datavorous/entrance-exam-dataset huggingface dataset:
The standard electrode potentials, E0 of Fe3+/Fe2+ and Fe2+/Fe at 300 K are +0.77 V and -0.44 V, respectively. The E0 of Fe 3 + /Fe at the same temperature is

Options:
A 1.21 V
B 0.33 V
C -0.036 V
D 0.036

The correct answer is:
-0.036 V

Solution:
Fe 3+ â+â e â â Fe 2+ ;âââ ÎG 1 o â=ââ1âÃâFâÃâ0.77
Fe 2+ â+â2â e â âFe;âââ ÎG 2 o â=â2âÃâFâÃâ0.44
Fe 3+ â+â3â e â âFe;âââ ÎG 3 o =â3ÃFÃ E o
=0.11âV
∴  E o = −0.11 3 =−0.036 V

Topic: Chemistry, Electrochemistry, JEE Main
The standard electrode potentials E M + / M o of four metals A, B, C and D are −1.2 V, 0.6 V , 0.85 V and −0.76 V , respectively. The sequence of deposition of metals on applying potential is

Options:
A B > D > C > A
B A > C > B > D
C C > B > D > A
D D > A > B > C

The correct answer is:
C > B > D > A

Solution:
In complete cation electrolysis at cathode, the species having higher reduction potential is preferably reduced at cathode
E C ∘ > E B ∘ > E D ∘ > E A ∘
Hence, C > B > D > A
will be the preference of deposition of these metals on competing.

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2014 (09 Apr Online)
The standard electrode potentials E M + / M o of four metals A, B, C and D are −1.2 V, , 0.6 V , 0.85 V and −0.76 V , respectively. The sequence of deposition of metals on applying potential is

Options:
A B > D > C > A
B A > C > B > D
C C > B > D > A
D D > A > B > C

The correct answer is:
C > B > D > A

Solution:
In complete cation electrolysis at cathode, the species having higher reduction potential is preferably reduced at cathode
E C â > E B â > E D â > E A â
Hence, C > B > D > A
will be the preference of deposition of these metals on competeting.

Topic: Chemistry, Electrochemistry, JEE Main
The standard electrode potentials of $Z n^{2+} / Z n$ and $A g^{+} / A g$ are $-0.763 \mathrm{~V}$ and $+0.799 \mathrm{~V}$ respectively. The standard potential of the cell is

Options:
A $1.56 \mathrm{~V}$
B $0.036 \mathrm{~V}$
C $-1.562 V$
D $0.799 \mathrm{~V}$

The correct answer is:
$1.56 \mathrm{~V}$

Solution:
$E_{\text {cell }}^o=E_{\text {cathode }}^o-E_{\text {thode }}^o=0.799-(-0.763)=1.562 \mathrm{~V}$

Topic: Chemistry, Electrochemistry, JEE Main
The standard electrode potentials of

Zn2+ | Zn , Cu2+ | Cu and Ag+ | Ag are respectively - 0.76, 0.34 and 0.8 V. The following cells were constructed

I - Zn | Zn2+ || Cu2+ | Cu

II - Zn | Zn2+ || Ag+ | Ag

III - Cu | Cu2+ || Ag+ | Ag

What is the correct order of EcellÂ° of these cells?

Options:
A II > III > I
B II > I > III
C I > II > III
D III > I > II

The correct answer is:
II > I > III

Solution:
Given,
Zn 2+ â+âââ 2e â â Zn , E Â° =â0.76 V
Cu 2+ +ââ 2e â â Cu , E Â° = 0.34 V
Ag + +ââ e â â Ag , E Â° = 0.8 V
Cell reaction of (I) is
Zn + Cu2+ â Zn2+ + Cu
Ecello=Eredo(cathode)-Eredo(anode)
=0.34â â0.76 =1.10âV
Cell reaction of (II) is
Zn + 2Ag + â Zn 2+ + 2Ag
E Cell o =+0.80â â0.76 =+1.56âV
Cell reaction of (III) is
Cu + 2Ag + â Cu 2+ + 2Ag
E Cell o =+0.80â 0.34 =+0.46âV
So, the correct order of EcellÂ° of these cell is
II > I > III.

