Electrochemistry · Chemistry · JEE Advanced

MCQ (More than One Correct Answer)

An aqueous solution of hydrazine $\left(\mathrm{N}_2 \mathrm{H}_4\right)$ is electrochemically oxidized by $\mathrm{O}_2$, thereby releasing chemical energy in the form of electrical energy. One of the products generated from the electrochemical reaction is $\mathrm{N}_2(\mathrm{~g})$.

Choose the correct statement(s) about the above process

JEE Advanced 2024 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)

JEE Advanced 2021 Paper 2 Online

In a galvanic cell, the salt bridge

JEE Advanced 2014 Paper 1 Offline

For the reduction of NO$$_3^ - $$ ion in an aqueous solution, E$$^0$$ is + 0.96 V. Values of E$$^0$$ for some metal ions are given below:

$$\matrix{ {{V^{2 + }}(aq.) + 2{e^ - } \to V} & {{E^0} = - 1.19\,V} \cr {F{e^{3 + }}(aq.) + 3{e^ - } \to Fe} & {{E^0} = - 0.04\,V} \cr {A{u^{3 + }}(aq) + 3{e^ - } \to Au} & {{E^0} = + 1.40\,V} \cr {H{g^{2 + }}(aq) + 2{e^ - } \to Hg} & {{E^0} = + 0.86\,V} \cr } $$

The pair(s) of metals that is (are) oxidized by NO$$_3^ - $$ in aqueous solution is(are)

IIT-JEE 2009 Paper 2 Offline

MCQ (Single Correct Answer)

In a conductometric titration, small volume of titrant of higher concentration is added stepwise to a larger volume of titrate of much lower concentration, and the conductance is measured after each addition.

The limiting ionic conductivity $\left(\Lambda_0\right)$ values (in $\mathrm{mS} \mathrm{m}{ }^2 \mathrm{~mol}^{-1}$ ) for different ions in aqueous solutions are given below:

$$ \begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline \text { Ions } & \mathrm{Ag}^{+} & \mathrm{K}^{+} & \mathrm{Na}^{+} & \mathrm{H}^{+} & \mathrm{NO}_3^{-} & \mathrm{Cl}^{-} & \mathrm{SO}_4^{2-} & \mathrm{OH}^{-} & \mathrm{CH}_3 \mathrm{COO}^{-} \\ \hline \Lambda_0 & 6.2 & 7.4 & 5.0 & 35.0 & 7.2 & 7.6 & 16.0 & 19.9 & 4.1 \\ \hline \end{array} $$

For different combinations of titrates and titrants given in List-I, the graphs of 'conductance' versus 'volume of titrant' are given in List-II.

Match each entry in List-I with the appropriate entry in List-II and choose the correct option.

LIST-I	LIST-II
(P) Titrate: KCl Titrant: AgNO$_3$
(Q) Titrate: AgNO$_3$ Titrant: KCl
(R) Titrate: NaOH Titrant: HCl
(S) Titrate: NaOH Titrant: CH$_3$COOH

JEE Advanced 2024 Paper 1 Online

Plotting $1 / \Lambda_{\mathrm{m}}$ against $\mathrm{c} \Lambda_{\mathrm{m}}$ for aqueous solutions of a monobasic weak acid $(\mathrm{HX})$ resulted in a straight line with $\mathrm{y}$-axis intercept of $\mathrm{P}$ and slope of $\mathrm{S}$. The ratio $\mathrm{P} / \mathrm{S}$ is

$$ \begin{aligned} & {\left[\Lambda_{\mathrm{m}}=\right.\text { molar conductivity }} \\\\ & \Lambda_{\mathrm{m}}^{\mathrm{o}}=\text { limiting molar conductivity } \\\\ & \mathrm{c}=\text { molar concentration } \\\\ & \left.\mathrm{K}_{\mathrm{a}}=\text { dissociation constant of } \mathrm{HX}\right] \end{aligned} $$

JEE Advanced 2023 Paper 1 Online

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)

