datavorous/entrance-exam-dataset huggingface dataset:
The standard enthalpy of atomization of ethane according to the equation $\mathrm{C}_2 \mathrm{H}_6(\mathrm{~g}) \rightarrow 2 \mathrm{C}(\mathrm{g})+6 \mathrm{H}(\mathrm{g})$ is $622 \mathrm{~kJ} \mathrm{~mol}^{-1}$. If standard mean $\mathrm{C}-\mathrm{H}$ bond dissociation enthalpy is $90 \mathrm{~kJ}$ $\mathrm{mol}^{-1}$, the standard mean dissociation enthalpy of $\mathrm{C}-\mathrm{C}$ bond (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) is

Options:
A 540
B 90
C 85
D 82

The correct answer is:
82

Solution:
$\begin{aligned} & \Delta_{\mathrm{a}} \mathrm{H}^{\circ}=\Delta_{\mathrm{C}-\mathrm{H}} \mathrm{H}^{\circ}+\Delta_{\mathrm{C}-\mathrm{C}} \mathrm{H}^{\circ} \\ & 622=[6 \times 90]+\left[\Delta_{\mathrm{C}-\mathrm{C}} \mathrm{H}^{\circ}\right]=540+\Delta_{\mathrm{C}-\mathrm{C}} \dot{\mathrm{H}}^{\circ} \\ & \Rightarrow \Delta_{\mathrm{C}-\mathrm{C}} \mathrm{H}^{\circ}=622-540=82 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}$

Topic: Chemistry, Thermodynamics (C), AP EAMCET, AP EAMCET 2023 (19 May Shift 1)
The standard enthalpy of formation for carbon is zero for-

Options:
A C(Diamond)
B C(Graphite)
C C(g)
D All of these

The correct answer is:
C(Graphite)

Solution:
Correct Option is : 2
C(Graphite)

Topic: Chemistry, Thermodynamics (C), NEET
The standard enthalpy of formation is zero for-

Options:
A I2(s)
B H2O(l)
C C(diamond)
D SO2(g)

The correct answer is:
I2(s)

Solution:
Correct Option is : 1
I2(s)

Topic: Chemistry, Thermodynamics (C), NEET
The standard enthalpy of formation is zero for-

Options:
A H2(l)
B CH4(g)
C H2(g)
D HBr(g)

The correct answer is:
H2(g)

Solution:
Correct Option is : 3
H2(g)

Topic: Chemistry, Thermodynamics (C), NEET
The standard enthalpy of formation of $\mathrm{CO}_2(\mathrm{~g})$, $\mathrm{CaO}(\mathrm{s})$ and $\mathrm{CaCO}_3(\mathrm{~s})$ are $-393,-634,-1210 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. If all the substances are in standard state, the standard enthalpy of decomposition of calcium carbonate to $\mathrm{CaO}(\mathrm{s})$ and $\mathrm{CO}_2$ (g) (in $\mathrm{kJ} \mathrm{mol}^{-1}$ ) is

Options:
A $969$
B $183$
C $-969$
D $-183$

The correct answer is:
$183$

Solution:
$\mathrm{CaCO}_3(\mathrm{~s}) \stackrel{\Delta}{\longrightarrow} \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})$
$\Delta \mathrm{H}_{\mathrm{r}}^{\circ}=\left[\Delta_{\mathrm{f}} \mathrm{H}^{\circ}(\mathrm{CaO})+\Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{CO}_2\right)\right]-\left[\Delta_{\mathrm{f}} \mathrm{H}^{\circ}\left(\mathrm{CaCO}_3\right)\right]$
$\begin{aligned} & =[(-634)+(-393)]-[-1210] \\ & =+183 \mathrm{~kJ} \mathrm{~mol}^{-1}\end{aligned}$

Topic: Chemistry, Thermodynamics (C), AP EAMCET, AP EAMCET 2023 (18 May Shift 1)
The standard enthalpy of formation of $\mathrm{CO}(g)$, $\mathrm{CO}_2(g), \mathrm{N}_2 \mathrm{O}(g)$ and $\mathrm{N}_2 \mathrm{O}_4(g)$ are respectively -10 , $-393,81$ and $-10 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Enthalpy change (in $\mathrm{kJ}$ ) of the following reaction is
$$
\mathrm{N}_2 \mathrm{O}_4(\mathrm{~g})+3 \mathrm{CO}(\mathrm{g}) \longrightarrow \mathrm{N}_2 \mathrm{O}(\mathrm{g})+3 \mathrm{CO}_2(\mathrm{~g})
$$