Topic: Chemistry, Electrochemistry, JEE Main
The standard e.m.f of a cell, involving one electron change is found to be $0.591 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. The equilibrium constant of the reaction is $\left(\mathrm{F}=96,500 \mathrm{C} \mathrm{mol}^{-1}: \mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)$

Options:
A
$1.0 \times 10^1$

B
$1.0 \times 10^{30}$

C
$1.0 \times 10^{10}$

D
$1.0 \times 10^5$

The correct answer is:
$1.0 \times 10^{10}$

Solution:
$\therefore \log \mathrm{K}_{\mathrm{eq}}=\frac{\mathrm{nE}^{\circ}}{0.0591}
=\frac{1 \times 0.591}{0.0591}$ $\Rightarrow \mathrm{K}_{\mathrm{eq}}=10^{10}$

Topic: Chemistry, Electrochemistry, JEE Main
The standard e.m.f of a cell, involving one electron change is found to be $0.591 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. The equilibrium constant of the reaction is $\left(\mathrm{F}=96,500 \mathrm{C} \mathrm{mol}^{-1}: \mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)$

Options:
A
$1.0 \times 10^1$

B
$1.0 \times 10^{30}$

C
$1.0 \times 10^{10}$

D
$1.0 \times 10^5$

The correct answer is:
$1.0 \times 10^{10}$

Solution:
$\therefore \log \mathrm{K}_{\mathrm{eq}}=\frac{\mathrm{nE}^{\circ}}{0.0591}
=\frac{1 \times 0.591}{0.0591}$ $\Rightarrow \mathrm{K}_{\mathrm{eq}}=10^{10}$

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2004
The standard EMF. of a cell, involving one electron change is found to be 0.591 V at 25â. The equilibrium constant of the reaction is 10 x . Find the value of x .
F = 96500 C equivalent-1 , R = 8.314 J K -1 mol -1

Options:

The correct answer is:
10

Solution:
ÎG=ÎGÂ°â0.591nlogQ
at equilibrium
ÎG=0,Ecell=0,âQ=KC,ÎGÂ°=â0.5911logQ
E cell â=â E cell o â 0.0591 n log K c
0 = 0.591â 0.0591 1 log K c
⇒ −0.591 = −0.0591 log K c
â log Kc = 0.5910.0591 = 10
â´ K c = antilog 10 = 1 Ã 10 10 .

Topic: Chemistry, Electrochemistry, JEE Main
The standard enthalpies of formation of Al2O3 and CaO are -1675 kJ mol-1 and -635 kJ mol-1 respectively.

For the reaction

3CaO+2Al→3Ca+Al2O3 the standard reaction enthalpy ΔrH0= ________ kJ.

(Round off to the Nearest Integer).

Options:

The correct answer is:
230

Solution:
Given reaction:

3CaO+Al→Al2O3+3Ca

Now, ΔrH°=ΣΔfHProducts º-ΣΔfHReactantsº

=[1×(-1675)+3×0]-[3×(-635)+2×0]

=+230 kJ mol-1

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2021 (17 Mar Shift 1)
The standard enthalpy of formation $\left(\Delta_f \mathrm{H}^{\circ}{ }_{298}\right)$ for methane, $\mathrm{CH}_4$ is $-74.9 \mathrm{~kJ} \mathrm{~mol}^{-1}$. In order to calculate the average energy given out in the formation of a $\mathrm{C}-\mathrm{H}$ bond from this it is necessary to know which one of the following?

Options:
A
The dissociation energy of the hydrogen molecule, $\mathrm{H}_2$.

B
The first four ionisation energies of carbon.

C
The dissociation energy of $\mathrm{H}_2$ and enthalpy and sublimation of carbon (graphite).

