Electrochemistry
348 Questions
2021
JEE Advanced
Numerical
JEE Advanced 2021 Paper 2 Online
At 298 K, the limiting molar conductivity of a weak monobasic acid is 4 $\times$ 102 S cm2 mol$-$1. At 298 K, for an aqueous solution of the acid the degree of dissociation is $\alpha$ and the molar conductivity is y $\times$ 102 S cm2 mol$-$1. At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes 3y $\times$ 102 S cm2 mol$-$1.
The value of $\alpha$ is __________.
The value of $\alpha$ is __________.
Correct Answer: 0.22
Explanation:
Degree of dissociation = $\alpha$
Limiting molar conductivity, $\Lambda _m^o$ = 4 $\times$ 102 S cm2 mol$-$1
Molar conductivity, $\Lambda _m$ = y $\times$ 102 S cm2 mol$-$1
Molar conductivity of dilution, $\Lambda _m^o$ = 3y $\times$ 102 S cm2 mol$-$1
Concentration before dilution = C
Concentration after dilution = ${C \over {20}}$
Using relation, $\alpha = {{\Lambda _m^{}} \over {\Lambda _m^o}}$ .... (i)
Dissociation constant, ${K_a} = {{C{\alpha ^2}} \over {(1 - \alpha )}}$
Putting Eq. (i),
${K_a} = {{C\Lambda _m^2} \over {\Lambda _m^{o2}\left( {1 - {{\Lambda _m^{}} \over {\Lambda _m^o}}} \right)}} = {{C\Lambda _m^2} \over {\Lambda _m^o(\Lambda _m^o - \Lambda _m^{})}}$
Dissociation constant before dillution,
${K_a} = {{C{{(y \times {{10}^2})}^2}} \over {(4 \times {{10}^2})(4 \times {{10}^2} - y \times {{10}^2})}}$ ..... (ii)
Dissociation constant after dilution,
${K_a} = {{{C \over {20}}{{(3y \times {{10}^2})}^2}} \over {(4 \times {{10}^2})(4 \times {{10}^2} - 3y \times {{10}^2})}}$ ..... (iii)
Comparing Eqs. (ii) and (iii),
${{C{{(y \times {{10}^2})}^2}} \over {(4 \times {{10}^2})(4 \times {{10}^2} - y \times {{10}^2})}} = {{C{{(3y \times {{10}^2})}^2}} \over {20(4 \times {{10}^2})(4 \times {{10}^2} - 3y \times {{10}^2})}}$
${{{y^2} \times {{10}^4}} \over {{{10}^2}(4 - y)}} = {{9{y^2} \times {{10}^4}} \over {{{10}^2} \times 20(4 - 3y)}}$
${1 \over {(4 - y)}} = {9 \over {20(4 - 3y)}}$
$80 - 60y = 36 - 9y$
$80 - 36 = 60y - 9y$
$44 = 51y$
$ \Rightarrow y = {{44} \over {51}}$ .... (iv)
Putting in Eq. (i),
$\alpha = {{\Lambda _m^{}} \over {\Lambda _m^o}} = {{y \times {{10}^2}} \over {4 \times {{10}^2}}}$
$\alpha = {{44} \over {51 \times 4}} = {{11} \over {51}}$
$\alpha = 0.22$
The value of $\alpha$ is 0.22.
Limiting molar conductivity, $\Lambda _m^o$ = 4 $\times$ 102 S cm2 mol$-$1
Molar conductivity, $\Lambda _m$ = y $\times$ 102 S cm2 mol$-$1
Molar conductivity of dilution, $\Lambda _m^o$ = 3y $\times$ 102 S cm2 mol$-$1
Concentration before dilution = C
Concentration after dilution = ${C \over {20}}$
Using relation, $\alpha = {{\Lambda _m^{}} \over {\Lambda _m^o}}$ .... (i)
Dissociation constant, ${K_a} = {{C{\alpha ^2}} \over {(1 - \alpha )}}$
Putting Eq. (i),
${K_a} = {{C\Lambda _m^2} \over {\Lambda _m^{o2}\left( {1 - {{\Lambda _m^{}} \over {\Lambda _m^o}}} \right)}} = {{C\Lambda _m^2} \over {\Lambda _m^o(\Lambda _m^o - \Lambda _m^{})}}$
Dissociation constant before dillution,
${K_a} = {{C{{(y \times {{10}^2})}^2}} \over {(4 \times {{10}^2})(4 \times {{10}^2} - y \times {{10}^2})}}$ ..... (ii)
Dissociation constant after dilution,
${K_a} = {{{C \over {20}}{{(3y \times {{10}^2})}^2}} \over {(4 \times {{10}^2})(4 \times {{10}^2} - 3y \times {{10}^2})}}$ ..... (iii)
Comparing Eqs. (ii) and (iii),
${{C{{(y \times {{10}^2})}^2}} \over {(4 \times {{10}^2})(4 \times {{10}^2} - y \times {{10}^2})}} = {{C{{(3y \times {{10}^2})}^2}} \over {20(4 \times {{10}^2})(4 \times {{10}^2} - 3y \times {{10}^2})}}$
${{{y^2} \times {{10}^4}} \over {{{10}^2}(4 - y)}} = {{9{y^2} \times {{10}^4}} \over {{{10}^2} \times 20(4 - 3y)}}$
${1 \over {(4 - y)}} = {9 \over {20(4 - 3y)}}$
$80 - 60y = 36 - 9y$
$80 - 36 = 60y - 9y$
$44 = 51y$
$ \Rightarrow y = {{44} \over {51}}$ .... (iv)
Putting in Eq. (i),
$\alpha = {{\Lambda _m^{}} \over {\Lambda _m^o}} = {{y \times {{10}^2}} \over {4 \times {{10}^2}}}$
$\alpha = {{44} \over {51 \times 4}} = {{11} \over {51}}$
$\alpha = 0.22$
The value of $\alpha$ is 0.22.