Options:
A -1058
B +1058
C -957
D +957

The correct answer is:
-1058

Solution:
Given, standard enthalpy of formation of $\mathrm{CO}=-10 \mathrm{~kJ} / \mathrm{mol}$
Standard enthalpy of formation of $\mathrm{CO}_2$
$$
=-393 \mathrm{~kJ} / \mathrm{mol}
$$
Standard enthalpy of formation of $\mathrm{N}_2 \mathrm{O}$
$$
=81 \mathrm{~kJ} / \mathrm{mol}
$$
Standard enthalpy of formation of $\mathrm{N}_2 \mathrm{O}_4$
$$
=-10 \mathrm{~kJ} / \mathrm{mol}
$$
$$
\begin{gathered}
\mathrm{N}_2 \mathrm{O}_4+3 \mathrm{CO} \longrightarrow \mathrm{N}_2 \mathrm{O}+3 \mathrm{CO}_2 \\
\Delta H(\text { enthalpy change })=\Sigma \Delta H_{\mathrm{N}_2 \mathrm{O}}+3 \Delta H_{\mathrm{CO}_2} \\
-\Sigma \Delta H_{\mathrm{N}_2 \mathrm{O}_4}+3 \Delta H_{\mathrm{CO}} \\
=[\Sigma 81+(3 \times-393)-\Sigma(-10+(3 \times-10)] \\
\Delta H=-1058 \mathrm{~kJ} / \mathrm{mol}
\end{gathered}
$$

Topic: Chemistry, Thermodynamics (C), TS EAMCET, TS EAMCET 2018 (05 May Shift 1)
The standard enthalpy of formation of octane (C8H18) is −250  kJmol −1 . The enthalpy of formation of CO2(g) and H2O(l) are −394  kJmol −1 and −286  kJmol −1 respectively. The enthalpy of combustion of C8H18(g) is

Options:
A −5200  kJmol −1
B −5726  kJmol −1
C −5476  kJmol −1
D −5310  kJmol −1

The correct answer is:
−5476  kJmol −1

Solution:
C8H18(g)+252O2(g)→8CO2(g)+9H2O(l)
ΔH°=8×(-394)+9×-286-(-250) =-5476

Topic: Chemistry, Thermodynamics (C), NEET
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), NEET
The standard free energy change (in $J)$ for the reaction $3 \mathrm{Fe}^{2+}(\mathrm{aq})+2 \mathrm{Cr}(\mathrm{s})=2 \mathrm{Cr}^{3+}(\mathrm{aq})+3 \mathrm{Fe}(\mathrm{s})$ given

Options:
A 57,900
B $-57,900$
C $-173,700$
D 173,700

The correct answer is:
$-173,700$

Solution:
$\mathrm{E}_{\mathrm{Cell}}^{\circ}=\mathrm{SRP}$ cathode - SRP anode $=-0.44-(-0.74)=0.3 \mathrm{~V}$
$\underset{Y C}{\Delta} \mathrm{G}=-\mathrm{nF} \mathrm{E}_{\mathrm{Cel|}}^{\circ}\left(\mathrm{n}=6\right.$ ele $\left.^{-}\right)$

Topic: Chemistry, Electrochemistry, KVPY, KVPY 2014 (SB/SX)
The standard free energy change of a reaction is $\Delta \mathrm{G}^{\circ}=-115 \mathrm{~kJ}$ at $298 \mathrm{~K}$. Calculate the equilibrium constant $\mathrm{k}_{\mathrm{p}}$ in $\log \mathrm{k}_{\mathrm{p}}$ $\left(\mathrm{R}=8.314 \mathrm{Jk}^{-1} \mathrm{~mol}^{-1}\right)$