JEE Advanced 2019 Paper 1 Offline

For the following cell,

$$Zn\left( s \right)\left| {ZnS{O_4}\left( {aq} \right)} \right|\left| {CuS{O_4}\left( {aq} \right)} \right|Cu\left( s \right)$$

when the concentration of $$Z{n^{2 + }}$$ is $$10$$ times the concentration of $$C{u^{2 + }},$$ the expression for $$\Delta G$$ (in $$J\,mo{l^{ - 1}}$$) is [$$F$$ is Faraday constant; $$R$$ is gas constant; $$T$$ is temperature; $${E^0}$$ (cell)$$=1.1$$ $$V$$]

JEE Advanced 2017 Paper 2 Offline

For the following electrochemical cell at 298 K

Pt(s) | H₂ (g, 1 bar) | H⁺ (aq, 1 M) || M⁴⁺ (aq), M²⁺ (aq) | Pt (s)

E_cell = 0.092 V when $${{\left[ {{M^{2 + }}(aq)} \right]} \over {\left[ {{M^{4 + }}(aq)} \right]}}$$ = 10^x

Give, $$E_{{M^{4+}}/{M^{2 + }}}^o$$ = 0.151 V; 2.303 RT/F = 0.059 V

The value of x is

JEE Advanced 2016 Paper 2 Offline

The standard reduction potential data at 25^oC is given below:
E^o (Fe³⁺ , Fe²⁺) = +0.77V;
E^o (Fe²⁺ , Fe) = -0.44V;
E^o (Cu²⁺ , Cu) = +0.34V;
E^o (Cu⁺ , Cu) = +0.52V;
E^o [O₂(g) + 4H⁺ + 4e^- $$\to$$ 2H₂O] = +1.23V;
E^o [O₂(g) + 2H₂O + 4e^- $$\to$$ 4OH^-] = +0.40 V
E^o (Cr³⁺ , Cr) = -0.74V;
E^o (Cr²⁺ , Cr) = -0.91V;

Match E^o of the redox pair in List – I with the values given in List – II and select the correct answer using the code given below the lists:

List - I
P. E^o (Fe³⁺ , Fe)
Q. E^o (4H₂O $$\leftrightharpoons$$ 4H⁺ + 4OH^-)
R. E^o (Cu²⁺ + Cu $$\to$$ 2Cu⁺)
S. E^o (Cr³⁺, Cr²⁺)

List - II
1. -0.18 V
2. -0.4 V
3. -0.04 V
4. -0.83 V

JEE Advanced 2013 Paper 2 Offline

An aqueous solution of X is added slowly to an aqueous solution of Y as shown in List – I. The variation in conductivity of these reactions in List – II. Match List – I with List – II and select the correct answer using the code given below the lists:

List - I
P. $$\mathop {(C{}_2{H_5}){}_3N}\limits_X $$ + $$\mathop {C{H_3}COOH}\limits_Y $$
Q. $$\mathop {KI(0.1M)}\limits_X $$ + $$\mathop {AgN{O_3}(0.01M)}\limits_Y $$
R. $$\mathop {C{H_3}COOH}\limits_X $$ + $$\mathop {KOH}\limits_Y $$
S. $$\mathop {NaOH}\limits_X $$ + $$\mathop {HI}\limits_Y $$

List - II
1. Conductivity decreases then increases
2. Conductivity decreases then does not change much
3. Conductivity increases then does not change much
4. Conductivity does not change much then increases

JEE Advanced 2013 Paper 2 Offline

The electrochemical cell shown below is a concentration cell. M | M²⁺ (saturated solution of a sparingly soluble salt, MX₂) || M²⁺ (0.001 mol dm^–3) | M The emf of the cell depends on the difference in concentrations of M²⁺ ions at the two electrodes. The emf of the cell at 298 K is 0.059 V.