2021
JEE Advanced
Numerical
JEE Advanced 2021 Paper 2 Online
At 298 K, the limiting molar conductivity of a weak monobasic acid is 4 $\times$ 102 S cm2 mol$-$1. At 298 K, for an aqueous solution of the acid the degree of dissociation is $\alpha$ and the molar conductivity is y $\times$ 102 S cm2 mol$-$1. At 298 K, upon 20 times dilution with water, the molar conductivity of the solution becomes 3y $\times$ 102 S cm2 mol$-$1.
The value of y is __________.
The value of y is __________.
Correct Answer: 0.86
Explanation:
Degree of dissociation = $\alpha$
Limiting molar conductivity, $\Lambda _m^o$ = 4 $\times$ 102 S cm2 mol$-$1
Molar conductivity, $\Lambda _m$ = y $\times$ 102 S cm2 mol$-$1
Molar conductivity of dilution, $\Lambda _m^o$ = 3y $\times$ 102 S cm2 mol$-$1
Concentration before dilution = C
Concentration after dilution = ${C \over {20}}$
Using relation, $\alpha = {{\Lambda _m^{}} \over {\Lambda _m^o}}$ .... (i)
Dissociation constant, ${K_a} = {{C{\alpha ^2}} \over {(1 - \alpha )}}$
Putting Eq. (i),
${K_a} = {{C\Lambda _m^2} \over {\Lambda _m^{o2}\left( {1 - {{\Lambda _m^{}} \over {\Lambda _m^o}}} \right)}} = {{C\Lambda _m^2} \over {\Lambda _m^o(\Lambda _m^o - \Lambda _m^{})}}$
Dissociation constant before dillution,
${K_a} = {{C{{(y \times {{10}^2})}^2}} \over {(4 \times {{10}^2})(4 \times {{10}^2} - y \times {{10}^2})}}$ ..... (ii)
Dissociation constant after dilution,
${K_a} = {{{C \over {20}}{{(3y \times {{10}^2})}^2}} \over {(4 \times {{10}^2})(4 \times {{10}^2} - 3y \times {{10}^2})}}$ ..... (iii)
Comparing Eqs. (ii) and (iii),
${{C{{(y \times {{10}^2})}^2}} \over {(4 \times {{10}^2})(4 \times {{10}^2} - y \times {{10}^2})}} = {{C{{(3y \times {{10}^2})}^2}} \over {20(4 \times {{10}^2})(4 \times {{10}^2} - 3y \times {{10}^2})}}$
${{{y^2} \times {{10}^4}} \over {{{10}^2}(4 - y)}} = {{9{y^2} \times {{10}^4}} \over {{{10}^2} \times 20(4 - 3y)}}$
${1 \over {(4 - y)}} = {9 \over {20(4 - 3y)}}$
$80 - 60y = 36 - 9y$
$80 - 36 = 60y - 9y$
$44 = 51y$
$ \Rightarrow y = {{44} \over {51}}$ = 0.86
Limiting molar conductivity, $\Lambda _m^o$ = 4 $\times$ 102 S cm2 mol$-$1
Molar conductivity, $\Lambda _m$ = y $\times$ 102 S cm2 mol$-$1
Molar conductivity of dilution, $\Lambda _m^o$ = 3y $\times$ 102 S cm2 mol$-$1
Concentration before dilution = C
Concentration after dilution = ${C \over {20}}$
Using relation, $\alpha = {{\Lambda _m^{}} \over {\Lambda _m^o}}$ .... (i)
Dissociation constant, ${K_a} = {{C{\alpha ^2}} \over {(1 - \alpha )}}$
Putting Eq. (i),
${K_a} = {{C\Lambda _m^2} \over {\Lambda _m^{o2}\left( {1 - {{\Lambda _m^{}} \over {\Lambda _m^o}}} \right)}} = {{C\Lambda _m^2} \over {\Lambda _m^o(\Lambda _m^o - \Lambda _m^{})}}$
Dissociation constant before dillution,
${K_a} = {{C{{(y \times {{10}^2})}^2}} \over {(4 \times {{10}^2})(4 \times {{10}^2} - y \times {{10}^2})}}$ ..... (ii)
Dissociation constant after dilution,