D
The first four ionisation energies of carbon and electron affinity of hydrogen.

The correct answer is:
The dissociation energy of the hydrogen molecule, $\mathrm{H}_2$.

Solution:
To calculate average enthalpy of $\mathrm{C}-\mathrm{H}$ bond in methane following informations are needed
(i) dissociation energy of $\mathrm{H}_2$ i.e.
$$
\frac{1}{2} \mathrm{H}_2(g) \longrightarrow \mathrm{H}(\mathrm{g}) ; \Delta \mathrm{H}=x(\text { suppose } \quad)
$$
(ii) Sublimation energy of $\mathrm{C}$ (graphite) to $\mathrm{C}(g)$ $\mathrm{C}($ graphite $) \longrightarrow \mathrm{C}(g) ; \Delta \mathrm{H}=y$ (Suppose) Given
$\mathrm{C}($ graphite $)+2 \mathrm{H}_2(\mathrm{~g}) \longrightarrow \mathrm{CH}_4(\mathrm{~g}) ; \Delta \mathrm{H}=75 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2014 (12 Apr Online)
The standard enthalpy of formation $\left(\Delta_{\mathrm{f}} \mathrm{H}^{\circ}\right)$ at $298 \mathrm{~K}$ for methane, $\mathrm{CH}_4(\mathrm{~g})$, is $-74.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$. The additional information required to determine the average energy for $\mathrm{C}-\mathrm{H}$ bond formation would be

Options:
A
the dissociation energy of $\mathrm{H}_2$ and enthalpy of sublimation of carbon

B
latent heat of vapourization of methane

C
the first four ionization energies of carbon and electron gain enthalpy of hydrogen

D
the dissociation energy of hydrogen molecule, $\mathrm{H}_2$

The correct answer is:
the dissociation energy of $\mathrm{H}_2$ and enthalpy of sublimation of carbon

Solution:
No Solution Available

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2006
The standard enthalpy of formation of gaseous H2O at 298âK is â241.82âkJ mol-1. Calculate ÎH f o at 373âK, given the following values of the molar heat capacities at constant pressure.
Molar heat capacity of H2g=33.58 JK-1mol-1
Molar heat capacity of H2g=28.84 JK-1mol-1
Molar heat capacity of O2g=29.37 JK-1mol-1
Assume that the heat capacities are independent of temperature.

Options:
A -242.6 kJ mol-1
B -485.2Â kJÂ mol-1
C -121.3Â kJÂ mol-1
D -286.4Â kJÂ mol-1

The correct answer is:
-242.6 kJ mol-1

Solution:
The reaction is H2g+12O2gâH2Og
ÎCpo=Cp,moH2O,g-Cp,moH2,g+12Cp,moO2
=33.58-28.84+1229.37
=-9.94 JK-1mol-1
Using Kirchhoff's equation,
ÎHÂ°373 K=ÎHÂ°298 K+T2-T1ÎCpo
=-241.82+373-298Ã-9.94
=-242.6 kJ mol-1

Topic: Chemistry, Thermodynamics (C), JEE Main
The standard enthalpy of formation of NH3 is - 46.0 kJ/mol. If bond enthalpy of H2 is -436 kJ/mol and that of N2 is -712 kJ/mol , the average bond enthalpy of N-H bond in NH3 is :

Options:
A - 964 kJ/mol
B - 1102 kJ/mol
C + 1056 kJ/mol
D + 352 kJ/mol

The correct answer is:
+ 352 kJ/mol

Solution:
For The Reaction
1 2 N 2 g + 3 2 H 2 g â NH 3 g ; Î H f â = - 4 6 KJ mol - 1
Î H f â = - 4 6 = â B.E - â B.E
Reactants Product
â B.E = â B.E + 4 6
Product Reactants
= 1 2 7 1 2 + 3 2 4 3 6 + 4 6
= 356 + 654 + 46 = 1056
Averge B.E for N-H Bond = 1 0 5 6 3 = 3 5 2 KJ mol - 1

