Electrochemistry
218 Questions
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 28th June Evening Shift
For the given reactions
Sn2+ + 2e$-$ $\to$ Sn
Sn4+ + 4e$-$ $\to$ Sn
the electrode potentials are ; $E_{S{n^{2 + }}/Sn}^o = - 0.140$ V and $E_{S{n^{4 + }}/Sn}^o = + 0.010$ V. The magnitude of standard electrode potential for $S{n^{4 + }}/S{n^{2 + }}$ i.e. $E_{S{n^{4 + }}/S{n^{2 + }}}^o$ is _____________ $\times$ 10$-$2 V. (Nearest integer)
Correct Answer: 16
Explanation:
$\mathrm{Sn} \longrightarrow \mathrm{Sn}^{2+}+2 \mathrm{e}^{-} \quad \mathrm{E}_{1}^{0}=0.140 \mathrm{~V}$
$ \mathrm{Sn}^{4+}+4 \mathrm{e}^{-} \longrightarrow \mathrm{Sn} \quad \mathrm{E}_{2}^{0}=0.010 \mathrm{~V} $
$ \begin{aligned} & \mathrm{Sn}^{4+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Sn}^{2+} \quad \mathrm{E}_{\text {cell }}^{0} \\\\ & \mathrm{E}_{\mathrm{cell}}^{\mathrm{O}}=\frac{\mathrm{n}_{2} \mathrm{E}_{2}^{\mathrm{o}}+\mathrm{n}_{1} \mathrm{E}_{1}^{0}}{\mathrm{n}}=\frac{4(0.010)+2(0.140)}{2} \\\\ & \mathrm{E}_{\text {cell }}^{0}=0.16 \mathrm{~V}=16 \times 10^{-2} \mathrm{~V} \end{aligned} $
$ \mathrm{Sn}^{4+}+4 \mathrm{e}^{-} \longrightarrow \mathrm{Sn} \quad \mathrm{E}_{2}^{0}=0.010 \mathrm{~V} $
$ \begin{aligned} & \mathrm{Sn}^{4+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Sn}^{2+} \quad \mathrm{E}_{\text {cell }}^{0} \\\\ & \mathrm{E}_{\mathrm{cell}}^{\mathrm{O}}=\frac{\mathrm{n}_{2} \mathrm{E}_{2}^{\mathrm{o}}+\mathrm{n}_{1} \mathrm{E}_{1}^{0}}{\mathrm{n}}=\frac{4(0.010)+2(0.140)}{2} \\\\ & \mathrm{E}_{\text {cell }}^{0}=0.16 \mathrm{~V}=16 \times 10^{-2} \mathrm{~V} \end{aligned} $
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 28th June Morning Shift
The quantity of electricity in Faraday needed to reduce 1 mol of Cr2O$_7^{2 - }$ to Cr3+ is ____________.
Correct Answer: 6
Explanation:
$
\overset{+6}{\mathrm{Cr}_2} \mathrm{O}_7^{2-} \longrightarrow 2 \mathrm{Cr}^{3+}
$
$\because$ Each $\mathrm{Cr}$ is converting from $+6$ to $+3$
$\therefore 6$ faradays of charge is required
$\because$ Each $\mathrm{Cr}$ is converting from $+6$ to $+3$
$\therefore 6$ faradays of charge is required
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 27th June Evening Shift
For the reaction taking place in the cell :
Pt (s)| H2 (g)|H+(aq) || Ag+(aq) |Ag (s)
E$_{cell}^o$ = + 0.5332 V.
The value of $\Delta$fG$^\circ$ is ______________ kJ mol$-$1. (in nearest integer)
Correct Answer: 51OR103
Explanation:
# At anode, oxidation occur
H2 $\to$ 2H+ + 2e$-$ ....... (1)
# At cathode, reduction occur
2Ag+ + 2e$-$ $\to$ 2Ag ...... (2)
Adding equation (1) and (2), we get n = 2, where n = cancelled out electron
Now,
$\Delta G^\circ = - nF\,E_{cell}^o$
$ = - 2 \times 96500 \times 0.5332$
$ = - 102907.6$
$ = - 102.9$ kJ/mol
$ = - 103$ kJ/mol
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 27th June Morning Shift
The limiting molar conductivities of NaI, NaNO3 and AgNO3 are 12.7, 12.0 and 13.3 mS m2 mol$-$1, respectively (all at 25$^\circ$C). The limiting molar conductivity of AgI at this temperature is ____________ mS m2 mol$-$1.
Correct Answer: 14
Explanation:
Given
(1) $\lambda_m^{\infty}(\mathrm{NaI})=12.7 \,\mathrm{mS} \mathrm{m}^2 \mathrm{~mol}^{-1}$
(2) $\lambda_{\mathrm{m}}^{\infty}\left(\mathrm{NaNO}_3\right)=12.0 \,\mathrm{mS} \mathrm{m}^2 \mathrm{~mol}^{-1}$
(3) $\lambda_{\mathrm{m}}^{\infty}\left(\mathrm{AgNO}_3\right)=13.3 \,\mathrm{mS} \mathrm{m}^2 \mathrm{~mol}^{-1}$
$\lambda_{\mathrm{m}}^{\infty}(\mathrm{Ag} \mathrm{I})=(1)+(3)-(2)$
$ =12.7+13.3-12.0 $
$ =26.0-12.0 $
$\lambda_{\mathrm{m}}^{\infty}(\mathrm{Ag} \mathrm{I})=14.0$
(1) $\lambda_m^{\infty}(\mathrm{NaI})=12.7 \,\mathrm{mS} \mathrm{m}^2 \mathrm{~mol}^{-1}$
(2) $\lambda_{\mathrm{m}}^{\infty}\left(\mathrm{NaNO}_3\right)=12.0 \,\mathrm{mS} \mathrm{m}^2 \mathrm{~mol}^{-1}$
(3) $\lambda_{\mathrm{m}}^{\infty}\left(\mathrm{AgNO}_3\right)=13.3 \,\mathrm{mS} \mathrm{m}^2 \mathrm{~mol}^{-1}$
$\lambda_{\mathrm{m}}^{\infty}(\mathrm{Ag} \mathrm{I})=(1)+(3)-(2)$
$ =12.7+13.3-12.0 $
$ =26.0-12.0 $
$\lambda_{\mathrm{m}}^{\infty}(\mathrm{Ag} \mathrm{I})=14.0$
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 26th June Evening Shift
Cu(s) + Sn2+ (0.001M) $\to$ Cu2+ (0.01M) + Sn(s)
The Gibbs free energy change for the above reaction at 298 K is x $\times$ 10$-$1 kJ mol$-$1. The value of x is __________. [nearest integer]
[Given : $E_{C{u^{2 + }}/Cu}^\Theta = 0.34\,V$ ; $E_{S{n^{2 + }}/Sn}^\Theta = - 0.14\,V$ ; F = 96500 C mol$-$1]
Correct Answer: 983
Explanation:
$
\begin{aligned}
\mathrm{Cu} &+\mathrm{Sn}^{+2} \longrightarrow \mathrm{Cu}^{+2}+\mathrm{Sn}(\mathrm{s}) \\\\
\mathrm{E}_{\text {cell }}^{\circ} &=\mathrm{E}_{\mathrm{OX}}^{\circ}+\mathrm{E}_{\text {Red }}^{\circ} \\\\
&=-0.34-0.14 \\\\
&=-0.48 \mathrm{~V} \\\\
\mathrm{E}=& ~\mathrm{E}^{\circ}-\frac{0.0591}{2} \log \frac{\left[\mathrm{Cu}^{+2}\right]}{\left[\mathrm{Sn}^{+2}\right]} \\\\
=&-0.48-0.0295 \log 10 \\\\
=&-0.5095 \mathrm{~V} \\\\
\Delta \mathrm{G}=&-\mathrm{nFE} \\\\
&=-2 \times 96500 \times-0.5095 \mathrm{~J} / \mathrm{mol} \\\\
&=98333.5 \times 10^{-3} \mathrm{~kJ} / \mathrm{mol} \\\\
&=983.3 \times 10^{-1} \mathrm{~kJ} / \mathrm{mol} \\\\
&=983 \times 10^{-1} \mathrm{~kJ} / \mathrm{mol}
\end{aligned}
$
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 25th June Evening Shift
A solution of Fe2(SO4)3 is electrolyzed for 'x' min with a current of 1.5 A to deposit 0.3482 g of Fe. The value of x is ___________. [nearest integer]
Given : 1 F = 96500 C mol$-$1
Atomic mass of Fe = 56 g mol$-$1
Correct Answer: 20
Explanation:
$\mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe}$
$ \text {Moles of Fe deposited }=\frac{0.3482}{56}=6.2 \times 10^{-3} $
For 1 mole $\mathrm{Fe}$, charge required is $3 \mathrm{~F}$
For $6.2 \times 10^{-3}$ mole $\mathrm{Fe}$, charge required is $3 \times 6.2 \times 10^{-3} \mathrm{~F}$
Since, charge required $=18.6 \times 10^{-3} \times 96500 \mathrm{C}$
$ =1794.9 \mathrm{C} $
And,
$ 1.5 \times \mathrm{t}=1794.9 $
$t=\frac{1794.9}{1.5 \times 60} \min$
$ \mathrm{t} \simeq 20 \mathrm{~min} $
$ \text {Moles of Fe deposited }=\frac{0.3482}{56}=6.2 \times 10^{-3} $
For 1 mole $\mathrm{Fe}$, charge required is $3 \mathrm{~F}$
For $6.2 \times 10^{-3}$ mole $\mathrm{Fe}$, charge required is $3 \times 6.2 \times 10^{-3} \mathrm{~F}$
Since, charge required $=18.6 \times 10^{-3} \times 96500 \mathrm{C}$
$ =1794.9 \mathrm{C} $
And,
$ 1.5 \times \mathrm{t}=1794.9 $
$t=\frac{1794.9}{1.5 \times 60} \min$
$ \mathrm{t} \simeq 20 \mathrm{~min} $
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 25th June Morning Shift
In a cell, the following reactions take place
$\matrix{ {F{e^{2 + }} \to F{e^{3 + }} + {e^ - }} & {E_{F{e^{3 + }}/F{e^{2 + }}}^o = 0.77\,V} \cr {2{I^ - } \to {I_2} + 2{e^ - }} & {E_{{I_2}/{I^ - }}^o = 0.54\,V} \cr } $
The standard electrode potential for the spontaneous reaction in the cell is x $\times$ 10$-$2 V 298 K. The value of x is ____________. (Nearest Integer)
Correct Answer: 23
Explanation:
$\underset{\text { Cathode }}{\mathrm{Fe}^{+3}}+\underset{\text { anode }}{\mathrm{I}^{-}} \longrightarrow \mathrm{I}_2+\mathrm{Fe}^{+2}$
$E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ}$
$ \begin{aligned} &=0.77-0.54 \\\\ &=0.23 \mathrm{~V} \\\\ &=23 \times 10^{-2} \mathrm{~V} \end{aligned} $
$E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ}$
$ \begin{aligned} &=0.77-0.54 \\\\ &=0.23 \mathrm{~V} \\\\ &=23 \times 10^{-2} \mathrm{~V} \end{aligned} $
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 24th June Evening Shift
The resistance of a conductivity cell containing 0.01 M KCl solution at 298 K is 1750 $\Omega$. If the conductivity of 0.01 M KCl solution at 298 K is 0.152 $\times$ 10$-$3 S cm$-$1, then the cell constant of the conductivity cell is ____________ $\times$ 10$-$3 cm$-$1.