${K_a} = {{{C \over {20}}{{(3y \times {{10}^2})}^2}} \over {(4 \times {{10}^2})(4 \times {{10}^2} - 3y \times {{10}^2})}}$ ..... (iii)
Comparing Eqs. (ii) and (iii),
${{C{{(y \times {{10}^2})}^2}} \over {(4 \times {{10}^2})(4 \times {{10}^2} - y \times {{10}^2})}} = {{C{{(3y \times {{10}^2})}^2}} \over {20(4 \times {{10}^2})(4 \times {{10}^2} - 3y \times {{10}^2})}}$
${{{y^2} \times {{10}^4}} \over {{{10}^2}(4 - y)}} = {{9{y^2} \times {{10}^4}} \over {{{10}^2} \times 20(4 - 3y)}}$
${1 \over {(4 - y)}} = {9 \over {20(4 - 3y)}}$
$80 - 60y = 36 - 9y$
$80 - 36 = 60y - 9y$
$44 = 51y$
$ \Rightarrow y = {{44} \over {51}}$ = 0.86
2021
JEE Advanced
MSQ
JEE Advanced 2021 Paper 2 Online
Some standard electrode potentials at 298 K are given below :
Pb2+ /Pb = $- $0.13 V
Ni2+ /Ni = $-$ 0.24 V
Cd2+ /Cd = $-$ 0.40 V
Fe2+ /Fe = $-$ 0.44 V
To a solution containing 0.001 M of X2+ and 0.1 M of Y2+, the metal rods X and Y are inserted (at 298 K) and connected by a conducting wire. This resulted in dissolution of X. The correct combination(s) of X and Y, respectively, is(are)
(Given : Gas constant, R = 8.314 J K$-$ mol$-$1, Faraday constant, F = 96500 C mol$-$1)
Pb2+ /Pb = $- $0.13 V
Ni2+ /Ni = $-$ 0.24 V
Cd2+ /Cd = $-$ 0.40 V
Fe2+ /Fe = $-$ 0.44 V
To a solution containing 0.001 M of X2+ and 0.1 M of Y2+, the metal rods X and Y are inserted (at 298 K) and connected by a conducting wire. This resulted in dissolution of X. The correct combination(s) of X and Y, respectively, is(are)
(Given : Gas constant, R = 8.314 J K$-$ mol$-$1, Faraday constant, F = 96500 C mol$-$1)
A.
Cd and Ni
B.
Cd and Fe
C.
Ni and Pb
D.
Ni and Fe
2021
AP-EAPCET
MCQ
AP EAPCET 2021 - 20th August Evening Shift
In the electrolysis of a CuSO$_4$ solution, how many grames of Cu are plated out on the cathode, in the time that is required to liberate 5.6 L of O$_2$(g), measured at 1 atm and 273 K, at the anode?
A.
31.75 g
B.
14.2 g
C.
4.32 g
D.
3.175 g
2021
AP-EAPCET
MCQ
AP EAPCET 2021 - 20th August Morning Shift
If hydrogen electrons dipped in two solutions of pH = 3 and pH = 6 are connected by a salt bridge, the emf of the resulting cell is
A.
0.177 V
B.
0.3 V
C.
0.052 V
D.
0.104 V
2021
AP-EAPCET
MCQ
AP EAPCET 2021 - 19th August Evening Shift
At $291 \mathrm{~K}$, saturated solution of $\mathrm{BaSO}_4$ was found to have a specific conductivity of $3.648 \times 10^{-6} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}$ and that of water being used is $1.25 \times 10^{-6} \mathrm{ohm}^{-1} \mathrm{~cm}^{-1}$. If the ionic conductances of $\mathrm{Ba}^{2+}$ and $\mathrm{SO}_4^{2-}$ are 110 and $136.6 \mathrm{ohm}^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ respectively. The solubility of $\mathrm{BaSO}_4$ at $291 \mathrm{~K}$ will be [Atomic masses of $\mathrm{Ba}=137, \mathrm{~S}=32, \mathrm{O}=16]$
A.
$1.435 \times 10^{-3} \mathrm{gL}^{-1}$
B.
$2.266 \times 10^{-3} \mathrm{gL}^{-1}$
C.
$2.843 \times 10^{-3} \mathrm{gL}^{-1}$
D.