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2014 (09 Apr Online)
The standard enthalpy of formation of $\mathrm{NH}_3$ is $-46.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$. If the enthalpy of formation of $\mathrm{H}_2$ from its atoms is $-436 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and that of $\mathrm{N}_2$ is $-712 \mathrm{~kJ} \mathrm{~mol}^{-1}$, the average bond enthalpy of $\mathrm{N}-\mathrm{H}$ bond in $\mathrm{NH}_3$ is

Options:
A
$-964 \mathrm{~kJ} \mathrm{~mol}^{-1}$

B
$+352 \mathrm{~kJ} \mathrm{~mol}^{-1}$

C
$+1056 \mathrm{~kJ} \mathrm{~mol}^{-1}$

D
$-1102 \mathrm{~kJ} \mathrm{~mol}^{-1}$

The correct answer is:
$+352 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution:
Enthalpy of formation of $\mathrm{NH}_3=-46 \mathrm{~kJ} / \mathrm{mole}$
$$
\therefore \mathrm{N}_2+3 \mathrm{H}_2 \rightarrow 2 \mathrm{NH}_3 \quad \Delta \mathrm{H}_{\mathrm{f}}=-2 \times 46 \mathrm{~kJ} \mathrm{~mol}
$$
Bond breaking is endothermic and Bond formation is exothermic
Assuming ' $\mathrm{x}$ ' is the bond energy of $\mathrm{N}-\mathrm{H}$ bond $\left(\mathrm{kJ} \mathrm{mol}^{-1}\right)$
$$
\begin{aligned}
& \therefore 712+(3 \times 436)-6 \mathrm{x}=-46 \times 2 \\
& \therefore \mathrm{x}=352 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}
$$

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2010
The standard enthalpy of formation of NH3 is - 46.0 kJ/mol. If bond enthalpy of H2 is - 436 kJ/mol and that of N2 is - 712 kJ/mol , the average bond enthalpy of N - H bond in NH3 is :

Options:
A - 964 kJ/mol
B - 1102 kJ/mol
C + 1056 kJ/mol
D + 352 kJ/mol

The correct answer is:
+ 352 kJ/mol

Solution:
For The Reaction
1 2 N 2 g + 3 2 H 2 g â NH 3 g ; Î H f â = - 4 6 KJ mol - 1
Î H f â = - 4 6 = â B.E - â B.E
Reactants Product
â B.E = â B.E + 4 6
Product Reactants
= 1 2 7 1 2 + 3 2 4 3 6 + 4 6
= 356 + 654 + 46 = 1056
Averge B.E for N-H Bond = 1 0 5 6 3 = 3 5 2 KJ mol - 1

Topic: Chemistry, Thermodynamics (C), JEE Main
The standard enthalpy of neutralization of strong acid and strong base is -57.3 kJ equiv −1 . If the enthalpy of neutralization of the first proton of aqueous H2S is -33.7 kJmol-1 then the pK a 1 of H2S is

Options:
A 23.6×103-TΔS°2.303RT
B 2.303RT23.6-TΔS°
C TΔS°-23.6RT
D 2.303TΔS°-23.6RT

The correct answer is:
23.6×103-TΔS°2.303RT

Solution:
H2S⇌H++HS-
ΔHion=23.6 kJ
∴ΔG°=23.6×103-TΔS°=-2.303RTlogKa1
∴PKa1=23.6×103-TΔS°2.303RT

Topic: Chemistry, Thermodynamics (C), JEE Main
The standard entropy change for the reaction

4Fes+3O2g→2Fe2O3s is -550 J K-1 at 298 K

[Given : The standard enthalpy change for the reaction is -165 kJ mol-1]. The temperature in K at which the reaction attains equilibrium is (Nearest Integer)

Options:

The correct answer is:
300

Solution:
4Fes+3O2g→3Fe2O3s ΔSr=-550 J/K

ΔSr=-165 kJ/mole

At equilibrium ΔG=0

T=ΔHΔS at equilibrium

=-165×103-550=300 K

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2022 (25 Jun Shift 1)
The standard Gibbs energy change at 300 K for the reaction 2A⇌B+C is 2494.2 J. At a given time, the composition of the reaction mixture is A=12, B=2 and C=12 . The reaction proceeds in the: [R=8.314 J/K-mol , e=2.718] {Given antilog (-0.44)=0.36}

Options:
A Reverse direction because Q<KC
B Forward direction because Q>KC
C Reverse direction because Q>KC
D Forward direction because Q<KC

The correct answer is:
Reverse direction because Q>KC

Solution:
Δ G ∘ = - RT ln K C
2494.2 = - 8.314 × 300 ln K C
2494.2 = - 8.314 × 300 × 2.303 log K C
- 2494.2 2.303 × 300 × 8.314 = - 0.44 = log K C
log KC=-0.44
K C = 0.36
Q=[B][C][A]2=2×12122=4
Q>KC, reaction will shift in reverse direction.

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2015 (04 Apr)
The standard Gibbs energy for the given cell reaction in kJ mol-1 at 298 K is:
Zns+Cu2+aq⟶Zn2+aq+Cu(s) ,
E0=2 V at 298 K
(Faraday's constant ,F=96000 C mol-1)

Options:
A -192
B 192
C 384
D -384

The correct answer is:
-384

Solution:
For the reaction,
Cu2+aq+Zns⟶Zn2+aq+Cus
n=2, two electrons are exchanged
∆Go=-nFEo
=-2×96000×2=-384000
=-384 kJ

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2019 (09 Apr Shift 1)
The standard Gibbs free energy change $\Delta G^{\prime}$ is related to equilibrium constant $K_p$ as

Options:
A $K_p=-R T \ln \Delta G^{\circ}$
B $K_p=\left(\frac{e}{R T}\right)^{\Delta G^e}$
C $K_p=-\frac{\Delta G^o}{R T}$
D $K_p=e^{-\frac{\Delta G^T}{R T}}$

The correct answer is:
$K_p=e^{-\frac{\Delta G^T}{R T}}$

Solution:
$K_p=e^{-\frac{\Delta C^T}{R T}}$
For equilibrium condition $\Delta \mathrm{G}=0$
So,
$0=\Delta G^{\mathrm{a}}+R T \ln K_p$
$\ln K_{\mathrm{p}}=-\Delta G^{\circ} / R T$
Now
$K_p=e^{-\frac{\Delta G^T}{R T}}$

Topic: Chemistry, Thermodynamics (C), JEE Main
The standard heat of formation ∆fH2980 of ethane (in kJ/mol ), if the heat of combustion of ethane, hydrogen and graphite are -1560,-393.5 and -286 kJ/mol , respectively is ________

Options:

The correct answer is:
-192

Solution:
2C( graphite )+3 H 2 (g)â C 2 H 6 (g)
Î f H C 2 H 6 =2Î H comb C gaphite +3Î H comb H 2
âÎ H comb C 2 H 6
=â(286Ã2)â(393.5Ã3)â(â1560)
=â572â1180.5+1560=â192.5 kJ/mole

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2020 (07 Jan Shift 2)
The standard molar heat of formation of ethane, CO2 and water are respectively -21.1,-94.1 and − 68.3 kcal. The standard molar heat of combustion of ethane will be:

Options:
A -372 kcal/mol
B 162 kcal/mol
C -240 kcal/mol
D 183.5 kcal/mol

The correct answer is:
-372 kcal/mol

Solution:
Given, 2C +  3H 2 → C 2 H 6 ;  ΔH = −21.1
Câ+â O 2 â CO 2 ; ÎH=â94.1
H 2 + 1 2 O 2 â H 2 O; âÎH=â68.3
Eqs. 2Ã(ii)+3Ã(iii)â(i)
C2H6+72O2â2CO2+3H2O
Îx=2 â94.1 +3 â68.3 â â21.1
=â372 kcal/mol