Correct Answer: 266
Explanation:
Molarity of $\mathrm{KCl}$ solution $=0.1 ~\mathrm{M}$
$ \begin{array}{ll} \text { Resistance } & =1750 ~\mathrm{ohm} \\\\ \text { Conductivity } & =0.152 \times 10^{-3} \mathrm{~S} \mathrm{~cm}^{-1} \\\\ \text { Conductivity } & =\frac{\text { Cell constant }}{\text { Resistance }} \\\\ \therefore \text { Cell constant } & =0.152 \times 10^{-3} \times 1750 \\\\ & =266 \times 10^{-3} \mathrm{~cm}^{-1} \end{array} $
$ \begin{array}{ll} \text { Resistance } & =1750 ~\mathrm{ohm} \\\\ \text { Conductivity } & =0.152 \times 10^{-3} \mathrm{~S} \mathrm{~cm}^{-1} \\\\ \text { Conductivity } & =\frac{\text { Cell constant }}{\text { Resistance }} \\\\ \therefore \text { Cell constant } & =0.152 \times 10^{-3} \times 1750 \\\\ & =266 \times 10^{-3} \mathrm{~cm}^{-1} \end{array} $
2022
JEE Mains
Numerical
JEE Main 2022 (Online) 24th June Morning Shift
The cell potential for the following cell
Pt |H2(g)|H+ (aq)|| Cu2+ (0.01 M)|Cu(s)
is 0.576 V at 298 K. The pH of the solution is __________. (Nearest integer)
(Given : $E_{C{u^{2 + }}/Cu}^o = 0.34$ V and ${{2.303\,RT} \over F} = 0.06$ V)
Correct Answer: 5
Explanation:
$E_{\text {cell }}=E_{\text {cell }}^{0}-\frac{0.06}{2} \log \frac{\left[\mathrm{H}^{\oplus}\right]^{2}}{\left[\mathrm{Cu}^{+2}\right]}$
$0.576=0.34-0.03 \log \frac{\left[\mathrm{H}^{\oplus}\right]^{2}}{[0.01]}$
$0.576-0.34=-0.03 \log \left[\mathrm{H}^{\oplus}\right]^{2}+0.03 \log (0.01)$
$ =0.06 \,\mathrm{pH}-0.06 $
$\mathrm{pH} \simeq 4.93 \simeq 5$
$0.576=0.34-0.03 \log \frac{\left[\mathrm{H}^{\oplus}\right]^{2}}{[0.01]}$
$0.576-0.34=-0.03 \log \left[\mathrm{H}^{\oplus}\right]^{2}+0.03 \log (0.01)$
$ =0.06 \,\mathrm{pH}-0.06 $
$\mathrm{pH} \simeq 4.93 \simeq 5$
2022
JEE Mains
MCQ
JEE Main 2022 (Online) 28th July Morning Shift
Match List - I with List - II.
| List - I | List - II | ||
|---|---|---|---|
| (A) | $Cd(s) + 2Ni{(OH)_3}(s) \to CdO(s) + 2Ni{(OH)_2}(s) + {H_2}O(l)$ | (I) | Primary battery |
| (B) | $Zn(Hg) + HgO(s) \to ZnO(s) + Hg(l)$ | (II) | Discharging of secondary battery |
| (C) | $2PbS{O_4}(s) + 2{H_2}O(l) \to Pb(s) + Pb{O_2}(s) + 2{H_2}S{O_4}(aq)$ | (III) | Fuel cell |
| (D) | $2{H_2}(g) + {O_2}(g) \to 2{H_2}O(l)$ | (IV) | Charging of secondary battery |
Choose the correct answer from the options given below:
A.
$(\mathrm{A})-(\mathrm{I}),(\mathrm{B})-(\mathrm{II}),(\mathrm{C})-(\mathrm{III}),(\mathrm{D})-(\mathrm{IV})$
B.
$(\mathrm{A})-(\mathrm{IV}),(\mathrm{B})-(\mathrm{I}),(\mathrm{C})-(\mathrm{II}),(\mathrm{D})-(\mathrm{III})$
C.
$(\mathrm{A})-(\mathrm{II}),(\mathrm{B})-(\mathrm{I}),(\mathrm{C})-(\mathrm{IV}),(\mathrm{D})-(\mathrm{III})$
D.
$(\mathrm{A})-(\mathrm{II}),(\mathrm{B})-(\mathrm{I}),(\mathrm{C})-(\mathrm{III}),(\mathrm{D})-(\mathrm{IV})$
2022
JEE Mains
MCQ
JEE Main 2022 (Online) 27th July Evening Shift
Given below are two statements :
Statement I : For KI, molar conductivity increases steeply with dilution
Statement II : For carbonic acid, molar conductivity increases slowly with dilution
In the light of the above statements, choose the correct answer from the options given below :
A.
Both Statement I and Statement II are true
B.
Both Statement I and Statement II are false
C.
Statement I is true but Statement II is false
D.
Statement I is false but Statement II is true
2022
JEE Mains
MCQ
JEE Main 2022 (Online) 25th July Evening Shift
The molar conductivity of a conductivity cell filled with 10 moles of 20 mL NaCl solution is ${\Lambda _{m1}}$ and that of 20 moles another identical cell heaving 80 mL NaCl solution is ${\Lambda _{m2}}$. The conductivities exhibited by these two cells are same. The relationship between ${\Lambda _{m2}}$ and ${\Lambda _{m1}}$ is
A.
${\Lambda _{m2}}$ = 2${\Lambda _{m1}}$
B.
${\Lambda _{m2}}$ = ${\Lambda _{m1}}$ / 2
C.
${\Lambda _{m2}}$ = ${\Lambda _{m1}}$
D.
${\Lambda _{m2}}$ = 4${\Lambda _{m1}}$
2022
JEE Mains
MCQ
JEE Main 2022 (Online) 30th June Morning Shift
In which of the following half cells, electrochemical reaction is pH dependent?
A.
$Pt\,|\,F{e^{3 + }},\,F{e^{2 + }}$
B.
$MnO_4^ - \,|M{n^{2 + }}$
C.
$Ag\,|\,AgCl\,|C{l^{ - 1}}$
D.
${1 \over 2}{F_2}\,|{F^ - }$
2022
JEE Mains
MCQ
JEE Main 2022 (Online) 27th June Evening Shift
In 3d series, the metal having the highest M2+/M standard electrode potential is :
A.
Cr
B.
Fe
C.
Cu
D.
Zn
2022
JEE Mains
MCQ
JEE Main 2022 (Online) 26th June Morning Shift
The ${\left( {{{\partial E} \over {\partial T}}} \right)_P}$ of different types of half cells are as follows:
| A | B | C | D |
|---|---|---|---|
| $1 \times {10^{ - 4}}$ | $2 \times {10^{ - 4}}$ | $0.1 \times {10^{ - 4}}$ | $0.2 \times {10^{ - 4}}$ |
(Where E is the electromotive force)
Which of the above half cells would be preferred to be used as reference electrode?
A.
A
B.
B
C.
C
D.
D
2022
JEE Mains
MCQ
JEE Main 2022 (Online) 25th June Evening Shift
The correct order of reduction potentials of the following pairs is
A. Cl2/Cl$-$
B. I2/I$-$
C. Ag+/Ag
D. Na+/Na
E. Li+/Li
Choose the correct answer from the options given below.
A.
A > C > B > D > E
B.
A > B > C > D > E
C.
A > C > B > E > D
D.