$1.768 \times 10^{-3} \mathrm{gL}^{-1}$
2021
AP-EAPCET
MCQ
AP EAPCET 2021 - 19th August Evening Shift
Find the emf of the following cell reaction. Given, $E_{\mathrm{Cr}^{3+} / \mathrm{Cr}^{2+}}^{\Upsilon}=-0.72 \mathrm{~V}$ and $E_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\Upsilon}= -0.42 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$ is $\mathrm{Cr}\left|\mathrm{Cr}^{3+}(0.1 \mathrm{M})\right| \mid \mathrm{Fe}^{2+} (0.1 \mathrm{M}) \mid \mathrm{Fe}$
A.
0.30 V
B.
0.25 V
C.
1.14 V
D.
1.56 V
2021
AP-EAPCET
MCQ
AP EAPCET 2021 - 19th August Evening Shift
For $\mathrm{C{r_2}O_7^{2 - } + 14{H^ + } + 6{e^ - }\buildrel {Yields} \over \longrightarrow 2C{r^{3 + }} + 7{H_2}O,{E^\Upsilon } = 1.33}$ V at $[C{r_2}O_7^{2 - }] = 4.5$ millimole, $[C{r^{3 + }}] = 1.5$ millimole and $E = 1.067$ V, then calculate the pH of the solution.
A.
2
B.
3
C.
2.5
D.
1.5
2021
AP-EAPCET
MCQ
AP EAPCET 2021 - 19th August Morning Shift
Assertion (A) Sodium acetate on Kolbe’s electrolysis gives ethane.
Reason (R) Methyl free radical is formed at cathode.
A.
Both A and R are true and R is a correct explanation of A.
B.
Both A and R are true but R is not a correct explanation of A.
C.
A is true but R is false.
D.
A is false but R is true.
2021
AP-EAPCET
MCQ
AP EAPCET 2021 - 19th August Morning Shift
When a current of 10 A is passes through molten AlCl$_3$ for 1.608 minutes. The mass of Al deposited will be
[Atomic mass of Al = 27 g]
A.
0.09 g
B.
0.81 g
C.
1.35 g
D.
0.27 g
2021
AP-EAPCET
MCQ
AP EAPCET 2021 - 19th August Morning Shift
The molar conductivities $\left(\lambda_{\mathrm{m}}^{\Upsilon}\right)$ at infinite dilution of $\mathrm{KBr}, \mathrm{HBr}$ and $\mathrm{KNH}_2$ are 120.5, 420.6 and $90.48 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$ respectively. Find the value of $\lambda_{\mathrm{m}}^\Upsilon$ for $\mathrm{NH}_3$.
A.
$511.0 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
B.
$390.5 \mathrm{~s} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
C.
$256.2 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
D.
$240.9 \mathrm{~s} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$
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
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 4th September Morning Slot
$E_{C{u^{2 + }}|Cu}^0$ = +0.34 V
$E_{Z{n^{2 + }}|Zn}^0$ = -0.76 V
Identify the incorrect statement from the option below for the above cell :
A.
If Eext < 1.1 V, Zn dissolves at anode and Cu
deposits at cathode
B.
If Eext = 1.1 V, no flow of e– or current
occurs
C.
If Eext > 1.1 V, e– flows from Cu to Zn
D.
If Eext > 1.1 V, Zn dissolves at Zn electrode
and Cu deposits at Cu electrode
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Morning Slot
Let CNaCl
and CBaSO4 be the conductances (in S) measured for saturated aqueous solutions of NaCl
and BaSO4, respectively, at a temperature T.
Which of the following is false?
A.
Ionic mobilities of ions from both salts increase with T.
B.
CNaCl(T2) > CNaCl(T1) for T2 > T1
C.
CBaSO4(T2) > CBaSO4(T1) for T2 > T1
D.
CNaCl >> CBaSO4 at a given T
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Evening Slot
The equation that is incorrect is :
A.
${\left( {\Lambda _m^0} \right)_{KCl}} - {\left( {\Lambda _m^0} \right)_{NaCl}} = {\left( {\Lambda _m^0} \right)_{KBr}} - {\left( {\Lambda _m^0} \right)_{NaBr}}$
B.
${\left( {\Lambda _m^0} \right)_{NaBr}} - {\left( {\Lambda _m^0} \right)_{NaI}} = {\left( {\Lambda _m^0} \right)_{KBr}} - {\left( {\Lambda _m^0} \right)_{NaBr}}$
C.
${\left( {\Lambda _m^0} \right)_{NaBr}} - {\left( {\Lambda _m^0} \right)_{NaCl}} = {\left( {\Lambda _m^0} \right)_{KBr}} - {\left( {\Lambda _m^0} \right)_{KCl}}$
D.