Options:
A $20.16$
B $2.303$
C $2.016$
D $13.83$

The correct answer is:
$20.16$

Solution:
$\begin{aligned} \Delta \mathrm{G}^{\circ} &=-115 \mathrm{~kJ} \text { at } 298 \mathrm{~K} . \\ \mathrm{Now}, \Delta \mathrm{G}^{\circ} &=-2.303 \mathrm{RT} \log \mathrm{k}_{\mathrm{p}} \\ \mathrm{R} &=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \& \mathrm{~T}=298 \mathrm{~K} . \\ \Delta \mathrm{G}^{\circ} &=-2.303 \times 8.314 \times 298 \times \log \mathrm{k}_{\mathrm{p}} \\ \log \mathrm{k}_{\mathrm{P}} &=\frac{-115 \times 10^{3}}{-2.303 \times 8.314 \times 298}=20.155 \\ &=20.16 \end{aligned}$

Topic: Chemistry, Thermodynamics (C), VITEEE, VITEEE 2008
The standard Gibbs' energy change for Daniel cell reaction is
$$
\begin{aligned}
& \mathrm{Zn}(s)+\mathrm{Cu}^{2+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Cu}(s) \\
& E_{\text {cell }}^{\circ}=1.1 \mathrm{~V} \text {. }
\end{aligned}
$$

Options:
A $-212.3 \mathrm{~kJ}$
B $106.15 \mathrm{~kJ}$
C $+212.3 \mathrm{~kJ}$
D $100 \mathrm{~kJ}$

The correct answer is:
$-212.3 \mathrm{~kJ}$

Solution:
Given,
$$
\begin{aligned}
n, E_{\text {cell }}^{\circ} & =1.1 \mathrm{~V} \\
n & =2 \\
F & =96500 \mathrm{C} / \mathrm{mol} \\
\Delta G & =-n F E^{\circ} \\
& =-2 \times 96500 \times 1.1 \\
& =-212300 \mathrm{~J} \\
& =-212.3 \mathrm{~kJ}
\end{aligned}
$$

Topic: Chemistry, Electrochemistry, AP EAMCET, AP EAMCET 2022 (06 Jul Shift 2)
The standard Gibb's energy $\left(\Delta G^{\circ}\right)$ for the following reaction is
$\begin{aligned} & A(s)+B^{2+}(a q) \rightleftharpoons A^{2+}(a q)+B(s), \\ & K_C=10^{12} \text { at } 25^{\circ} \mathrm{C}\left(K_C=\text { equilibrium constant }\right)\end{aligned}$

Options:
A $-150 \mathrm{~kJ}$
B $-96.80 \mathrm{~kJ}$
C $-68.47 \mathrm{~kJ}$
D $-100 \mathrm{~kJ}$

The correct answer is:
$-68.47 \mathrm{~kJ}$

Solution:
The standard Gibb's free energy change $\left(\Delta G^{\circ}\right)$ at standard conditions is given by the formula $\Delta G^{\circ}=-R T \ln K_{\text {eq }}$
where, $R$ is gas constant $=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$, $T=(25+273) \mathrm{K}=298 \mathrm{~K}$ and $K_{\mathrm{eq}}$ is the equilibrium constant
$$
\begin{aligned}
\Delta G^{\circ} & =-8.314 \times 298 \times 2.303 \log 10^{12} \\
& =-68,47018 \mathrm{~J}=-68.47 \mathrm{~kJ}
\end{aligned}
$$

Topic: Chemistry, Thermodynamics (C), AP EAMCET, AP EAMCET 2022 (07 Jul Shift 2)
The standard Gibbs free energy change $\left(\Delta \mathrm{G}^{\circ}\right.$ in $\mathrm{kJ} \mathrm{mol}^{-1}$ ), in a Daniel cell $\left(\mathrm{E}_{\text {cell }}^{\circ}=1.1 \mathrm{~V}\right.$ ), when 2 moles of $\mathrm{Zn}(\mathrm{s})$ is oxidized at $298 \mathrm{~K}$, is closest to

Options:
A $-212.3$
B $-106.2$
C $-424.6$
D $-53.1$

The correct answer is:
$-424.6$

Solution:
$\mathrm{Zn}+\mathrm{Cu}^{+2} \rightarrow \mathrm{Zn}^{+2}+\mathrm{Cu}$
$2 \mathrm{Zn}+2 \mathrm{Cu}^{+2} \rightarrow 2 \mathrm{Zn}^{+2}+2 \mathrm{Cu}$
For 2 moles of $\mathrm{Zn}, \mathrm{n}=4$
$\Delta \mathrm{G}^{\circ}=-\mathrm{nFE}_{\mathrm{Cell}}^{\circ}=-4 \times 96500 \times 1.1=-424.6 \mathrm{~kJ}$