A > B > C > E > D
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 1st September Evening Shift
If the conductivity of mercury at 0$^\circ$C is 1.07 $\times$ 106 S m$-$1 and the resistance of a cell containing mercury is 0.243$\Omega$, then the cell constant of the cell is x $\times$ 104 m$-$1. The value of x is ____________. (Nearest integer)
Correct Answer: 26
Explanation:
Conductance (G) is reciprocal of resistance (R)
$R = {1 \over G}$ or $G = {1 \over R} = {1 \over {0.243\Omega }} = 4.115{\Omega ^{ - 1}}$
Relation between conductance (G),
conductivity ($\kappa $) and cell constant $\left( {{l \over A}} \right)$ is given as
$\kappa = {{Gl} \over A}$
$ \Rightarrow {l \over A} = {\kappa \over G} = {{1.07 \times {{10}^6}S{m^{ - 1}}} \over {4.115{\Omega ^{ - 1}}}} = 26 \times {10^4}{m^{ - 1}} \Rightarrow x = 26$
$R = {1 \over G}$ or $G = {1 \over R} = {1 \over {0.243\Omega }} = 4.115{\Omega ^{ - 1}}$
Relation between conductance (G),
conductivity ($\kappa $) and cell constant $\left( {{l \over A}} \right)$ is given as
$\kappa = {{Gl} \over A}$
$ \Rightarrow {l \over A} = {\kappa \over G} = {{1.07 \times {{10}^6}S{m^{ - 1}}} \over {4.115{\Omega ^{ - 1}}}} = 26 \times {10^4}{m^{ - 1}} \Rightarrow x = 26$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 31st August Morning Shift
Consider the following cell reaction :
$C{d_{(s)}} + H{g_2}S{O_{4(s)}} + {9 \over 5}{H_2}{O_{(l)}}$ $\rightleftharpoons$ $CdS{O_4}.{9 \over 5}{H_2}{O_{(s)}} + 2H{g_{(l)}}$
The value of $E_{cell}^0$ is 4.315 V at 25$^\circ$C. If $\Delta$H$^\circ$ = $-$825.2 kJ mol$-$1, the standard entropy change $\Delta$S$^\circ$ in J K$-$1 is ___________. (Nearest integer) [Given : Faraday constant = 96487 C mol$-$1]
$C{d_{(s)}} + H{g_2}S{O_{4(s)}} + {9 \over 5}{H_2}{O_{(l)}}$ $\rightleftharpoons$ $CdS{O_4}.{9 \over 5}{H_2}{O_{(s)}} + 2H{g_{(l)}}$
The value of $E_{cell}^0$ is 4.315 V at 25$^\circ$C. If $\Delta$H$^\circ$ = $-$825.2 kJ mol$-$1, the standard entropy change $\Delta$S$^\circ$ in J K$-$1 is ___________. (Nearest integer) [Given : Faraday constant = 96487 C mol$-$1]
Correct Answer: 25
Explanation:
$\Delta G^\circ = - nFE^\circ = \Delta H^\circ - T\Delta S^\circ $
$ \therefore $ $\Delta$S$^\circ$ $ = {{\Delta H^\circ + nFE^\circ } \over T}$
$ = {{( - 825.2 \times {{10}^3}) + (2 \times 96487 \times 4.315)} \over {298}}$
$ = {{ - 825.2 \times {{10}^3} + 832.682 \times {{10}^3}} \over {298}}$
$ = {{7.483 \times {{10}^3}} \over {298}} = 25.11$ JK$-$1 mol$-$1
$\therefore$ Nearest integer answer is 25.
$ \therefore $ $\Delta$S$^\circ$ $ = {{\Delta H^\circ + nFE^\circ } \over T}$
$ = {{( - 825.2 \times {{10}^3}) + (2 \times 96487 \times 4.315)} \over {298}}$
$ = {{ - 825.2 \times {{10}^3} + 832.682 \times {{10}^3}} \over {298}}$
$ = {{7.483 \times {{10}^3}} \over {298}} = 25.11$ JK$-$1 mol$-$1
$\therefore$ Nearest integer answer is 25.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th August Evening Shift
The resistance of a conductivity cell with cell constant 1.14 cm$-$1, containing 0.001 M KCl at 298 K is 1500 $\Omega$. The molar conductivity of 0.001 M KCl solution at 298 K in S cm2 mol$-$1 is ____________. (Integer answer)
Correct Answer: 760
Explanation:
$K = {1 \over R} \times {l \over A} = \left( {\left( {{1 \over {1500}}} \right) \times 1.14} \right)$ S cm$-$1
$ \Rightarrow { \wedge _m} = 1000 \times {{\left( {{{1.14} \over {1500}}} \right)} \over {0.001}}$ S cm2 mol$-$1
= 760 S cm2 mol$-$1
$\Rightarrow$ 760
$ \Rightarrow { \wedge _m} = 1000 \times {{\left( {{{1.14} \over {1500}}} \right)} \over {0.001}}$ S cm2 mol$-$1
= 760 S cm2 mol$-$1
$\Rightarrow$ 760
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th August Evening Shift
For the galvanic cell,
Zn(s) + Cu2+ (0.02 M) $\to$ Zn2+ (0.04 M) + Cu(s),
Ecell = ______________ $\times$ 10$-$2 V. (Nearest integer)
[Use : $E_{Cu/C{u^{2 + }}}^0$ = $-$ 0.34 V, $E_{Zn/Z{n^{2 + }}}^0$ = + 0.76 V, ${{2.303RT} \over F} = 0.059\,V$]
Zn(s) + Cu2+ (0.02 M) $\to$ Zn2+ (0.04 M) + Cu(s),
Ecell = ______________ $\times$ 10$-$2 V. (Nearest integer)
[Use : $E_{Cu/C{u^{2 + }}}^0$ = $-$ 0.34 V, $E_{Zn/Z{n^{2 + }}}^0$ = + 0.76 V, ${{2.303RT} \over F} = 0.059\,V$]
Correct Answer: 109
Explanation:
Galvanic cell :
$Z{n_{(s)}} + \mathop {Cu_{(aq.)}^{ + 2}}\limits_{0.02\,M} \to \mathop {Zn_{}^{ + 2}}\limits_{0.04\,M} + Cu(s)$
Nernst equation = ${F_{cell}} = E_{cell}^o - {{0.059} \over 2}\log {{[2{n^{ + 2}}]} \over {[C{u^{ + 2}}]}}$
$ \Rightarrow {E_{cell}}\left[ {E_{cell}^o - E_{Z{n^{ + 2}}/Zn}^o} \right] - {{0.059} \over 2}\log {{0.04} \over {0.02}}$
$ \Rightarrow {E_{cell}}[0.34 - ( - 0.76)] - {{0.059} \over 2}{\log ^2}$
$ \Rightarrow {E_{cell}}1 - 1 - {{0.059} \over 2} \times 0.3010$
= 1.0911 = 109.11 $\times$ 10$-$2
= 109
$Z{n_{(s)}} + \mathop {Cu_{(aq.)}^{ + 2}}\limits_{0.02\,M} \to \mathop {Zn_{}^{ + 2}}\limits_{0.04\,M} + Cu(s)$
Nernst equation = ${F_{cell}} = E_{cell}^o - {{0.059} \over 2}\log {{[2{n^{ + 2}}]} \over {[C{u^{ + 2}}]}}$
$ \Rightarrow {E_{cell}}\left[ {E_{cell}^o - E_{Z{n^{ + 2}}/Zn}^o} \right] - {{0.059} \over 2}\log {{0.04} \over {0.02}}$
$ \Rightarrow {E_{cell}}[0.34 - ( - 0.76)] - {{0.059} \over 2}{\log ^2}$
$ \Rightarrow {E_{cell}}1 - 1 - {{0.059} \over 2} \times 0.3010$
= 1.0911 = 109.11 $\times$ 10$-$2
= 109
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th August Morning Shift
These are physical properties of an element
(A) Sublimation enthalpy
(B) Ionisation enthalpy
(C) Hydration enthalpy
(D) Electron gain enthalpy
The total number of above properties that affect the reduction potential is ____________ (Integer answer)
(A) Sublimation enthalpy
(B) Ionisation enthalpy
(C) Hydration enthalpy
(D) Electron gain enthalpy
The total number of above properties that affect the reduction potential is ____________ (Integer answer)
Correct Answer: 3
Explanation:
Sublimation enthalpy, Ionisation enthalpy and hydration enthalpy affect the reduction potential.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th July Evening Shift
For the cell
Cu(s) | Cu2+ (aq) (0.1 M) || Ag+(aq) (0.01 M) | Ag(s)
the cell potential E1 = 0.3095 V
For the cell
Cu(s) | Cu2+ (aq) (0.01 M) || Ag+(aq) (0.001 M) | Ag(s)
the cell potential = ____________ $\times$ 10$-$2 V. (Round off the nearest integer).
[Use : ${{2.303RT} \over F}$ = 0.059]
Cu(s) | Cu2+ (aq) (0.1 M) || Ag+(aq) (0.01 M) | Ag(s)
the cell potential E1 = 0.3095 V
For the cell
Cu(s) | Cu2+ (aq) (0.01 M) || Ag+(aq) (0.001 M) | Ag(s)
the cell potential = ____________ $\times$ 10$-$2 V. (Round off the nearest integer).