${\left( {\Lambda _m^0} \right)_{{H_2}O}} = {\left( {\Lambda _m^0} \right)_{HCl}} + {\left( {\Lambda _m^0} \right)_{NaOH}} - {\left( {\Lambda _m^0} \right)_{NaCl}}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Morning Slot
Given that the standard potentials (Eo) of Cu2+/Cu and Cu+/Cu are 0.34 V and 0.522 V respectively, the Eo of Cu2+/Cu+ :
A.
- 0.182 V
B.
- 0.158 V
C.
0.182 V
D.
+0.158 V
2020
JEE Advanced
Numerical
JEE Advanced 2020 Paper 1 Offline
Consider a 70% efficient hydrogen-oxygen fuel cell working under standard conditions at 1 bar and 298 K. Its cell reaction is
${H_2}(g) + {1 \over 2}{O_2}(g)\buildrel {} \over \longrightarrow {H_2}O(l)$
The work derived from the cell on the consumption of 1.0 $ \times $ 10$-$3 mole of H2(g) is used to compress 1.00 mole of a monoatomic ideal gas in a thermally insulated container. What is the change in the temperature (in K) of the ideal gas?
The standard reduction potentials for the two half-cells are given below :
${O_2}(g) + 4{H^ + }(aq) + 4{e^ - }\buildrel {} \over \longrightarrow 2{H_2}O(l),$
${E^o} = 1.23V$
$2{H^ + }(aq) + 2{e^ - }\buildrel {} \over \longrightarrow {H_2}(g),$
${E^o} = 0.00\,V$
Use, $F = 96500\,C\,mo{l^{ - 1}}$, $R = 8.314\,J\,mo{l^{ - 1}}\,{K^{ - 1}}$.
${H_2}(g) + {1 \over 2}{O_2}(g)\buildrel {} \over \longrightarrow {H_2}O(l)$
The work derived from the cell on the consumption of 1.0 $ \times $ 10$-$3 mole of H2(g) is used to compress 1.00 mole of a monoatomic ideal gas in a thermally insulated container. What is the change in the temperature (in K) of the ideal gas?
The standard reduction potentials for the two half-cells are given below :
${O_2}(g) + 4{H^ + }(aq) + 4{e^ - }\buildrel {} \over \longrightarrow 2{H_2}O(l),$
${E^o} = 1.23V$
$2{H^ + }(aq) + 2{e^ - }\buildrel {} \over \longrightarrow {H_2}(g),$
${E^o} = 0.00\,V$
Use, $F = 96500\,C\,mo{l^{ - 1}}$, $R = 8.314\,J\,mo{l^{ - 1}}\,{K^{ - 1}}$.
Correct Answer: 13.32
Explanation:
Vessel is insulated, thus q = 0
For the given reaction : $\eqalign{ & {H_2}(g) + 1/2{O_2}(g)\buildrel {} \over \longrightarrow {H_2}O(l); \cr & {E^o} = 1.23 - 0.00 = 1.23V \cr} $
$\Delta $Go = $-$nFEo
= $-$2 $ \times $ 96500 $ \times $ 1.23 J/mol
Therefore, work derived from this fuel cell using 70% efficiency and on consumption of 1.0 $ \times $ 10$-$3 mol of H2(g)
= 2 $ \times $ 96500 $ \times $ 1.23 $ \times $ 0.7 $ \times $ 1 $ \times $ 10$-$3
= 166.17 J
This work done = change in internal energy (for monoatomic gas,
${C_{V'm}} = 3R/2$),
$166.17 = n{C_V}{,_m}\Delta T$
$ \Rightarrow \Delta T = {{166.17 \times 2} \over {1 \times 3 \times 8.314}}$
$ \Rightarrow 13.32\,K$
For the given reaction : $\eqalign{ & {H_2}(g) + 1/2{O_2}(g)\buildrel {} \over \longrightarrow {H_2}O(l); \cr & {E^o} = 1.23 - 0.00 = 1.23V \cr} $
$\Delta $Go = $-$nFEo
= $-$2 $ \times $ 96500 $ \times $ 1.23 J/mol
Therefore, work derived from this fuel cell using 70% efficiency and on consumption of 1.0 $ \times $ 10$-$3 mol of H2(g)
= 2 $ \times $ 96500 $ \times $ 1.23 $ \times $ 0.7 $ \times $ 1 $ \times $ 10$-$3
= 166.17 J
This work done = change in internal energy (for monoatomic gas,
${C_{V'm}} = 3R/2$),
$166.17 = n{C_V}{,_m}\Delta T$
$ \Rightarrow \Delta T = {{166.17 \times 2} \over {1 \times 3 \times 8.314}}$
$ \Rightarrow 13.32\,K$
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 14th September Evening Shift
A solution of $\mathrm{Fe}^{2+}$ is titrated potentiometrically using $\mathrm{Ce}^{4+}$ solution. When $80 \% \mathrm{Fe}^{2+}$ is titrated, the EMF of the system in $V$ is
(Given, $E^{\circ} \mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}=0.77 \mathrm{~V}$ and $\left.\mathrm{Fe}^{2+}+\mathrm{Ce}^{4+} \longrightarrow \mathrm{Fe}^{3+}+\mathrm{Ce}^{3+}\right) (\log 2=0.3, \log 3=0.5, \log 4=0.6)$
A.