Topic: Chemistry, Thermodynamics (C), JEE Main
The standard oxidation potential of \( \mathrm{Ni} / \mathrm{Ni}^{2+} \) electrode is \( 0.236 \mathrm{~V} \). If this is combined with a hydrogen electrode in acid solution, at what \( \mathrm{pH} \) of the solution will the measured emf be zero at \( 25{ }^{\circ} \mathrm{C} ? \) Assume \( \left[\mathrm{Ni}^{2+}\right]=1 \mathrm{M} \) and \( \mathrm{p}_{\mathrm{H}_{2}}=1 \) atm

Options:
A \( \mathrm{pH}=4 \)
B \( \mathrm{pH}=2 \)
C \( \mathrm{pH}=3 \)
D \( \mathrm{pH}=1 \)

The correct answer is:
\( \mathrm{pH}=4 \)

Solution:
E Ni 2 + / Ni s ∘ = - 0.236 V
The cell is represented as:
Ni s Ni aq 2 + 1 M H aq + , pH = ?? Pt s , H 2 g 1 atm
Ni s → Ni 2 + + 2 e - 2 H + + 2 e - → H 2 g 1 atm
--------------------------------------------------------------------
Ni s + 2 H aq + → n = 2 Ni aq 2 + + H 2 g (1 atm)
E cell = 0 = E cell ∘ - 0.059 2 log Ni 2 + H + 2
0 = 0 - - 0.236 - 0.059 2 log 1 H + 2
0.236 × 2 0.059 = log 1 H + 2
= 0.472 0.059 = 8 = log 1 H + 2
1 H + 2 = 1 0 8 ⇒ 1 H + = 1 0 + 4
H + = 1 0 - 4 ∴ pH = 4

Topic: Chemistry, Electrochemistry, JEE Main
The standard oxidation potential of zinc and silver in water at $298 \mathrm{~K}$ are
$\mathrm{Zn}(\mathrm{s}) \rightarrow \mathrm{Zn}^{2+}+2 \mathrm{e}^{-} ; E=0.76 \mathrm{~V}$
$\mathrm{Ag}(\mathrm{s}) \rightarrow \mathrm{Ag}^{2+}+2 \mathrm{e}^{-} ; E=-0.80 \mathrm{~V}$
Which of the following reactions actually take place

Options:
A $\mathrm{Zn}(\mathrm{s})+2 \mathrm{Ag}^{+}(a q) \rightarrow Z \mathrm{n}^{++}(a q)+2 \mathrm{Ag}(\mathrm{s})$
B $Z n^{++}(a q)+2 A g(s) \rightarrow 2 A g^{+}(a q)+Z n(s)$
C $\mathrm{Zn}(\mathrm{s})+\mathrm{Ag}(\mathrm{s}) \rightarrow \mathrm{Zn}^{++}(a q)+\mathrm{Ag}^{+}(a q)$
D $Z n^{++}(a q)+A g^{+}(a q) \rightarrow Z n(s)+A g(s)$

The correct answer is:
$\mathrm{Zn}(\mathrm{s})+2 \mathrm{Ag}^{+}(a q) \rightarrow Z \mathrm{n}^{++}(a q)+2 \mathrm{Ag}(\mathrm{s})$

Solution:
In this reaction zinc act as a anode and Ag act as a cathode.