Topic: Chemistry, Electrochemistry, KVPY, KVPY 2011 (SB/SX)
The standard Gibb's free energy change, $\Delta \mathrm{G}^{\circ}$ is related to equilibrium constant, $\mathrm{K}_{\mathrm{p}}$ as

Options:
A $K_{p}=-R T \ln \Delta G^{\circ}$
B $K_{p}=\left[\frac{e}{R T}\right]^{\Delta G^{\circ}}$
C $\mathrm{K}_{\mathrm{p}}=-\frac{\Delta \mathrm{G}}{\mathrm{RT}}$
D $K_{p}=e^{-\Delta G^{0} / R T}$

The correct answer is:
$K_{p}=e^{-\Delta G^{0} / R T}$

Solution:
$\Delta G^{\circ}$ and $K_{p}$ are related as
$$
\begin{array}{l}
\Delta G^{\circ}=-R T \ln K_{p} \\
\Rightarrow \ln K_{p}=\frac{\Delta G^{\circ}}{R T} \Rightarrow K_{p}=e^{-\Delta G^{\circ} / R T}
\end{array}
$$

Topic: Chemistry, Electrochemistry, VITEEE, VITEEE 2010
The standard Gibbs free energy change $\left(\Delta G^{o}\right)$ at $25^{\circ} \mathrm{C}$ for the dissociation of $\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g})$ to $\mathrm{NO}_{2}(g)$ is (given, equilibrium constant $=0.15, R=8.314 \mathrm{JK} / \mathrm{mol})$

Options:
A $1.1 \mathrm{KJ}$
B $4.7 \mathrm{KJ}$
C $8.1 \mathrm{KJ}$
D $38.2 \mathrm{KJ}$

The correct answer is:
$4.7 \mathrm{KJ}$

Solution:
$$
\begin{array}{l}
=-2.303 \times 8.314 \times 298 \times \log 0.15 \\
=-2.303 \times 8.314 \times 298 \times(-0.82) \\
=4678.7 \mathrm{J}=4.67 \mathrm{kJ}
\end{array}
$$

Topic: Chemistry, Thermodynamics (C), WBJEE, WBJEE 2013
The standard half-cell reduction potential for $\mathrm{Ag}^{+} \mid \mathrm{Ag}$ is $0.7991 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. Given the experimental value $K_{s p}=1.56 \times 10^{-10}$ for $\mathrm{AgCl}$, calculate the standard half-cell reduction potential for the $\mathrm{Ag} \mid \mathrm{AgCl}$ electrode.

Options:
A $0.2192 \mathrm{~V}$
B $-0.2192 \mathrm{~V}$
C $-1.2192 \mathrm{~V}$
D $1.2192 \mathrm{~V}$

The correct answer is:
$0.2192 \mathrm{~V}$

Solution:
For the desired reaction,
$$
\mathrm{AgCl}+1 e^{-} \rightarrow \mathrm{Ag}+\mathrm{Cl}^{-}, \quad E^{\circ}=\frac{-\Delta_r G^{\circ}}{(1) F}
$$
The needed $\Delta_r G^{\circ}$ can be obtained by adding the values of $\Delta_r G^{\circ}$ for the reactions
$$
\begin{aligned}
& \mathrm{Ag}^{+}+1 e^{-} \rightarrow \mathrm{Ag} ; \Delta G^{\circ}=-n F E^{\circ} \\
& \mathrm{AgCl} \rightarrow \mathrm{Ag}^{+}+\mathrm{Cl}^{-} ; \Delta G^{\circ}=-R T \ln K_{s p}, \text { giving } \\
& \Delta_r G^{\circ}=-(1 \mathrm{~mol})\left(9.648 \times 10^4 \mathrm{~J} \mathrm{~mol}{ }^{-1} \mathrm{~V}^{-1}\right)(0.7991 \mathrm{~V}) \\
& \quad-\left(8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\right)(298 \mathrm{~K}) \ln \left(1.56 \times 10^{-10}\right) \\
& =-77.10 \mathrm{~kJ}+55.95 \mathrm{~kJ}=-21.15 \mathrm{~kJ}
\end{aligned}
$$
The potential is $E^{\circ}=\frac{-(-21.15)}{(1)(96.485)}=+0.2192 \mathrm{~V}$.