0.806
B.
0.532
C.
0.734
D.
0.756
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 14th September Evening Shift
$\mathrm{Mg}^{2+}$ displaces hydrogen from acids but copper does not. A galvanic cell prepared by combining $\mathrm{Cu} / \mathrm{Cu}^{2+}$ and $\mathrm{Mg} / \mathrm{Mg}^{2+}$ has an EMF of 2.71 V at 298 K . If the potential of copper electrode is 0.34 V , what is the reduction potential of Mg electrode?
A.
+3.05 V
B.
-2.37 V
C.
+2.37 V
D.
2 V
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 10th September Evening Shift
The maximum work that can be obtained from the following cells is
$ X\left|X^{2+}(a q) \| Y^{+}(a q)\right| Y $
Given, $E_{X^{2+} / X}^{\circ}=-1.7 \mathrm{~V}, E_{Y^{2+} / Y}^{\circ}=0.8 \mathrm{~V}$
A.
$579 \mathrm{~kJ} / \mathrm{mol}$
B.
$482.5 \mathrm{~kJ} / \mathrm{mol}$
C.
$289.5 \mathrm{~kJ} / \mathrm{mol}$
D.
$301.8 \mathrm{~kJ} / \mathrm{mol}$
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 10th September Morning Shift
The standard electrode potentials of $\mathrm{Ag}^{+} / \mathrm{Ag}$ is +0.80 V and $\mathrm{Cu}^{+} / \mathrm{Cu}$ is +0.34 V . If these electrodes are connected through a salt-bridge, which of the following statements is correct?
A.
Silver electrode acts as anode and $E_{\text {cell }}^{\circ}$ is -0.34 V .
B.
Copper electrode acts as anode and $E_{\text {cell }}^{\circ}$ is +0.46 V .
C.
Silver electrode acts as a cathode and $E_{\text {cell }}^{\circ}$ is -0.34 V .
D.
Copper electrode acts as cathode and $E_{\text {cell }}^{\circ}$ is +0.46 V .
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Morning Slot
Given
CO3+ + e– $ \to $ CO2+ ; Eo = + 1.81 V
Pb4+ + 2e– $ \to $ Pb2+ ; Eo = + 1.67 V
Ce4+ + e– $ \to $ Ce3+ ; Eo = + 1.61 V
Bi3+ + 3e– $ \to $ Bi ; Eo = + 0.20 V
Oxidizing power of the species will increase in the order :
CO3+ + e– $ \to $ CO2+ ; Eo = + 1.81 V
Pb4+ + 2e– $ \to $ Pb2+ ; Eo = + 1.67 V
Ce4+ + e– $ \to $ Ce3+ ; Eo = + 1.61 V
Bi3+ + 3e– $ \to $ Bi ; Eo = + 0.20 V
Oxidizing power of the species will increase in the order :
A.
Co3+ < Ce4+
< Bi3+ < Pb4+
B.
Co3+ < Pb4+ < Ce4+
< Bi3+
C.
Ce4+
< Pb4+ < Bi3+ < Co3+
D.
Bi3+ < Ce4+
< Pb4+ < Co3+
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Evening Slot
Which one of the following graphs between molar conductivity (${\Lambda _m}$) versus $\sqrt C $ is correct ?
A.
B.
C.
D.
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Morning Slot
Consider the statements S1 and S2
S1 : Conductivity always increases with decrease in the concentration of electrolyte.
S2 : Molar conductivity always increases with decrease in the concentration of electrolyte.
The correct option among the following is :
S1 : Conductivity always increases with decrease in the concentration of electrolyte.
S2 : Molar conductivity always increases with decrease in the concentration of electrolyte.
The correct option among the following is :
A.
Both S1 and S2 are wrong
B.
S1 is correct and S2 is wrong
C.
Both S1 and S2 are correct
D.
S1 is wrong and S2 is correct
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Evening Slot
A solution of Ni(NO3)2 is electrolysed between
platinum electrodes using 0.1 Faraday
electricity. How many mole of Ni will be
deposited at the cathode?
A.
0.10
B.
0.15
C.
0.20
D.