Topic: Chemistry, Electrochemistry, JEE Main
The standard potential of the cell formed by combining the Cl2/Cl– (aq) half-cell with the standard hydrogen electrode is +1.36V and ∂ E o ∂ T P = - 1.2 × 1 0 - 3 VK -1 . What is the value of Δ S reaction o for reaction H 2 g + Cl 2 g → 2 H aq + + 2 Cl aq -

Options:
A - 1 × 1 0 - 2 JK -1
B 2.3 × 1 0 - 2 JK -1
C - 2.3 × 1 0 - 2 JK -1
D 1 JK -1

The correct answer is:
- 2.3 × 1 0 - 2 JK -1

Solution:
Δ S reaction o = nF ∂ E o ∂ T P
= 2 × 9 6 4 8 0 C mol -1 × - 1.2 × 1 0 - 3 VK -1
= - 2.3 × 1 0 - 2 JK -1

Topic: Chemistry, Electrochemistry, JEE Main
The standard potentials of $\mathrm{Ag}^{+} / \mathrm{Ag}, \mathrm{Hg}_2{ }^{2+} / 2 \mathrm{Hg}$, $\mathrm{Cu}^{2+} / \mathrm{Cu}$ and $\mathrm{Mg}^{2+} / \mathrm{Mg}$ electrodes are $0.80,0.79$, $0.34$ and $-2.37 \mathrm{~V}$, respectively. An aqueous solution which contains one mole per litre of the salts of each of the four metals is electrolyzed. With increasing voltage, the correct sequence of deposition of the metals at the cathode is

Options:
A
$\mathrm{Ag}, \mathrm{Hg}, \mathrm{Cu}, \mathrm{Mg}$

B
$\mathrm{Cu}, \mathrm{Hg}$, Ag only

C
Ag, Hg, Cu only

D
$\mathrm{Mg}, \mathrm{Cu}, \mathrm{Hg}, \mathrm{Ag}$

The correct answer is:
Ag, Hg, Cu only

Solution:
More the reduction potential, more is the deposition of metals at cathode. Cation having $\mathrm{E}^0$ value less than $-0.83 \mathrm{~V}$ (reduction potential of $\mathrm{H}_2 \mathrm{O}$ ) will not deposit from aqueous solution. Hence correct order of A deposition of the metal at the cathode is

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2012 (19 May Online)
The standard reaction Gibbs energy for a chemical reaction at an absolute temperature $\mathrm{T}$ is given by $\Delta \mathrm{G}^{\circ}=\mathrm{A}-\mathrm{BT}$ where A and $B$ are non-zero constants. Which of the following is true about this reaction?

Options:
A Exothermic if $B < 0$
B Endothermic if $\mathrm{A}>0$
C Endothermic if $\mathrm{A} < 0$ and $\mathrm{B}>0$
D Exothermic if $\mathrm{A}>0$ and $\mathrm{B} < 0$

The correct answer is:
Endothermic if $\mathrm{A}>0$

Solution:
$\Delta G^{\circ}=A-B T$

A and B are non-zero constants

$\therefore \Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}=A-B T$

$\therefore$ Reaction will be endothermic if $\Delta \mathrm{G}^{\circ}>0 .$

Hence, $\mathrm{A}>0$ and $\mathrm{B} < 0$

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2019 (11 Jan Shift 2)
The standard reduction potential for Cu 2+ /Cu is + 0.34 V. Calculate the reduction potential at pH=14 for the above couple. K sp of Cu OH 2 =1Ã 10 â19

Options:
A -0.22 V
B +0.22Â V
C -0.44Â V
D +0.44Â V

The correct answer is:
-0.22 V

Solution:
When âpH=14, [ H + ]= 10 â14 âandâ[ O H â ]=1 M
K sp =[ C u 2+ ] [ O H â ] 2 = 10 â19
â´[ C u 2+ ]= 10 â19 [ O H â ] 2 = 10 â19
The half cell reaction
C u 2+ +2 e â âCu
E= E O â 0.059 2 log 1 [ C u 2+ ]
=0.34â 0.059 2 log 1 10 â19 =â0.22 V

Topic: Chemistry, Electrochemistry, JEE Main
The standard reduction potentials at 295 K for the following half cells are given below:

NO3-+4H++3e-→NOg+2H2O	Eo=0.97 V

V2+aq+2e-→Vs	Eo=-1.19 V

Fe3+aq+3e-→Fes	Eo=-0.04 V

Ag+aq+e-→Ags	Eo=0.80 V

Au3+aq+3e-→Aus	Eo=1.40 V

The number of metal(s) which will be oxidised by NO3- in aqueous solution is _____.

