell at $298 K$ is $0.059V$. The following questions are multiple choice questions. Choose the most appropriate answer:

The solubility product $(K_{sp'}\ mol^3\ dm^{-9})$ of $MX_2$ at $298 K$ based on the information available for the given concentration cell is $(\text{take }2.303 \times \text{R}\times \frac{298}{\text{F}} = 0.059)$

$2 \times 10^{-15}$
$4 \times 10^{-15}$
$3 \times 10^{-12}$
$1 \times 10^{-12}$

The value of $\triangle\text{G}\ ($in $kJ\ mol^{-1})$ for the given cell is $($take $1 F = 96500 C\ mol^{-1})$

$3.7$
$-3.7$
$10.5$
$-11.4$

The equilibrium constant for the following reaction is:

$\text{Fe}^{2+}+\text{Ce}^{4+}\rightleftharpoons\text{Ce}^{3+}+\text{Fe}^{3+}$
(Given, $\text{E}^\circ_\frac{\text{Ce}^{4+}}{\text{Ce}^{3+}}=1.44\text{V}$ and $\text{E}^\circ_\frac{\text{Fe}^{3+}}{\text{Fe}^{2+}}=0.68\text{V})$

$7.6 \times 10^{12}$
$6.5 \times 10^{10}$
$5.2 \times 10^9$
$3.4 \times 10^{12}$

The solubility product of a saturated solution of $\ce{Ag2CrO4}$ in water at $298 K$ if the emf of the cell

$Ag|Ag^+ ($satd. $\ce{Ag2CrO4}$ soln$) || Ag^+ (0.1 M) | Ag$
is $0.164V$ at $298 K,$ is :

$3.359 \times 10^{-12}\ mol^3\ L^{-3}$
$2.287 \times 10^{-12}\ mol^3\ L^{-3}$
$1.158 \times 10^{-12}\ mol^3\ L^{-3}$
$4.135 \times 10^{-12}\ mol^3\ L^{-3}$

To calculate the emf of the cell, which of the following options is correct?

$emf = E_{cathode }- E_{anode}$
$emf = E_{anode }- E_{cathode}$
$emf = E_{anode} + E_{cathode}$
None of these.

Answer

$(b)\ 4 \times 10^{-15}$

$0.059=\frac{+0.059}{2}\log\frac{0.001}{[\text{M}^{2+}]}$
$\log\frac{0.001}{[\text{M}^{2+}]}=2$ or $[\text{M}^{2+}]=10^{-5}$
Let solubility of sah be S mol/ litre.
Thus, $\text{MX}_2\xrightarrow{\ \ \ \ }\text{M}^{2+}+2\text{X}^-\\\ \ \text{S}\ \ \ \ \ \ \ \ \ \ \ \ \text{S}\ \ \ \ \ \ \ \ \ \ 2\text{S}$
$\therefore\text{K}_\text{sp}=4\text{S}^3=4\times(10^{-5})^3=4\times10^{-15}$

$(d)\ -11.4$

$\triangle\text{G}=\text{nFE}=-2\times96500\times0.059$
$= -11387\ J\ mol^{-1} = -11.4\ kJ\ mol^{-1}$

$(a)\ 7.6 \times 10^{12}$

$\text{E}^\circ_\text{cell}=\frac{0.059}{1}\log\text{K}_\text{C}$
$\text{E}^\circ_\text{cell}=\text{E}^\circ_\frac{{\text{Fe}^{2+}}}{{\text{Fe}^{3+}}}+\text{E}^\circ_\frac{\text{Ce}^{4+}}{\text{Ce}^{3+}}$
$= -0.68 + 1.44 = 0.76 \text{V}$
$\log_{10}\text{K}_\text{C}=\frac{0.76}{0.059}=12.88$
$\text{K}_\text{C}=7.6\times10^{12}$

$(b)\ 2.287 \times 10^{-12}\ mol^3 L^{-3}$

$\text{E}_\text{cell}=\frac{0.059}{1}\log\frac{[\text{Ag}^+]_\text{RHS}}{[\text{Ag}^+]_\text{LHS}}$
$0.164=\frac{0.059}{1}\log\frac{0.1}{[\text{Ag}^+]_\text{LHS}}$
$[\text{Ag}^+]_\text{LHS}=1.66\times10^{-4}\text{M}$
So, $[\text{CrO}^{2-}_4]=\frac{1.66\times10^{-4}}{2}$
$\text{K}_\text{Sp}(\text{Ag}_2\text{CrO}_4)=[\text{Ag}^+]^2[\text{CrO}_4^{2-}]$
$= (1.66\times10^{-4} )^2\Big(\frac{1.66\times10^{-4}}{2}\Big)$
$=2.287\times10^{-12} \text{mol}^{3} \text{L}^{-3}$

$(a)\ emf = E_{cathode }- E_{anode}$

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Chemistry Case study (4 Marks)

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