[Use : ${{2.303RT} \over F}$ = 0.059]
Correct Answer: 28
Explanation:
Cell reaction is :
$Cu(s) + 2A{g^ + }(aq) \to C{u^{2 + }}(aq) + 2Ag(s)$
Now, ${E_{cell}} = E_{cell}^o - {{0.059} \over 2}\log {{[C{u^{2 + }}]} \over {{{[A{g^ + }]}^2}}}$ .... (1)
$\therefore$ ${E_1} = 0.3095 = E_{cell}^o - {{0.059} \over 2}.\log {{0.01} \over {{{(0.001)}^2}}}$ ....(2)
From (1) and (2), E2 = 0.28 V = 28 $\times$ 10$-$2 V
$Cu(s) + 2A{g^ + }(aq) \to C{u^{2 + }}(aq) + 2Ag(s)$
Now, ${E_{cell}} = E_{cell}^o - {{0.059} \over 2}\log {{[C{u^{2 + }}]} \over {{{[A{g^ + }]}^2}}}$ .... (1)
$\therefore$ ${E_1} = 0.3095 = E_{cell}^o - {{0.059} \over 2}.\log {{0.01} \over {{{(0.001)}^2}}}$ ....(2)
From (1) and (2), E2 = 0.28 V = 28 $\times$ 10$-$2 V
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 27th July Morning Shift
The conductivity of a weak acid HA of concentration 0.001 mol L$-$1 is 2.0 $\times$ 10$-$5 S cm$-$1. If $\Lambda _m^o$(HA) = 190 S cm2 mol$-$1, the ionization constant (Ka) of HA is equal to ______________ $\times$ 10$-$6. (Round off to the Nearest Integer)
Correct Answer: 12
Explanation:
$\Lambda _m^{} = 1000 \times {\kappa \over M}$
$ = 1000 \times {{2 \times {{10}^{ - 5}}} \over {0.001}} = 20$ S cm2 mol$-$1
$ \Rightarrow \alpha = {{\Lambda _m^{}} \over {\Lambda _m^\infty }} = {{20} \over {190}} = \left( {{2 \over {19}}} \right)$
HA $\rightleftharpoons$ H+ + A$-$
$0.001(1 - \alpha )0.001\alpha 0.001\alpha $
$ \Rightarrow {k_a} = 0.001\left( {{{{\alpha ^2}} \over {1 - \alpha }}} \right) = {{0.001 \times {{\left( {{2 \over {19}}} \right)}^2}} \over {1 - \left( {{2 \over {19}}} \right)}}$
$ = 12.3 \times {10^{ - 6}}$
$ = 1000 \times {{2 \times {{10}^{ - 5}}} \over {0.001}} = 20$ S cm2 mol$-$1
$ \Rightarrow \alpha = {{\Lambda _m^{}} \over {\Lambda _m^\infty }} = {{20} \over {190}} = \left( {{2 \over {19}}} \right)$
HA $\rightleftharpoons$ H+ + A$-$
$0.001(1 - \alpha )0.001\alpha 0.001\alpha $
$ \Rightarrow {k_a} = 0.001\left( {{{{\alpha ^2}} \over {1 - \alpha }}} \right) = {{0.001 \times {{\left( {{2 \over {19}}} \right)}^2}} \over {1 - \left( {{2 \over {19}}} \right)}}$
$ = 12.3 \times {10^{ - 6}}$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th July Morning Shift
Consider the cell at 25$^\circ$C
Zn | Zn2+ (aq), (1M) || Fe3+ (aq), Fe2+ (aq) | Pt(s)
The fraction of total iron present as Fe3+ ion at the cell potential of 1.500 V is x $\times$ 10$-$2. The value of x is ______________. (Nearest integer)
(Given : $E_{F{e^{3 + }}/F{e^{2 + }}}^0 = 0.77V$, $E_{Z{n^{2 + }}/Zn}^0 = - 0.76V$)
Zn | Zn2+ (aq), (1M) || Fe3+ (aq), Fe2+ (aq) | Pt(s)
The fraction of total iron present as Fe3+ ion at the cell potential of 1.500 V is x $\times$ 10$-$2. The value of x is ______________. (Nearest integer)
(Given : $E_{F{e^{3 + }}/F{e^{2 + }}}^0 = 0.77V$, $E_{Z{n^{2 + }}/Zn}^0 = - 0.76V$)
Correct Answer: 24
Explanation:
$Zn\mathrel{\mathop{\kern0pt\longrightarrow}
\limits_{}} Z{n^{2 + }} + 2{e^ - }$
$2F{e^{3 + }}\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} 2{e^ - } + 2{e^{2 + }}$
$Zn + 2F{e^{3 + }}\buildrel {} \over \longrightarrow Z{n^{2 + }} + 2F{e^{2 + }}$
$E_{cell}^0 = 0.77 - (0.76)$
$ = 1.53$ V
$1.50 = 1.53 - {{0.06} \over 2}\log {\left( {{{F{e^{2 + }}} \over {F{e^{3 + }}}}} \right)^2}$
$\log \left( {{{F{e^{2 + }}} \over {F{e^{3 + }}}}} \right) = {{0.03} \over {0.06}} = {1 \over 2}$
${{[F{e^{2 + }}]} \over {[F{e^{3 + }}]}} = {10^{1/2}} = \sqrt {10} $
${{[F{e^{3 + }}]} \over {[F{e^{2 + }}]}} = {1 \over {\sqrt {10} }}$
${{[F{e^{3 + }}]} \over {[F{e^{2 + }}] + [F{e^{3 + }}]}} = {1 \over {1 + \sqrt {10} }} = {1 \over {4.16}}$
= 0.2402
= 24 $\times$ 10-2
$2F{e^{3 + }}\mathrel{\mathop{\kern0pt\longrightarrow} \limits_{}} 2{e^ - } + 2{e^{2 + }}$
$Zn + 2F{e^{3 + }}\buildrel {} \over \longrightarrow Z{n^{2 + }} + 2F{e^{2 + }}$
$E_{cell}^0 = 0.77 - (0.76)$
$ = 1.53$ V
$1.50 = 1.53 - {{0.06} \over 2}\log {\left( {{{F{e^{2 + }}} \over {F{e^{3 + }}}}} \right)^2}$
$\log \left( {{{F{e^{2 + }}} \over {F{e^{3 + }}}}} \right) = {{0.03} \over {0.06}} = {1 \over 2}$
${{[F{e^{2 + }}]} \over {[F{e^{3 + }}]}} = {10^{1/2}} = \sqrt {10} $
${{[F{e^{3 + }}]} \over {[F{e^{2 + }}]}} = {1 \over {\sqrt {10} }}$
${{[F{e^{3 + }}]} \over {[F{e^{2 + }}] + [F{e^{3 + }}]}} = {1 \over {1 + \sqrt {10} }} = {1 \over {4.16}}$
= 0.2402
= 24 $\times$ 10-2
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 22th July Evening Shift
Assume a cell with the following reaction
$C{u_{(s)}} + 2A{g^ + }(1 \times {10^{ - 3}}M) \to C{u^{2 + }}(0.250M) + 2A{g_{(s)}}$
$E_{cell}^\Theta = 2.97$ V
Ecell for the above reaction is ______________ V. (Nearest integer)
[Given : log 2.5 = 0.3979, T = 298 K]
$C{u_{(s)}} + 2A{g^ + }(1 \times {10^{ - 3}}M) \to C{u^{2 + }}(0.250M) + 2A{g_{(s)}}$
$E_{cell}^\Theta = 2.97$ V
Ecell for the above reaction is ______________ V. (Nearest integer)
[Given : log 2.5 = 0.3979, T = 298 K]
Correct Answer: 3
Explanation:
$E = {E^o} - {{0.059} \over 2}\log {{[C{u^{ + 2}}]} \over {{{[A{g^ + }]}^2}}}$
$ = 2.97 - {{0.059} \over 2}\log {{0.25} \over {{{({{10}^{ - 3}})}^2}}} = 2.81V$
$ = 2.97 - {{0.059} \over 2}\log {{0.25} \over {{{({{10}^{ - 3}})}^2}}} = 2.81V$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 20th July Evening Shift
Potassium chlorate is prepared by electrolysis of KCl in basic solution as shown by following equation.
6OH$-$ + Cl$-$ $\to$ ClO3$-$ + 3H2O + 6e$-$
A current of xA has to be passed for 10h to produce 10.0g of potassium chlorate. The value of x is ____________. (Nearest integer)
(Molar mass of KClO3 = 122.6 g mol$-$1, F = 96500 C)
6OH$-$ + Cl$-$ $\to$ ClO3$-$ + 3H2O + 6e$-$
A current of xA has to be passed for 10h to produce 10.0g of potassium chlorate. The value of x is ____________. (Nearest integer)
(Molar mass of KClO3 = 122.6 g mol$-$1, F = 96500 C)
Correct Answer: 1
Explanation:
$W = {E \over F} \times I \times t$
$10 = {{122.6} \over {96500 \times 6}} \times x \times 10 \times 3600$
$x = 1.311$
Ans. (1)
$10 = {{122.6} \over {96500 \times 6}} \times x \times 10 \times 3600$
$x = 1.311$
Ans. (1)
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 18th March Evening Shift
The molar conductivities at infinite dilution of barium chloride, sulphuric acid and hydrochloric acid are 280, 860 and 426 S cm2 mol$-$1 respectively. The molar conductivity at infinite dilution of barium sulphate is _________ S cm2 mol$-$1. (Round off to the Nearest Integer ).