Options:

The correct answer is:
3

Solution:
For feasibility, check , Eocell=Ecathode(reduction)o-Eanode(oxidation)o>0

The Eo values when metal acts as anode and NO3- reaction is cathodic reaction.

For Vanadium metal, Ecello=0.97+1.19=2.16V

For Iron metal, Ecello=0.97+0.04=1.01V

For silver metal, Ecello=0.97-0.80=0.17V

For Gold metal, Ecello=0.97-1.140=-0.17V

For electrodes having oxidation potential greater than -0.97 V, Eocell>0

∴ Ag, Fe and V can be oxidised

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2023 (06 Apr Shift 2)
The standard reduction potentials at 298 K for single electrodes are given below.

Electrode	electrode potential (volt)

Mg 2+ /Mg	−2.34

Zn 2+ /Zn	−0.76

Fe 2+ /Fe	−0.44

From this we can tell that

Options:
A Zn can reduce both Mg2+ and Fe2+
B Fe can reduce both Mg2+ and Zn2+
C Mg can reduce both Zn2+ and Fe2+
D Mg can reduce Zn2+ but not Fe2+

The correct answer is:
Mg can reduce both Zn2+ and Fe2+

Solution:
Metals with lower SRP can displace the metal with higher SRP from their solutions. Mg with lower SRP (-2.34) can reduce both Zn2+ (-0.76) and Fe2+ (-0.44).

Topic: Chemistry, Electrochemistry, JEE Main
The standard reduction potentials at $298 \mathrm{~K}$ for the following half reactions are given against each
$\begin{aligned}
& Z n^{2+}(a q .)+2 e \rightleftharpoons Z n(s) ; \quad-0.762 \\
& \mathrm{Cr}^{3+}(\mathrm{aq})+3 e \rightleftharpoons \mathrm{Cr}(\mathrm{s}) ; \quad-0.740 \\
& 2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e} \rightleftharpoons \mathrm{H}_2(g) ; \quad 0.00 \\
& \mathrm{Fe}^{3+}(a q)+e \rightleftharpoons \mathrm{Fe}^{2+}(a q) ; \quad 0.770 \\
&
\end{aligned}$
Which is the strongest reducing agent

Options:
A $\mathrm{Zn}(\mathrm{s})$
B $\mathrm{Cr}(\mathrm{s})$
C $H_2(g)$
D $\mathrm{Fe}^{2+}(a q)$

The correct answer is:
$\mathrm{Zn}(\mathrm{s})$

Solution:
More negative is the reduction potential, higher will be the reducing property, i.e. the power to give up electrons.

Topic: Chemistry, Electrochemistry, JEE Main
The standard reduction potentials at \(298 \mathrm{~K}\) for the following half cells are given below :
\(\begin{array}{ll}
\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}, \mathrm{E}^{\circ}=1.33 \mathrm{~V} \\
\mathrm{Fe}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} \qquad \mathrm{E}^{\circ}=-0.04 \mathrm{~V} \\
\mathrm{Ni}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Ni} \qquad \mathrm{E}^{\circ}=-0.25 \mathrm{~V} \\
\mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Ag} \quad\qquad \mathrm{E}^{\circ}=0.80 \mathrm{~V} \\
\mathrm{Au}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Au} \qquad \mathrm{E}^{\circ}=1.40 \mathrm{~V}
\end{array}\)
Consider the given electrochemical reactions,
The number of metal(s) which will be oxidized be \(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\), in aqueous solution is ______

Options:

The correct answer is:
3

Solution:
$\mathrm{Fe}, \mathrm{Ni}, \mathrm{Ag}$ will be oxidized due to lower S.R.P.

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2024 (09 Apr Shift 1)