0.05
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Morning Slot
The standard Gibbs energy for the given cell
reaction in kJ mol–1 at 298 K is :
Zn(s) + Cu2+ (aq) $ \to $ Zn2+ (aq) + Cu (s),
E° = 2 V at 298 K
(Faraday's constant, F = 96000 C mol–1)
Zn(s) + Cu2+ (aq) $ \to $ Zn2+ (aq) + Cu (s),
E° = 2 V at 298 K
(Faraday's constant, F = 96000 C mol–1)
A.
384
B.
–192
C.
–384
D.
192
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Evening Slot
Calculate the standard cell potential in (V) of the
cell in which following reaction takes place :
Fe2+(aq) + Ag+(aq) $ \to $ Fe3+(aq) + Ag (s)
Given that
$E_{A{g^ + }/Ag}^o = xV$
$E_{Fe^{2+ }/Fe}^o = yV$
$E_{Fe^{3+ }/Fe}^o = zV$
Fe2+(aq) + Ag+(aq) $ \to $ Fe3+(aq) + Ag (s)
Given that
$E_{A{g^ + }/Ag}^o = xV$
$E_{Fe^{2+ }/Fe}^o = yV$
$E_{Fe^{3+ }/Fe}^o = zV$
A.
x + 2y - 3z
B.
x - z
C.
x - y
D.
x + y - z
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Morning Slot
Given that ${E^\Theta }_{{O_2}/{H_2}O} = 1.23\,V$ ;
${E^\Theta }_{{S_2}O_8^{2 - }/SO_4^{2 - }} = 2.05\,V$
${E^\Theta }_{B{r_2}/B{r^ - }} = 1.09\,V$
${E^\Theta }_{A{u^{3 + }}/Au} = 1.4\,V$
The strongest oxidizing agent is :
${E^\Theta }_{{S_2}O_8^{2 - }/SO_4^{2 - }} = 2.05\,V$
${E^\Theta }_{B{r_2}/B{r^ - }} = 1.09\,V$
${E^\Theta }_{A{u^{3 + }}/Au} = 1.4\,V$
The strongest oxidizing agent is :
A.
O2
B.
Au3+
C.
Br2
D.
${S_2}O_8^{2 - }$
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th January Evening Slot
$ \wedge _m^ \circ $ for NaCl, HCl and NaA are 126.4, 425.9 and 100.5 S cm2 mol–1, respectively. If the conductivity of 0.001 M HA is-
5 $ \times $ 10–5 S cm–1, degree of dissociation of HA is -
5 $ \times $ 10–5 S cm–1, degree of dissociation of HA is -
A.
0.50
B.
0.125
C.
0.25
D.
0.75
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th January Morning Slot
The standard electrode potential ${E^o }$ and its temperature coefficient $\left( {{{d{E^o }} \over {dT}}} \right)$ for a cell are 2V and $-$ 5 $ \times $ 10$-$4 VK$-$1 at 300 K respectively.
The cell reaction is
Zn(s) + Cu2+ (aq) $\buildrel \, \over \longrightarrow $ Zn2+ (aq) + Cu(s)
The standard reaction enthalpy ($\Delta $rH${^o }$) at 300 K in kJ mol–1 is, [Use R = 8 JK–1 mol–1 and F = 96,000C mol–1]
The cell reaction is
Zn(s) + Cu2+ (aq) $\buildrel \, \over \longrightarrow $ Zn2+ (aq) + Cu(s)
The standard reaction enthalpy ($\Delta $rH${^o }$) at 300 K in kJ mol–1 is, [Use R = 8 JK–1 mol–1 and F = 96,000C mol–1]
A.
$-$ 412.8
B.
$-$ 384.0
C.
192.0
D.
206.4
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 11th January Evening Slot
Given the equilibrium constant:
KC of the reaction :
Cu(s) + 2Ag+ (aq) $ \to $ Cu2+ (aq) + 2Ag(s) is
10 $ \times $ 1015, calculate the E$_{cell}^0$ of this reaciton at 298 K
[2.303 ${{RT} \over F}$ at 298 K = 0.059V]
KC of the reaction :
Cu(s) + 2Ag+ (aq) $ \to $ Cu2+ (aq) + 2Ag(s) is
10 $ \times $ 1015, calculate the E$_{cell}^0$ of this reaciton at 298 K
[2.303 ${{RT} \over F}$ at 298 K = 0.059V]
A.
0.4736 mV
B.
0.04736 V
C.
0.4736 V
D.
0.04736 mV
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 11th January Morning Slot
For the cell Zn(s) |Zn2+ (aq)| |Mx+ (aq)| M(s), different half cells and their standard electrode potentials are given below :
If $E_{z{n^{2 + }}/zn}^0$ = $-$ 0.76 V, which cathode will give maximum value of Eocell per electron transferred?
| Mx+ (aq)/M(s) | Au3+(aq)/Au(s) | Ag+(aq)/Ag(s) | Fe3+(aq)/Fe2+ (aq) | Fe2+(aq)/Fe(s) |
|---|---|---|---|---|
| E0Mx+/M/(V) | 1.40 | 0.80 | 0.77 | $-$0.44 |
If $E_{z{n^{2 + }}/zn}^0$ = $-$ 0.76 V, which cathode will give maximum value of Eocell per electron transferred?