Correct Answer: 288
Explanation:
From Kohlrausch's law
$\Lambda _m^\infty (BaS{O_4}) = \lambda _m^\infty (B{a^{2 + }}) + \lambda _m^\infty (SO_4^{2 - })$
$\Lambda _m^\infty (BaS{O_4}) = \Lambda _m^\infty (BaC{l_2}) + \Lambda _m^\infty ({H_2}S{O_4}) - 2\Lambda _m^\infty (HCl)$
$ = 280 + 860 - 2(426)$
$ = 288$ S cm2 mol$-$1
$\Lambda _m^\infty (BaS{O_4}) = \lambda _m^\infty (B{a^{2 + }}) + \lambda _m^\infty (SO_4^{2 - })$
$\Lambda _m^\infty (BaS{O_4}) = \Lambda _m^\infty (BaC{l_2}) + \Lambda _m^\infty ({H_2}S{O_4}) - 2\Lambda _m^\infty (HCl)$
$ = 280 + 860 - 2(426)$
$ = 288$ S cm2 mol$-$1
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 18th March Morning Shift
For the reaction
2Fe3+(aq) + 2I$-$(aq) $ \to $ 2Fe2+(aq) + I2(s)
the magnitude of the standard molar Gibbs free energy change, $\Delta$rG$_m^o$ = $-$ ___________ kJ (Round off to the Nearest Integer).
$\left[ {\matrix{ {E_{F{e^{2 + }}/Fe(s)}^o = - 0.440V;} & {E_{F{e^{3 + }}/Fe(s)}^o = - 0.036V} \cr {E_{{I_2}/2{I^ - }}^o = 0.539V;} & {F = 96500C} \cr } } \right]$
2Fe3+(aq) + 2I$-$(aq) $ \to $ 2Fe2+(aq) + I2(s)
the magnitude of the standard molar Gibbs free energy change, $\Delta$rG$_m^o$ = $-$ ___________ kJ (Round off to the Nearest Integer).
$\left[ {\matrix{ {E_{F{e^{2 + }}/Fe(s)}^o = - 0.440V;} & {E_{F{e^{3 + }}/Fe(s)}^o = - 0.036V} \cr {E_{{I_2}/2{I^ - }}^o = 0.539V;} & {F = 96500C} \cr } } \right]$
Correct Answer: 46
Explanation:
$E_{F{e^{3 + }}/Fe}^o = 0.036V$
$F{e^{3 + }} + 3{e^ - } \to Fe$
$ \therefore $ $\Delta G_1^o = - nF{E^o}$
$ = - 3F( - 0.036)$
$E_{F{e^{2 + }}/Fe}^o = 0.440V$
$F{e^{2 + }} + 2{e^ - } \to Fe;{E^o} = - 0.440V$
$Fe \to F{e^{2 + }} + 2{e^ - };{E^o} = 0.440V$
$ \therefore $ $\Delta G_2^o = - 2F(0.440)$
$E_{{I_2}/2{I^ - }}^o = 0.539V$
${I_2} + 2{e^ - } \to 2{I^ - };{E^o} = 0.539V$
$2{I^ - } \to {I_2} + 2{e^ - };{E^o} = - 0.539V$
$\Delta G_3^o = - 2F( - 0.539)$
$ \therefore $ $\Delta {G^o} = 2\left[ {\Delta G_1^o + \Delta G_2^o} \right] + \Delta G_3^o$
$ = 2\left[ {3F(0.036) - 2F(0.440)} \right] + 2F(0.539)$
$ = - 45934 = - 45.9KJ \simeq - 46KJ$
$F{e^{3 + }} + 3{e^ - } \to Fe$
$ \therefore $ $\Delta G_1^o = - nF{E^o}$
$ = - 3F( - 0.036)$
$E_{F{e^{2 + }}/Fe}^o = 0.440V$
$F{e^{2 + }} + 2{e^ - } \to Fe;{E^o} = - 0.440V$
$Fe \to F{e^{2 + }} + 2{e^ - };{E^o} = 0.440V$
$ \therefore $ $\Delta G_2^o = - 2F(0.440)$
$E_{{I_2}/2{I^ - }}^o = 0.539V$
${I_2} + 2{e^ - } \to 2{I^ - };{E^o} = 0.539V$
$2{I^ - } \to {I_2} + 2{e^ - };{E^o} = - 0.539V$
$\Delta G_3^o = - 2F( - 0.539)$
$ \therefore $ $\Delta {G^o} = 2\left[ {\Delta G_1^o + \Delta G_2^o} \right] + \Delta G_3^o$
$ = 2\left[ {3F(0.036) - 2F(0.440)} \right] + 2F(0.539)$
$ = - 45934 = - 45.9KJ \simeq - 46KJ$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 17th March Evening Shift
A KCl solution of conductivity 0.14 S m$-$1 shows a resistance of 4.19$\Omega$ in a conductivity cell. If the same cell is filled with an HCl solution, the resistance drops to 1.03$\Omega$. The conductivity of the HCl solution is ____________ $\times$ 10$-$2 S m$-$1. (Round off to the Nearest Integer).
Correct Answer: 57
Explanation:
For KCl solution,
$R = \left( {{1 \over K}} \right)\left( {{l \over A}} \right) \Rightarrow {l \over A} = R \times K = 4.19 \times 0.14$
= 0.58
For HCl solution,
$R = \left( {{1 \over K}} \right)\left( {{l \over A}} \right)$
$ \Rightarrow K = {{(l/A)} \over R} = {{0.58} \over {1.03}} = 0.56 = 56 \times {10^{ - 2}}$ Sm$-$1
$R = \left( {{1 \over K}} \right)\left( {{l \over A}} \right) \Rightarrow {l \over A} = R \times K = 4.19 \times 0.14$
= 0.58
For HCl solution,
$R = \left( {{1 \over K}} \right)\left( {{l \over A}} \right)$
$ \Rightarrow K = {{(l/A)} \over R} = {{0.58} \over {1.03}} = 0.56 = 56 \times {10^{ - 2}}$ Sm$-$1
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 16th March Evening Shift
A 5.0 m mol dm$-$3 aqueous solution of KCl has a conductance of 0.55 mS when measured in a cell of cell constant 1.3 cm$-$1. The molar conductivity of this solution is ___________ mSm2 mol$-$1. (Round off to the Nearest Integer).
Correct Answer: 14
Explanation:
Conductance = ${{Conductivity} \over {Cell\,cons\tan t}}$
$ \therefore $ Conductivity = 0.55 $\times$ 10$-$3 $\times$ 1.3 S cm$-$1
Molar conductivity = ${{Conductivity\,(S\,c{m^{ - 1}}) \times 1000} \over {Molarity\,(mol/L)}}$
$ = {{0.55 \times {{10}^{ - 3}} \times 1.3 \times 100} \over {5 \times {{10}^{ - 3}}}}$
= 143 S cm2 mol$-$1
= 14.3 mS m2 mol$-$1
$ \approx $ 14 mS m2 mol$-$1
$ \therefore $ Conductivity = 0.55 $\times$ 10$-$3 $\times$ 1.3 S cm$-$1
Molar conductivity = ${{Conductivity\,(S\,c{m^{ - 1}}) \times 1000} \over {Molarity\,(mol/L)}}$
$ = {{0.55 \times {{10}^{ - 3}} \times 1.3 \times 100} \over {5 \times {{10}^{ - 3}}}}$
= 143 S cm2 mol$-$1
= 14.3 mS m2 mol$-$1
$ \approx $ 14 mS m2 mol$-$1
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th February Evening Shift
Emf of the following cell at 298K in V is x $\times$ 10$-$2.
Zn|Zn2+(0.1 M)||Ag+ (0.01 M)|Ag
The value of x is _________. (Rounded off to the nearest integer)
[Given : $E_{Z{n^{2 + }}/Zn}^\theta = - 0.76V;E_{A{g^{2 + }}/Ag}^\theta = + 0.80V;{{2.303RT} \over F} = 0.059$]
Zn|Zn2+(0.1 M)||Ag+ (0.01 M)|Ag
The value of x is _________. (Rounded off to the nearest integer)
[Given : $E_{Z{n^{2 + }}/Zn}^\theta = - 0.76V;E_{A{g^{2 + }}/Ag}^\theta = + 0.80V;{{2.303RT} \over F} = 0.059$]
Correct Answer: 147
Explanation:
Zn | Zn2+(0.1 M) || Ag+ (0.01 M) | Ag
Zn(s) + 2Ag+ $ \rightleftharpoons $ 2Ag(s) + Zn+2
$E_{cell}^0 = E_{A{g^ + }/Ag}^0 - E_{Z{n^{2 + }}/Zn}^0$
$ = 0.80 - ( - 0.76)$
$ = 1.56V$
${E_{cell}} = 1.56 - {{ 0.059} \over 2}\log {{[Z{n^{2 + }}]} \over {{{[A{g^ + }]}^2}}}$
$ = 1.56 - {{0.059} \over 2}\log {{0.1} \over {{{(0.01)}^2}}}$
$ = 1.56 - {{0.059} \over 2} \times 3$
$ = 1.56 - 0.0885$
$ = 1.4715$
$ = 147.15 \times {10^{ - 2}}$
Zn(s) + 2Ag+ $ \rightleftharpoons $ 2Ag(s) + Zn+2
$E_{cell}^0 = E_{A{g^ + }/Ag}^0 - E_{Z{n^{2 + }}/Zn}^0$
$ = 0.80 - ( - 0.76)$
$ = 1.56V$
${E_{cell}} = 1.56 - {{ 0.059} \over 2}\log {{[Z{n^{2 + }}]} \over {{{[A{g^ + }]}^2}}}$
$ = 1.56 - {{0.059} \over 2}\log {{0.1} \over {{{(0.01)}^2}}}$
$ = 1.56 - {{0.059} \over 2} \times 3$
$ = 1.56 - 0.0885$
$ = 1.4715$
$ = 147.15 \times {10^{ - 2}}$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 26th February Morning Shift
Consider the following reaction
$MnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{ + 2}} + 4{H_2}O,{E^o} = 1.51V$.