Topic: Chemistry, Ionic Equilibrium, AIIMS, AIIMS 2010
The standard heat of formation (ΔfH2980) 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:
A 192.5
B – 192.5
C 96.25
D +96.25

The correct answer is:
192.5

Solution:
2C(graphite)+3H2(g)→1C2H6+ΔHf(C2H6)
(A) C2H6+72O2→2CO2+3H2O
ΔHC=–1560
(B) H2+12O2→H2O
ΔHC=–393.5
(C) C(graphite) +O2→CO2
ΔHC=–286kJ/mol
ΔHf(C2H6)=3(–393.5)+2(–286)+1560
=–1180.5+(–572)+1560
=192.5
2Cgraphite+3H2→C2H6 (1) eqn
C2H6+72O2→2CO2+3H2O (2) eqn
H2+12O2→2CO2+3H2O (3) eqn
Cgraphite+H2O→CO2 (4) eqn
To get equation (1) eqn (2) + 2 × eqn (3) 2 × eqn
∆Hf of ethane =1560+2×-286+3(-393.5)
=1560-572-1180.5=192.5

Topic: Chemistry, Thermodynamics (C), NEET
The standard molar enthalpies of formation of cycohexane (l) and benzene (l) at 25°C are -156 and +49 KJ mol-1 respectively. The standard enthalpy of hydrogenation of cyclo hexene (l) at 25°C is -116 KJ mol-1 . Use these data to estimate the magnitude of the resonance energy of benzene.

Options:
A 152 kJ  mol −1
B 1.52 kJ  mol −1
C 15.2 kJ  mol −1
D None of these

The correct answer is:
152 kJ  mol −1

Solution:
Enthalpy of hydrogenation of these double bond in benzene
=3Ãâ119=â357âkJâ mol â1
ΔH(reaction) = ΣH(products)−ΔΣH(reactants)
=â156â 49+0 =â205âkJâ mol â1
Resonance energy = Calculated heat of hydrogenation â Actual heat of hydrogenation
=357â205=152âkJâ mol â1 .

Topic: Chemistry, Thermodynamics (C), NEET
The standard molar enthalpy of vaporisation $\left(\Delta_{\text {vap }} H^{\circ}\right)$ of $A, B$ and $C$ liquids is $23.3,41$ and $29 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. The correct order of diople-dipole attractive forces in these liquids is

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

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

Solution:
( The enthalpy of vaporisation is the amount of heat that must be added to a liquid substance to transform it into a gas. Dipole-Dipole interactions are intermolecular force of attraction between polar molecules. Stronger the dipole-dipole interactions, higher will be the heat energy required to break these interactions and vapourise the liquid. Since, the liquids $A, B$ and $C$ have molar enthalpy of vaporisation as $23.3,41$ and $29 \mathrm{~kJ} \mathrm{~mol}^{-1}$, the order of dipole-dipole attractive interactions will be $B>C>A$.

Topic: Chemistry, Thermodynamics (C), AP EAMCET, AP EAMCET 2022 (07 Jul Shift 2)
The standard molar entropy of H2O (l) is 70 J K–1mol–1. Will the standard molar entropy of H2O(s) be more, or less than 70 J K–1mol–1?

Options:
A Less, because H2O(l) is more ordered than ice.
B More, because ice is more ordered than H2O(l).
C Less, because ice is more ordered than H2O(l).
D More, because H2O(l) is more ordered than ice.

The correct answer is:
Less, because ice is more ordered than H2O(l).

Solution:
Correct Option is : (C)
Less, because ice is more ordered than H2O(l).

Topic: Chemistry, Thermodynamics (C), NEET
The standard molar heat of formation of ethane, $\mathrm{CO}_{2}$ and water (1) are respectively $-21.1,-94.1$ and $-68.3 \mathrm{kcal}$. The standard molar heat of combustion of ethane will be

Options:
A $-372 \mathrm{kcal}$
B $162 \mathrm{kcal}$
C $-240 \mathrm{kcal}$
D $183.5 \mathrm{kcal}$