The solubility product (K_sp; mol³ dm^–9) of MX₂ at 298 K based on the information available for the given concentration cell is (take 2.303 $$\times$$ R $$\times$$ 298/F = 0.059 V)

IIT-JEE 2012 Paper 2 Offline

The electrochemical cell shown below is a concentration cell. M | M²⁺ (saturated solution of a sparingly soluble salt, MX₂) || M²⁺ (0.001 mol dm^–3) | M The emf of the cell depends on the difference in concentrations of M²⁺ ions at the two electrodes. The emf of the cell at 298 K is 0.059 V.

The value of ∆G (kJ mol^–1) for the given cell is (take 1F = 96500 C mol^–1)

IIT-JEE 2012 Paper 2 Offline

AgNO₃(aq.) was added to an aqueous KCl solution gradually and the conductivity of the solution was measured. The plot of conductance ($$\Lambda $$) versus the volume of AgNO₃ is

IIT-JEE 2011 Paper 1 Offline

Consider the following cell reaction:
2Fe(s) + O₂(g) + 4H⁺(aq) $$\to$$ 2Fe²⁺ (aq) + 2H₂O (l); E^o = 1.67 V
At [Fe²⁺] = 10^-3 M, P(O₂) = 0.1 atm and pH = 3, the cell potential at 25^oC is

IIT-JEE 2011 Paper 2 Offline

The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is :
M(s) | M⁺ (aq ; 0.05 molar) || M⁺ (aq ; 1 molar) | M(s)
For the above electrolytic cell the magnitude of the cell potential | E_cell | = 70 mV.

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

IIT-JEE 2010 Paper 1 Offline

Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time required to liberate 0.01 mol of H$$_2$$ gas at the cathode is (1 Faraday = 96500 C mol$$^{-1}$$].

IIT-JEE 2008 Paper 2 Offline

Numerical

Consider the strong electrolytes $Z_{m} X_{n}, U_{m} Y_{p}$ and $V_{m} X_{n}$. Limiting molar conductivity ( $\Lambda^{0}$ ) of $\mathrm{U}_{\mathrm{m}} \mathrm{Y}_{\mathrm{p}}$ and $\mathrm{V}_{\mathrm{m}} \mathrm{X}_{\mathrm{n}}$ are 250 and $440 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$, respectively. The value of $(\mathrm{m}+\mathrm{n}+\mathrm{p})$ is

Given:

Ion	$\mathrm{Z}^{\mathrm{n}+}$	$\mathrm{U}^{\mathrm{p}+}$	$\mathrm{V}^{\mathrm{n}+}$	$\mathrm{X}^{\mathrm{m}-}$	$\mathrm{Y}^{\mathrm{m}-}$
$\lambda^{0}\left(\mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\right)$	$50.0$	$25.0$	$100.0$	$80.0$	$100.0$

$\lambda^{0}$ is the limiting molar conductivity of ions

The plot of molar conductivity ( $\Lambda$ ) of $\mathrm{Z}_{\mathrm{m}} \mathrm{X}_{\mathrm{n}} v s\, \mathrm{c}^{1 / 2}$ is given below.

JEE Advanced 2022 Paper 2 Online

The reduction potential $\left(E^{0}\right.$, in $\left.\mathrm{V}\right)$ of $\mathrm{MnO}_{4}^{-}(\mathrm{aq}) / \mathrm{Mn}(\mathrm{s})$ is __________.

[Given: $E_{\left(\mathrm{MnO}_{4}^{-}(\mathrm{aq}) / \mathrm{MnO}_{2}(\mathrm{~s})\right)}^{0}=1.68 \mathrm{~V} ; E_{\left(\mathrm{MnO}_{2}(\mathrm{~s}) / \mathrm{Mn}^{2+}(\mathrm{aq})\right)}^{0}=1.21 \mathrm{~V} ; E_{\left(\mathrm{Mn}^{2+}(\mathrm{aq}) / \mathrm{Mn}(\mathrm{s})\right)}^{0}=-1.03 \mathrm{~V}$ ]

JEE Advanced 2022 Paper 1 Online