The quantity of electricity required in Faraday to reduce five moles of $MnO_4^ - $ is ___________. (Integer answer)
$MnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{ + 2}} + 4{H_2}O,{E^o} = 1.51V$.
The quantity of electricity required in Faraday to reduce five moles of $MnO_4^ - $ is ___________. (Integer answer)
Correct Answer: 25
Explanation:
$MnO_4^ - + 8{H^ + } + 5{e^ - } \to M{n^{ + 2}} + 4{H_2}O,{E^o} = 1.51V$
1 mole of MnO4- required 5 moles of electrons or 5 F electricity.
$ \therefore $ 5 moles of MnO4- required 25 F electricity.
1 mole of MnO4- required 5 moles of electrons or 5 F electricity.
$ \therefore $ 5 moles of MnO4- required 25 F electricity.
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 25th February Evening Shift
Copper reduces NO$_3^ - $ into NO and NO2 depending upon the concentration of HNO3 in solution. (Assuming fixed [Cu2+] and PNO = PNO2), the HNO3 concentration at which the thermodynamic tendency for reduction of NO$_3^ - $ into NO and NO2 by copper is same is 10x M. The value of 2x is _______. (Rounded off to the nearest integer)
[Given, $E_{C{u^{2 + }}/Cu}^o = 0.34$ V, $E_{NO_3^ - /NO}^o = 0.96$ V, $E_{NO_3^ - /N{O_2}}^o = 0.79$ V and at 298 K, ${{RT} \over F}$(2.303) = 0.059]
[Given, $E_{C{u^{2 + }}/Cu}^o = 0.34$ V, $E_{NO_3^ - /NO}^o = 0.96$ V, $E_{NO_3^ - /N{O_2}}^o = 0.79$ V and at 298 K, ${{RT} \over F}$(2.303) = 0.059]
Correct Answer: 1
Explanation:
Cell-I $(HN{O_3} \to NO)$
$3Cu + 2NO_3^ - + 8{H^ + } \to 3C{u^{2 + }} + 2NO + 4{H_2}O$
${Q_1} = {{{{[C{u^{2 + }}]}^3} \times {{({p_{NO}})}^2}} \over {{{[NO_3^ - ]}^2} \times {{[{H^ + }]}^8}}}$
$\because$ $E_1^o = 0.96 - ( - 0.34) = 1.3\,V$
${E_1} = 1.3 - {{0.059} \over 6}\log {Q_1}$
Cell-II $(HN{O_3} \to N{O_2})$
$Cu + 2NO_3^ - + 4{H^ + } \to C{u^{2 + }} + 2N{O_2} + 2{H_2}O$
${Q_2} = {{[C{u^2}] \times {{({p_{N{O_2}}})}^2}} \over {{{[NO_3^ - ]}^2} \times {{[{H^ + }]}^4}}}$
$\because$ $E_2^o = 0.79 - ( - 0.34)\,V = 1.13\,V$
${E_2} = 1.13 - {{0.059} \over 2}\log {Q_2}$
Now, ${E_1} = {E_2}$
$1.3 - {{0.059} \over 6}\log {Q_1} = 1.13 - {{0.059} \over 2}\log {Q_2}$
$0.17 = {{0.059} \over 6}[\log {Q_1} - 3\log {Q_2} - = {{0.059} \over 6}\log {{{Q_1}} \over {{Q_2}}}$
$ = {{0.059} \over 6}\log {{{{[C{u^{2 + }}]}^3} \times {{({p_{NO}})}^2}} \over {{{[NO_3^ - ]}^2} \times {{[{H^ + }]}^8}}} \times {{{{[NO_3^ - ]}^6} \times {{[{H^ + }]}^{12}}} \over {{{[C{u^{2 + }}]}^3} \times {{({p_{N{O_2}}})}^6}}}$
$ = {{0.059} \over 6}\log {{{{[{H^ + }]}^4} \times {{[NO_3^ - ]}^4}} \over {{{({p_{N{O_2}}})}^4}}}$ [$\because$ ${p_{NO}} = {p_{N{O_2}}}$]
$ = {{0.059} \over 6}\log {{{{[HN{O_3}]}^4}} \over {{{({p_{N{O_2}}})}^4}}}$
Now, ${p_{N{O_2}}} \equiv [HN{O_3}]$
So, $0.17 = {{0.059} \over 6}\log {[HN{O_3}]^8}$
$ = {{0.059} \over 6} \times 8\log [HN{O_3}]$
$\log [HN{O_3}] = 2.16$
$[HN{O_3}] = {10^{2.16}}M = {10^x}M$
$\therefore$ $x = 2.16$
$ \Rightarrow 2x = 2 \times 2.16 = 4.32 \simeq 4$
2021
JEE Mains
Numerical
JEE Main 2021 (Online) 24th February Evening Shift
The magnitude of the change in oxidising power of the $MnO_4^ - /M{n^{2 + }}$ couple is x $\times$ 10$-$4 V, if the H+ concentration is decreased from 1M to 10$-$4 M at 25$^\circ$C. (Assume concentration of $MnO_4^ - $ and $M{n^{2 + }}$ to be same on change in H+ concentration). The value of x is ___________. $\left[ {Given\,:{{2.303RT} \over F} = 0.059} \right]$
Correct Answer: 3776
Explanation:
Reaction,
$MnO_4^ - + {H^ + } + 5{e^ - } \to M{n^{2 + }} + 4{H_2}O$
n = 5
Applying Nernst equation, ${E_{cell}} = E_{cell}^o - {{0.0591} \over n}\log {{[P]} \over {[R]}}$
or ${E_{cell}} = E_{cell}^o - {{0.0591} \over n}\log {{[M{n^{2 + }}]} \over {[MnO_4^ - ]}}{\left[ {{1 \over {{H^ + }}}} \right]^8}$
(I) Given, [H+] = 1 M
${E_1} = E^\circ - {{0.0591} \over 5}\log {{[M{n^{2 + }}]} \over {[MnO_4^ - ]}}$
(II) Now, [H+] = 10$-$4 M
${E_2} = E^\circ - {{0.0591} \over 5}\log {{[M{n^{2 + }}]} \over {[MnO_4^ - ]}} \times {1 \over {{{({{10}^{ - 4}})}^8}}}$
$\therefore$ $\left| {{E_1} - {E_2}} \right|$
$\left| {{E_1} - {E_2}} \right| = {{0.0591} \over 5} \times 32 = 0.3776\,V = 3776 \times {10^{ - 4}}$
x = 3776
$MnO_4^ - + {H^ + } + 5{e^ - } \to M{n^{2 + }} + 4{H_2}O$
n = 5
Applying Nernst equation, ${E_{cell}} = E_{cell}^o - {{0.0591} \over n}\log {{[P]} \over {[R]}}$
or ${E_{cell}} = E_{cell}^o - {{0.0591} \over n}\log {{[M{n^{2 + }}]} \over {[MnO_4^ - ]}}{\left[ {{1 \over {{H^ + }}}} \right]^8}$
(I) Given, [H+] = 1 M
${E_1} = E^\circ - {{0.0591} \over 5}\log {{[M{n^{2 + }}]} \over {[MnO_4^ - ]}}$
(II) Now, [H+] = 10$-$4 M
${E_2} = E^\circ - {{0.0591} \over 5}\log {{[M{n^{2 + }}]} \over {[MnO_4^ - ]}} \times {1 \over {{{({{10}^{ - 4}})}^8}}}$
$\therefore$ $\left| {{E_1} - {E_2}} \right|$
$\left| {{E_1} - {E_2}} \right| = {{0.0591} \over 5} \times 32 = 0.3776\,V = 3776 \times {10^{ - 4}}$
x = 3776
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 31st August Evening Shift
Match List - I with List - II
Choose the most appropriate answer from the options given below :
| List - I (Parmeter) |
List - II (Unit) |
||
|---|---|---|---|
| (a) | Cell constant | (i) | $S\,c{m^2}mo{l^{ - 1}}$ |
| (b) | Molar conductivity | (ii) | Dimensionless |
| (c) | Conductivity | (iii) | ${m^{ - 1}}$ |
| (d) | Degree of dissociation of electrolyte | (iv) | ${\Omega ^{ - 1}}{m^{ - 1}}$ |
Choose the most appropriate answer from the options given below :
A.
(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
B.
(a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)
C.
(a)-(i), (b)-(iv), (c)-(iii), (d)-(ii)
D.
(a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th August Morning Shift
Given below are two statements :
Statement I : The limiting molar conductivity of KCl (strong electrolyte) is higher compared to that of CH3COOH (weak electrolyte).
Statement II : Molar conductivity decreases with decrease in concentration of electrolyte.
In the light of the above statements, choose the most appropriate answer from the options given below :
Statement I : The limiting molar conductivity of KCl (strong electrolyte) is higher compared to that of CH3COOH (weak electrolyte).
Statement II : Molar conductivity decreases with decrease in concentration of electrolyte.
In the light of the above statements, choose the most appropriate answer from the options given below :
A.
Statement I is true but Statement II is false.
B.
Statement I is false but Statement II is true.
C.
Both Statement I and Statement II are true.
D.
Both Statement I and Statement II are false.