The correct answer is:
$-372 \mathrm{kcal}$

Solution:
Given,
(a) $2 \mathrm{C}+3 \mathrm{H}_{2} \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6} ; \Delta \mathrm{H}=-21.1 \mathrm{kcal}$
(b) $\mathrm{C}+\mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2} ; \Delta \mathrm{H}=-94.1 \mathrm{kcal}$
(c) $\mathrm{H}_{2}+\frac{1}{2} \mathrm{O}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O} ; \Delta \mathrm{H}=-68.3 \mathrm{kcal}$
Now, eqs. $2 \times(\mathrm{b})+3 \times(\mathrm{c})-(\mathrm{a})$
$\mathrm{C}_{2} \mathrm{H}_{6}+\frac{3}{2} \mathrm{O}_{2} \longrightarrow 2 \mathrm{CO}_{2}+3 \mathrm{H}_{2} \mathrm{O}$
Heat of combustion of Ethane $\Delta \mathrm{x}=2(-94.1)$
$\begin{array}{l}
+3(-68.3)-(-21.1) \\
=(-188.2)+(-204.9)-21.1 \\
=-372 \mathrm{kcal}
\end{array}$

Topic: Chemistry, Thermodynamics (C), VITEEE, VITEEE 2014
The standard oxidation potential \(E^{\circ}\) for the half cell reaction are
\[
\begin{array}{ll}
\mathrm{Zn} \rightarrow \mathrm{Zn}^{2+}+2 e^{-} ; & E^{\circ}=+0.76 \mathrm{~V} \\
\mathrm{Fe} \rightarrow \mathrm{Fe}^{2+}+2 e^{-} ; & E^{\circ}=+0.41 \mathrm{~V}
\end{array}
\]
\(E M F\) of the cell reaction is
\[
\mathrm{Zn}+\mathrm{Fe}^{2+} \rightarrow \mathrm{Zn}^{2+}+\mathrm{Fe}
\]

Options:
A \(-0.35 \mathrm{~V}\)
B \(+0.35 \mathrm{~V}\)
C \(0.17 \mathrm{~V}\)
D \(1.17 \mathrm{~V}\).

The correct answer is:
\(+0.35 \mathrm{~V}\)

Solution:
$E_{\text {cell }}^{\circ}=0.76-0.41=0.35 \mathrm{~V}$

Topic: Chemistry, Electrochemistry, AIIMS, AIIMS 2008
The standard petal of the papilionaceous corolla is also called:

Options:
A Carina
B Pappus
C Vexillum
D Corona

The correct answer is:
Vexillum

Solution:
Papilionaceous corolla.

A single, large, upper petal is known as the banner also vexillum or standard petal of a papilionaceous flower.

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2016 (Phase 1)
The standard reduction potential at $298 \mathrm{~K}$ for the following half cell reaction
$Z n^{2+}(a q)+2 e^{-} \rightarrow Z n(s) ; \quad E^{\circ}=-0.762 V$
$C r^{3+}(a q)+3 e^{-} \rightarrow C r(s) ; \quad E^{\circ}=0.740 V$
$2 H^{+}(a q)+2 e^{-} \rightarrow H_{2}(g) ; \quad E^{\circ}=0.0 V$
$F_{2}(g)+2 e^{-} \rightarrow 2 F^{-}(a q) ; \quad E^{\circ}=2.87 V$
Which of the following is strongest reducing agent ?

Options:
A \( Z n(s) \)
B \( C r(s) \)
C \( H_{2}(g) \)
D \( F_{2}(g) \)

The correct answer is:
\( Z n(s) \)

Solution:
The more negative the \( E^{\circ} \) value. The more readily the metal loses electrons. The more negative the value the stronger
is the reducing agent. Therefore, strongest reducing agent is \( \mathrm{Zn} \)

Topic: Chemistry, Biomolecules, KCET, KCET 2017
The standard reduction potential EÂ° for the half reactions are as:

Zn 2+ + 2 e â âZn, E Â° =â0.76 V

Cu 2+ + 2 e â âCu, E Â° = 0.34 V

The standard cell voltage for the cell reaction is?