A.
Ag+/Ag
B.
Fe3+/Fe2+
C.
Au3+/Au
D.
Fe2+/Fe
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th January Evening Slot
In the cell
Pt$\left| {\left( s \right)} \right|$H2(g, 1 bar)$\left| {HCl\left( {aq} \right)} \right|$AgCl$\left| {\left( s \right)} \right|$Ag(s)|Pt(s)
the cell potential is 0.92 V when a 10–6 molal HCl solution is used. The standard electrode potential of (AgCl/ AgCl– ) electrode is :
$\left\{ {} \right.$Given, ${{2.303RT} \over F} = 0.06V$ at $\left. {298} \right\}$
Pt$\left| {\left( s \right)} \right|$H2(g, 1 bar)$\left| {HCl\left( {aq} \right)} \right|$AgCl$\left| {\left( s \right)} \right|$Ag(s)|Pt(s)
the cell potential is 0.92 V when a 10–6 molal HCl solution is used. The standard electrode potential of (AgCl/ AgCl– ) electrode is :
$\left\{ {} \right.$Given, ${{2.303RT} \over F} = 0.06V$ at $\left. {298} \right\}$
A.
0.94 V
B.
0.40 V
C.
0.76 V
D.
0.20 V
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th January Morning Slot
Consider the following reduction processes :
Zn2+ + 2e– $ \to $ Zn(s) ; Eo = – 0.76 V
Ca2+ + 2e– $ \to $ Ca(s); Eo = –2.87 V
Mg2+ + 2e– $ \to $ Mg(s) ; Eo = – 2.36 V
Ni2 + 2e– $ \to $ Ni(s) ; Eo = – 0.25
The reducing power of the metals increases in the order :
Zn2+ + 2e– $ \to $ Zn(s) ; Eo = – 0.76 V
Ca2+ + 2e– $ \to $ Ca(s); Eo = –2.87 V
Mg2+ + 2e– $ \to $ Mg(s) ; Eo = – 2.36 V
Ni2 + 2e– $ \to $ Ni(s) ; Eo = – 0.25
The reducing power of the metals increases in the order :
A.
Ca < Mg < Zn < Ni
B.
Ni < Zn < Mg < Ca
C.
Zn < Mg < Ni < Ca
D.
Ca < Zn < Mg < Ni
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th January Evening Slot
If the standard electrode potential for a cell is 2 V at 300 K, the equilibrium constant (K) for the reaction
Zn(s) + Cu2+ (aq) $\rightleftharpoons$ Zn2+(aq) + Cu(s)
at 300 K is approximately,
(R = 8 JK$-$1mol$-$1, F = 96000 C mol$-$1)
Zn(s) + Cu2+ (aq) $\rightleftharpoons$ Zn2+(aq) + Cu(s)
at 300 K is approximately,
(R = 8 JK$-$1mol$-$1, F = 96000 C mol$-$1)
A.
e$-$80
B.
e$-$160
C.
e320
D.
e160
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th January Morning Slot
The anodic half-cell of lead-acid battery is recharged using electricity of 0.05 Faraday. The amount of PbSO4 electrolyzed in g during the process is : (Molar mass of PbSO4 = 303 g mol$-$1)
A.
22.8
B.
15.2
C.
7.6
D.
11.4
2019
JEE Advanced
MCQ
JEE Advanced 2019 Paper 1 Offline
Molar conductivity ($\Lambda $m) of aqueous solution of sodium stearate, which behaves as a strong electrolyte, is recorded at varying concentrations (C) of sodium stearate. Which one of the following plots provides the correct representation of micelle formation in the solution?
(critical micelle concentration (CMC) is marked with an arrow in the figures)
(critical micelle concentration (CMC) is marked with an arrow in the figures)
A.
B.
C.
D.
2018
JEE Mains
MCQ
JEE Main 2018 (Online) 16th April Morning Slot
When 9.65 ampere current was passed for 1.0 hour into nitrobenzene in acidic medium, the amount of p-aminophenol produced is :
A.
9.81 g
B.
10.9 g
C.
98.1 g
D.
109.0 g
2018
JEE Mains
MCQ
JEE Main 2018 (Offline)
How long (approximate) should water be electrolysed by passing through 100 amperes current so that the
oxygen released can completely burn 27.66 g of diborane?
(Atomic weight of B = 10.8 u)
(Atomic weight of B = 10.8 u)
A.
1.6 hours
B.
6.4 hours
C.
0.8 hours
D.
3.2 hours