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 26th February Morning Shift
Compound A used as a strong oxidizing agent is amphoteric in nature. It is the part of lead storage batteries. Compound A is :
A.
PbO
B.
PbO2
C.
PbSO4
D.
Pb3O4
2021
JEE Mains
MCQ
JEE Main 2021 (Online) 24th February Morning Shift
The electrode potential of M2+/M of 3d-series elements shows positive value for :
A.
Zn
B.
Fe
C.
Cu
D.
Co
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 6th September Morning Slot
Potassium chlorate is prepared by the
electrolysis of KCl in basic solution
6OH- + Cl- $ \to $ ClO3- + 3H2O + 6e-
If only 60% of the current is utilized in the reaction, the time (rounded to the nearest hour) required to produce 10 g of KClO3 using a current of 2 A is_________.
(Given : F = 96,500 C mol–1; molar mass of KCIO3 = 122 g mol–1)
6OH- + Cl- $ \to $ ClO3- + 3H2O + 6e-
If only 60% of the current is utilized in the reaction, the time (rounded to the nearest hour) required to produce 10 g of KClO3 using a current of 2 A is_________.
(Given : F = 96,500 C mol–1; molar mass of KCIO3 = 122 g mol–1)
Correct Answer: 11
Explanation:
For synthesis of 1 mole of ClO3- , 6F of charge
is required.
$ \therefore $ To synthesise ${{10} \over {122}}$ moles of KClO3,
Charge required = ${{10} \over {122}} \times 6$ F
${{10} \over {122}} \times 6 = {{2 \times t(hr) \times 3600} \over {96500}} \times {{60} \over {100}}$
$ \Rightarrow $ t(hr) = ${{965 \times 100} \over {122 \times 2 \times 36}}$ = 10.98 hr $ \simeq $ 11 Hr
$ \therefore $ To synthesise ${{10} \over {122}}$ moles of KClO3,
Charge required = ${{10} \over {122}} \times 6$ F
${{10} \over {122}} \times 6 = {{2 \times t(hr) \times 3600} \over {96500}} \times {{60} \over {100}}$
$ \Rightarrow $ t(hr) = ${{965 \times 100} \over {122 \times 2 \times 36}}$ = 10.98 hr $ \simeq $ 11 Hr
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 5th September Morning Slot
An oxidation-reduction reaction in which
3 electrons are transferred has a $\Delta $Gº of 17.37 kJ mol–1 at
25 oC. The value of Eo
cell (in V) is ______ × 10–2.
(1 F = 96,500 C mol–1)
3 electrons are transferred has a $\Delta $Gº of 17.37 kJ mol–1 at
25 oC. The value of Eo
cell (in V) is ______ × 10–2.
(1 F = 96,500 C mol–1)
Correct Answer: -6
Explanation:
$\Delta $Gº = 17.37 kJ ; n = 3
$\Delta $Gº = -nF$E_{cell}^o$
$ \Rightarrow $ $E_{cell}^o$ = $ - {{17.37 \times 1000} \over {3 \times 96500}}$
= -0.06 = -6.00 $ \times $ 10-2
$\Delta $Gº = -nF$E_{cell}^o$
$ \Rightarrow $ $E_{cell}^o$ = $ - {{17.37 \times 1000} \over {3 \times 96500}}$
= -0.06 = -6.00 $ \times $ 10-2
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 3rd September Evening Slot
An acidic solution of dichromate is electrolyzed
for 8 minutes using 2A current. As per the
following equation
Cr2O72- + 14H+ + 6e– $ \to $ 2Cr3+ + 7H2O
The amount of Cr3+ obtained was 0.104 g. The efficiency of the process(in%) is (Take : F = 96000 C, At. mass of chromium = 52) ______.
Cr2O72- + 14H+ + 6e– $ \to $ 2Cr3+ + 7H2O
The amount of Cr3+ obtained was 0.104 g. The efficiency of the process(in%) is (Take : F = 96000 C, At. mass of chromium = 52) ______.
Correct Answer: 60
Explanation:
Cr2O72-
+ 14H+ + 6e– $ \to $ 2Cr3+ + 7H2O
I = 2 A, t = 8 min
From 1st law of faraday,
wCr+3 = z $ \times $ i $ \times $ t
$ \Rightarrow $ wCr+3 = ${{52} \over {96000 \times 3}} \times 2 \times 8 \times 60$
= ${{52} \over {300}}$
Mass of Cr3+ ions actually obtained = 0.104 gm
% efficiency =
= ${{0.104} \over {{{52} \over {300}}}}$ $ \times $ 100
= 60 %
I = 2 A, t = 8 min
From 1st law of faraday,
wCr+3 = z $ \times $ i $ \times $ t
$ \Rightarrow $ wCr+3 = ${{52} \over {96000 \times 3}} \times 2 \times 8 \times 60$
= ${{52} \over {300}}$
Mass of Cr3+ ions actually obtained = 0.104 gm
% efficiency =
Actual obtained Amt
Theo. obtained Amt
$ \times $ 100
= ${{0.104} \over {{{52} \over {300}}}}$ $ \times $ 100
= 60 %
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 3rd September Morning Slot
The photoelectric current from Na (Work function, w0
= 2.3 eV) is stopped by the output voltage of
the cell
Pt(s) | H2 (g, 1 Bar) | HCl (aq., pH =1) | AgCl(s) | Ag(s).
The pH of aq. HCl required to stop the photoelectric current form K(w0 = 2.25 eV), all other conditions remaining the same, is _______ $ \times $ 10-2 (to the nearest integer).
Given, 2.303${{RT} \over F}$ = 0.06 V;
$E_{AgCl|Ag|C{l^ - }}^0$ = 0.22 V
Pt(s) | H2 (g, 1 Bar) | HCl (aq., pH =1) | AgCl(s) | Ag(s).
The pH of aq. HCl required to stop the photoelectric current form K(w0 = 2.25 eV), all other conditions remaining the same, is _______ $ \times $ 10-2 (to the nearest integer).
Given, 2.303${{RT} \over F}$ = 0.06 V;
$E_{AgCl|Ag|C{l^ - }}^0$ = 0.22 V
Correct Answer: 142
Explanation:
${1 \over 2}$H2 + AgCl $ \to $ H+ + Ag + Cl-
From Nernst Equation,
Ecell = $E_{cell}^0$ - ${{0.06} \over 1}\log \left[ {{H^ + }} \right]\left[ {C{l^ - }} \right]$
= 0.22 – 0.06 log 10–2 = 0.34 V
Work function of Na metal = 2.3 eV
KE of photoelectron = 0.34 eV
Energy of incident radiation = 2.3 + 0.34 = 2.64 eV
For K atom,
Energy of incident radiation for K metal = 2.64 eV
Work function of K metal = 2.25 eV
KE of photoelectrons = 2.64 – 2.25 = 0.39 eV
$ \therefore $ Ecell = 0.39 V
Using Nernst Equation,
0.39 = 0.22 - ${{0.06} \over 1}\log \left[ {{H^ + }} \right]\left[ {C{l^ - }} \right]$
As $\left[ {{H^ + }} \right] = \left[ {C{l^ - }} \right]$
0.39 = 0.22 - ${{0.06} \over 1}\log {\left[ {{H^ + }} \right]^2}$
$ \Rightarrow $ 0.39 = 0.22 - $0.12 \times \log \left[ {{H^ + }} \right]$
$ \Rightarrow $ 0.39 = 0.22 + 0.12 $ \times $ pH
$ \Rightarrow $ pH = 1.42 = 142 $ \times $ 10-2
From Nernst Equation,
Ecell = $E_{cell}^0$ - ${{0.06} \over 1}\log \left[ {{H^ + }} \right]\left[ {C{l^ - }} \right]$
= 0.22 – 0.06 log 10–2 = 0.34 V
Work function of Na metal = 2.3 eV
KE of photoelectron = 0.34 eV
Energy of incident radiation = 2.3 + 0.34 = 2.64 eV
For K atom,
Energy of incident radiation for K metal = 2.64 eV
Work function of K metal = 2.25 eV
KE of photoelectrons = 2.64 – 2.25 = 0.39 eV
$ \therefore $ Ecell = 0.39 V
Using Nernst Equation,
0.39 = 0.22 - ${{0.06} \over 1}\log \left[ {{H^ + }} \right]\left[ {C{l^ - }} \right]$
As $\left[ {{H^ + }} \right] = \left[ {C{l^ - }} \right]$
0.39 = 0.22 - ${{0.06} \over 1}\log {\left[ {{H^ + }} \right]^2}$
$ \Rightarrow $ 0.39 = 0.22 - $0.12 \times \log \left[ {{H^ + }} \right]$
$ \Rightarrow $ 0.39 = 0.22 + 0.12 $ \times $ pH
$ \Rightarrow $ pH = 1.42 = 142 $ \times $ 10-2
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 2nd September Evening Slot
For the disproportionation reaction
2Cu+(aq) ⇌ Cu(s) + Cu2+(aq) at 298 K. ln K
(where K is the equilibrium constant) is
___________ × 10–1.
Given :
($E_{C{u^{2 + }}/C{u^ + }}^0 = 0.16V$
$E_{C{u^ + }/Cu}^0 = 0.52V$
${{RT} \over F} = 0.025$)