Zn + Cu2+ â Zn2++ Cu

Options:
A 0.42 V
B -0.42 V
C -1.1 V
D 1.10 V

The correct answer is:
1.10 V

Solution:
E cell o = E Cathode o - E Anode o
E cell o = E Cu +2 /Cu o - E Zn +2 /Zn o
=â0.34â(â0.76) âââââââââ=â1.10 V

Topic: Chemistry, Electrochemistry, NEET
The standard reduction potential for Cu2+/Cu is 0.34 V. The reduction potential at pH = 14 for the above couple (KspCu(OH)2=1×1019-) is

Options:
A -0.22 V
B -0.34 V
C +0.22 V
D +0.34 V

The correct answer is:
-0.22 V

Solution:
pH=14
pOH=0
OH − = 10 0 =1
Cu2+=Ksp[OH-]2=1Ã10-191=1Ã10-19
E Cu 2+ /Cu â=â E o Cu 2+ /Cu â 0.059 2 âlogâ 1 1Ã10 â19
=0.34âVâ 0.059 2 Ã19=â0.22âV

Topic: Chemistry, Electrochemistry, NEET
The standard reduction potential for $\mathrm{Cu}^{2+} / \mathrm{Cu}$ is $+0.34$. Calculate the reduction potential at $\mathrm{pH}$
$=14$ for the above couple. $\left(\mathrm{K}_{\text {sp }} \mathrm{Cu}(\mathrm{OH})_{2}\right.$ $\left.=1 \times 10^{-19}\right)$

Options:
A $-0.22 \mathrm{~V}$
B $+0.22 \mathrm{~V}$
C $-0.44 \mathrm{~V}$
D $+0.44 \mathrm{~V}$

The correct answer is:
$-0.22 \mathrm{~V}$

Solution:
When $\mathrm{pH}=14\left[\mathrm{H}^{+}\right]=10^{-14}$ and $\left[\mathrm{OH}^{-}\right]=1 \mathrm{M}$
$\begin{array}{l}
\mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Cu}^{2+}\right]\left[\mathrm{OH}^{-}\right]^{2}=10^{-19} \\
\therefore \quad\left[\mathrm{Cu}^{2+}\right]=\frac{10^{-19}}{\left[\mathrm{OH}^{-}\right]^{2}}=10^{-19}
\end{array}$
The half cell reaction
$\begin{array}{l}
\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu} \\
\mathrm{E}=\mathrm{E}^{\circ}-\frac{0.059}{2} \log \frac{1}{\left[\mathrm{Cu}^{2+}\right]} \\
=0.34-\frac{0.059}{2} \log \frac{1}{10^{-19}}=-0.22 \mathrm{~V}
\end{array}$

Topic: Chemistry, Electrochemistry, VITEEE, VITEEE 2017
The standard reduction potential for $\mathrm{Fe}^{2+} / \mathrm{Fe}$ and $\mathrm{Sn}^{2+} / \mathrm{Sn}$ electrodes are $-0.44$ and $-0.14 \mathrm{~V}$ respectively. For the cell reaction, $\mathrm{Fe}^{2+}+\mathrm{Sn} \longrightarrow \mathrm{Fe}+\mathrm{Sn}^{2+}$ the standard emf is

Options:
A $+0.30 \mathrm{~V}$
B $-0.58 \mathrm{~V}$
C $+0.58 \mathrm{~V}$
D $-0.30 \mathrm{~V}$

The correct answer is:
$-0.30 \mathrm{~V}$

Solution:
$\mathrm{Fe}^{2+}+\mathrm{Sn} \longrightarrow \mathrm{Fe}+\mathrm{Sn}^{2+}$
$$
=-0.44+(0.14)
$$
$=-0.30 \mathrm{~V}$

Topic: Chemistry, Electrochemistry, MHT CET, MHT CET 2011
The standard reduction potential for $\mathrm{Mg}^{2+} / \mathrm{Mg}$ is $-2.37 \mathrm{~V}$ and for $\mathrm{Cu}^{2+} / \mathrm{Cu}$ is $0.337$. The $E_{\text {cell }}^{\circ}$ for the following reaction is $\mathrm{Mg}+\mathrm{Cu}^{2+} \longrightarrow \mathrm{Mg}^{2+}+\mathrm{Cu}$

Options:
A $+2.03 \mathrm{~V}$
B $-2.03 \mathrm{~V}$
C $-2.7 \mathrm{~V}$
D $+2.7 \mathrm{~V}$

The correct answer is:
$+2.7 \mathrm{~V}$

Solution:
$E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ}$
$$
\begin{aligned}
=E_{\left(\mathrm{Cu}^{2+} / \mathrm{Cu}\right)}^{\circ}-E_{\left(\mathrm{Mg}^{2+} / \mathrm{Mg}\right)}^{\circ} &=0.337-(-2.37) \mathrm{V} \\
&=2.7 \mathrm{~V}
\end{aligned}
$$