2Cu+(aq) ⇌ Cu(s) + Cu2+(aq) at 298 K. ln K
(where K is the equilibrium constant) is
___________ × 10–1.
Given :
($E_{C{u^{2 + }}/C{u^ + }}^0 = 0.16V$
$E_{C{u^ + }/Cu}^0 = 0.52V$
${{RT} \over F} = 0.025$)
Correct Answer: 144
Explanation:
$E_{cell}^0$ = $E_{C{u^ + }/Cu}^0$ - $E_{C{u^{2 + }}/C{u^ + }}^0$
= 0.52 – 0.16
= 0.36 V
At equilibrium, Ecell = 0
$E_{cell}^0$ = ${{RT} \over {nF}}$ln K
$ \Rightarrow $ ln K = ${{E_{cell}^0 \times nF} \over {RT}}$
= ${{0.36 \times 1} \over {0.025}}$ = 14.4 = 144 $ \times $ 10-1
= 0.52 – 0.16
= 0.36 V
At equilibrium, Ecell = 0
$E_{cell}^0$ = ${{RT} \over {nF}}$ln K
$ \Rightarrow $ ln K = ${{E_{cell}^0 \times nF} \over {RT}}$
= ${{0.36 \times 1} \over {0.025}}$ = 14.4 = 144 $ \times $ 10-1
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 2nd September Morning Slot
The Gibbs change (in J) for the given reaction at
[Cu2+] = [Sn2+] = 1 M and 298K is :
Cu(s) + Sn2+(aq.) $ \to $ Cu2+(aq.) + Sn(s);
($E_{S{n^{2 + }}|Sn}^0 = - 0.16\,V$,
$E_{C{u^{2 + }}|Cu}^0 = 0.34\,V$)
Take F = 96500 C mol–1)
[Cu2+] = [Sn2+] = 1 M and 298K is :
Cu(s) + Sn2+(aq.) $ \to $ Cu2+(aq.) + Sn(s);
($E_{S{n^{2 + }}|Sn}^0 = - 0.16\,V$,
$E_{C{u^{2 + }}|Cu}^0 = 0.34\,V$)
Take F = 96500 C mol–1)
Correct Answer: 96500
Explanation:
$\Delta $G = $\Delta $Go + RTln $\left[ {{{S{n^{ + 2}}} \over {C{u^{ + 2}}}}} \right]$
= –2 × 96500 [(–0.16) – 0.34] + RT$\left[ {{1 \over 1}} \right]$
= 96500 J
= –2 × 96500 [(–0.16) – 0.34] + RT$\left[ {{1 \over 1}} \right]$
= 96500 J
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 9th January Morning Slot
108 g of silver (molar mass 108 g mol–1) is
deposited at cathode from AgNO3(aq) solution
by a certain quantity of electricity. The
volume (in L) of oxygen gas produced at
273 K and 1 bar pressure from water by the
same quantity of electricity is _______.
Correct Answer: 5.66to5.68
Explanation:
Cathode : Ag+(aq) + e- $ \to $ Ag(s)
Moles of Ag deposited = ${{108} \over {108}}$ = 1 mole
Anode : 2H2O $ \to $ O2 + 4H+ + 4e-
Here we have to find volume of O2 evolved.
Equivalance of Ag = Equivalance of O2
$ \Rightarrow $ 1 $ \times $ 1 = nO2 $ \times $ 4
$ \Rightarrow $ nO2 = ${1 \over 4}$ mol
$ \therefore $ Volume of O2 evolved
= ${1 \over 4}$ $ \times $ 22.4
= 5.6 lit
Moles of Ag deposited = ${{108} \over {108}}$ = 1 mole
Anode : 2H2O $ \to $ O2 + 4H+ + 4e-
Here we have to find volume of O2 evolved.
Equivalance of Ag = Equivalance of O2
$ \Rightarrow $ 1 $ \times $ 1 = nO2 $ \times $ 4
$ \Rightarrow $ nO2 = ${1 \over 4}$ mol
$ \therefore $ Volume of O2 evolved
= ${1 \over 4}$ $ \times $ 22.4
= 5.6 lit
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 8th January Evening Slot
For an electrochemical cell
Sn(s) | Sn2+ (aq,1M)||Pb2+ (aq,1M)|Pb(s)
the ratio ${{\left[ {S{n^{2 + }}} \right]} \over {\left[ {P{b^{2 + }}} \right]}}$ when this cell attains equilibrium is _________.
(Given $E_{S{n^{2 + }}|Sn}^0 = - 0.14V$,
$E_{P{b^{2 + }}|Pb}^0 = - 0.13V$, ${{2.303RT} \over F} = 0.06$)
Sn(s) | Sn2+ (aq,1M)||Pb2+ (aq,1M)|Pb(s)
the ratio ${{\left[ {S{n^{2 + }}} \right]} \over {\left[ {P{b^{2 + }}} \right]}}$ when this cell attains equilibrium is _________.
(Given $E_{S{n^{2 + }}|Sn}^0 = - 0.14V$,
$E_{P{b^{2 + }}|Pb}^0 = - 0.13V$, ${{2.303RT} \over F} = 0.06$)
Correct Answer: 2.13TO2.16
Explanation:
Cell reaction is :
Sn(s) + Pb+2(aq) $ \to $ Sn+2(aq) + Pb(s)
Apply Nernst equation :
Ecell = $E_{cell}^0$ - ${{0.06} \over 2}\log {{\left[ {S{n^{ + 2}}} \right]} \over {\left[ {P{b^{ + 2}}} \right]}}$ ....(1)
$ \Rightarrow $ $E_{cell}^0$ = -0.13 + 0.14 = 0.01 V
At equilibrium : Ecell = 0
Substituting in (1), we get
0 = 0.01 - ${{0.06} \over 2}\log {{\left[ {S{n^{ + 2}}} \right]} \over {\left[ {P{b^{ + 2}}} \right]}}$
$ \Rightarrow $ $\log {{\left[ {S{n^{ + 2}}} \right]} \over {\left[ {P{b^{ + 2}}} \right]}}$ = ${1 \over 3}$
$ \Rightarrow $ ${{\left[ {S{n^{2 + }}} \right]} \over {\left[ {P{b^{2 + }}} \right]}}$ = 2.15
Sn(s) + Pb+2(aq) $ \to $ Sn+2(aq) + Pb(s)
Apply Nernst equation :
Ecell = $E_{cell}^0$ - ${{0.06} \over 2}\log {{\left[ {S{n^{ + 2}}} \right]} \over {\left[ {P{b^{ + 2}}} \right]}}$ ....(1)
$ \Rightarrow $ $E_{cell}^0$ = -0.13 + 0.14 = 0.01 V
At equilibrium : Ecell = 0
Substituting in (1), we get
0 = 0.01 - ${{0.06} \over 2}\log {{\left[ {S{n^{ + 2}}} \right]} \over {\left[ {P{b^{ + 2}}} \right]}}$
$ \Rightarrow $ $\log {{\left[ {S{n^{ + 2}}} \right]} \over {\left[ {P{b^{ + 2}}} \right]}}$ = ${1 \over 3}$
$ \Rightarrow $ ${{\left[ {S{n^{2 + }}} \right]} \over {\left[ {P{b^{2 + }}} \right]}}$ = 2.15
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 8th January Morning Slot
What would be the electrode potential for the given half cell reaction at pH = 5?
______.
2H2O $ \to $ O2 + 4H$ \oplus $ + 4e– ; $E_{red}^0$ = 1.23 V
(R = 8.314 J mol–1 K–1 ; Temp = 298 k;
oxygen under std. atm. pressure of 1 bar)
2H2O $ \to $ O2 + 4H$ \oplus $ + 4e– ; $E_{red}^0$ = 1.23 V
(R = 8.314 J mol–1 K–1 ; Temp = 298 k;
oxygen under std. atm. pressure of 1 bar)
Correct Answer: 1.52TO1.53
Explanation:
E = E0 - ${{0.0591} \over 4}\log {\left[ {{H^ + }} \right]^4}$
$ \Rightarrow $ E = 1.23 + 0.0591 × pH
$ \Rightarrow $ E = 1.23 + 0.0591 × (5)
$ \Rightarrow $ E = 1.52
$ \Rightarrow $ E = 1.23 + 0.0591 × pH
$ \Rightarrow $ E = 1.23 + 0.0591 × (5)
$ \Rightarrow $ E = 1.52
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Evening Slot
For the given cell :
Cu(s) | Cu2+(C1M) || Cu2+(C2M) | Cu(s)
change in Gibbs energy ($\Delta $G) is negative, if :
Cu(s) | Cu2+(C1M) || Cu2+(C2M) | Cu(s)
change in Gibbs energy ($\Delta $G) is negative, if :
A.
C2 = $\sqrt 2 $C1
B.
C2 = ${{{C_1}} \over {\sqrt 2 }}$
C.
C1 = 2C2
D.
C1 = C2
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Evening Slot
The variation of molar conductivity with concentration of an electrolyte (X) in aqueous solution
is shown in the given figure.
The electrolyte X is :
The electrolyte X is :
A.
HCl
B.
CH3COOH
C.
NaCl
D.
KNO3
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Evening Slot
250 mL of a waste solution obtained from the
workshop of a goldsmith contains 0.1 M AgNO3
and 0.1 M AuCl. The solution was electrolyzed
at 2V by passing a current of 1A for 15
minutes. The metal/metals electrodeposited will
be
[ $E_{A{g^ + }/Ag}^0$ = 0.80 V, $E_{A{u^ + }/Au}^0$ = 1.69 V ]
[ $E_{A{g^ + }/Ag}^0$ = 0.80 V, $E_{A{u^ + }/Au}^0$ = 1.69 V ]
A.
Silver and gold in equal mass proportion
B.
Silver and gold in proportion to their atomic
weights
C.
Only gold
D.
Only silver