Topic: Chemistry, Electrochemistry, MHT CET, MHT CET 2009
The standard reduction potential for $\mathrm{Zn}^{2+} / \mathrm{Zn}, \mathrm{Ni}^{2+} / \mathrm{Ni}$ and $\mathrm{Fe}^{2+} / \mathrm{Fe}$ are -0.76 , -0.23 and -0.44 V , respectively.
The reaction $X+Y^{2+} \longrightarrow X^{2+}+Y$ will be spontaneous when

Options:
A X=Ni, Y=Fe
B X=Ni, Y=Zn
C X=Fe, Y=Zn
D X=Zn, Y=Ni

The correct answer is:
X=Zn, Y=Ni

Solution:
The reaction is spontaneous when $\Delta G \lt 0$ and $E_{\text {cell }}^{\circ}\gt0 .$
$\begin{aligned} & \because \Delta G=-n F E^{\circ} \text { and } E^{\circ}\gt0 \\ & \text { Also } E_{\text {cell }}^{\circ}=E_{\text {oxi }}^{\circ}+E_{\text {red }}^{\circ} \text { (or) } E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ}\end{aligned}$
Here,
Thus, $E_{\text {cell }}^{\circ}$ for
(a) -0.21 V , i.e. $E_{\text {cell }}^0 \lt 0$
(b) -0.53 V , i.e. $E_{\text {cell }}^0 \lt 0$
(c) -0.32 V , i.e. $E_{\text {cell }}^{\circ} \lt 0$
(d) 0.53 V , i.e. $E^{\circ}\gt0$

Topic: Chemistry, Electrochemistry, JIPMER, JIPMER 2017
The standard reduction potential \(\mathrm{E}^{\circ}\) for half reations are
\(\begin{array}{ll}
\mathrm{Zn}=\mathrm{Zn}^{+2}+\mathrm{Ze} & \mathrm{E}^{\circ}=+0.76 \mathrm{~V} \\
\mathrm{Fe}=\mathrm{Fe}^{+2}+\mathrm{Ze} & \mathrm{E}^{\circ}=+0.41 \mathrm{~V}
\end{array}\)
The EMF of hte cell reaction \(\mathrm{Fe}^{+2}+\mathrm{Zn}=\mathrm{Zn}^{+2}+\mathrm{Fe}\) is

Options:
A \(-0.35 \mathrm{~V}\)
B +0.35
C \(+1.17 \mathrm{~V}\)
D \(-1.17 \mathrm{~V}\)

The correct answer is:
+0.35

Solution:
\(\begin{aligned}
& \text {Hints: } \mathrm{E}_{\text {cell }}=\mathrm{E}_{\operatorname{Anu}([\mathrm{Qp})}^{\circ}-\mathrm{E}_{\text {cathode }(\mathrm{op})}^{\circ} \\
& =0.76-0.41 \\
& =+0.35 \mathrm{~V}
\end{aligned}\)

Topic: Chemistry, Electrochemistry, WBJEE, WBJEE 2011
The standard reduction potentials at $298 \mathrm{~K}$ for the following half reactions are given against each
$\begin{array}{ll}
\mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Zn}(\mathrm{s}) & \mathrm{E}_{0}=-0.762 \mathrm{~V} \\
\mathrm{Cr}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Cr}(\mathrm{s}) & \mathrm{E}_{0}=-0.740 \mathrm{~V} \\
2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2}(\mathrm{~g}) & \mathrm{E}_{0}=0.00 \mathrm{~V} \\
\mathrm{Fe}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+}(\mathrm{aq}) & \mathrm{E}_{0}=+0.762 \mathrm{~V}
\end{array}$

Options:
A $\mathrm{Zn}(\mathrm{s})$
B $\mathrm{Cr}(\mathrm{s})$
C $\mathrm{H}_{2}(\mathrm{~g})$
D $\mathrm{Fe}^{2+}(\mathrm{aq})$

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

Solution:
Lower the value of reduction potential, greater will be the reducing power of element.
Since $\mathrm{Zn}$ has lowest reduction potention hence $\mathrm{Zn}$ is the strongest reducing agent.

Topic: Chemistry, Electrochemistry, VITEEE, VITEEE 2007