CBSE 12th Standard Chemistry Subject Electrochemistry Case Study Questions With Solution 2021
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CBSE 12th Standard Chemistry Subject Electrochemistry Case Study Questions With Solution 2021
12th Standard CBSE
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Reg.No. :
Chemistry
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Read the passage given below and answer the following questions:
Molar conductivity of ions are given as product of charge on ions to their ionic mobilities and Faraday's constant.
\(\lambda_{A^{n+}}=n \mu_{A^{n+}} F\) (here \(\mu\) is the ionic mobility of An+).
For electrolytes say AXBy, molar conductivity is given by
\(\lambda_{m\left(A_{x} B_{y}\right)}=x_{n} \mu_{A^{n+}} F+y_{m} \lambda_{A^{m}-F}\)Ions Ionic mobility K+ 7.616 x 10- 4 Ca2+ 12.33 x 10-4 Br- 8.09 x 10- 4 \(\mathrm{SO}_{4}^{2-}\) 16.58 x 10- 4 The following questions are multiple choice questions. Choose the most appropriate answer
(i) At infinite dilution, the equivalent conductance of CaSO4 is(a) 256 x 10-4 (b) 279 (c) 23.7 (d) 2.0 x 10- 8 (ii) If the degree of dissociation of CaSO4 solution is 10% then equivalent conductance of CaSO4 is
(a) 3.59 (b) 36.9 (c) 27.9 (d) 30.6 (iii) What is the unit of equivalent conductivity?
(a) ohm-1 cm2 eq-1 (b) ohm cm2eq-1 (c) ohm-1 cm eq-1 (d) ohm cm2 eq-1 (iv) If the molar conductance value of Ca2+ and Cl- at infinite dilution are 118.88 x 10-4 m2 mho mol-1 and 77.33 x 10-4 m 2 mho mol-1 respectively then the molar conductance of CaCl2 (in m2 mho mol-1) will be
(a) 120.18 x 10- 4 (b) 135 x 10-4 (c) 273.54 x 10-4 (d) 192.1 x 10-4 (a) -
Read the passage given below and answer the following questions:
Standard electrode potentials are used for various processes:
(i) It is used to measure relative strengths of various oxidants and reductants.
(ii) It is used to calculate standard cell potential.
(iii) It is used to predict possible reactions.
A set of half-reactions (in acidic medium) along with their standard reduction potential, Eo (in volt) values are given below
\(\mathrm{I}_{2}+2 e^{-} \rightarrow 2 \mathrm{I}^{-} ; \quad E^{\circ}=0.54 \mathrm{~V}\)
\(\mathrm{Cl}_{2}+2 e^{-} \rightarrow 2 \mathrm{Cl}^{-} ; \quad E^{\circ}=1.36 \mathrm{~V}\)
\(\mathrm{Mn}^{3+}+e^{-} \rightarrow \mathrm{Mn}^{2+} ; \quad E^{\circ}=1.50 \mathrm{~V}\)
\(\mathrm{Fe}^{3+}+e^{-} \longrightarrow \mathrm{Fe}^{2+} ; \quad E^{\circ}=0.77 \mathrm{~V}\)
\(\mathrm{O}_{2}+4 \mathrm{H}^{+}+4 e^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O} ; E^{\circ}=1.23 \mathrm{~V}\)
The following questions are multiple choice questions. Choose the most appropriate answer:
(i) Which of the following statements is correct?(a) CI- is oxidised by O2 (b) Fe2+ is oxidised by iodine (c) I- is oxidised by chlorine. (d) Mn2+ is oxidised by chlorine (ii) Mn3+ is not stable in acidic medium, while Fe3+is stable because
(a) O2 oxidises Mn2+ to Mn3+ (b) O2 oxidises both Mn2+ to Mn3+ and Fe2+ to Fe3+ (c) Fe3-oxidises H2O to O2 (d) Mn3+ oxidises H2O to O2 (iii) The strongest reducing agent in the aqueous solution is
(a) I- (b) Cl- (c) Mn2+ (d) Fe2+ (iv) The emf for the following reaction is
\(\mathrm{I}_{2}+\mathrm{KCl} \rightleftharpoons 2 \mathrm{KI}+\mathrm{Cl}_{2}\)(a) -0.82 V (b) +0.82 V (c) -0.73 V (d) +0.73 V (a) -
Read the passage given below and answer the following questions :
All chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules are present in a few gram of any chemical compound varying with their atomic/molecular masses. To handle such large number conveniently, the mole concept was introduced. All electrochemical cell reactions are also based on mole concept. For example, a 4.0 molar aqueous solution of NaCI is prepared and 500 mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrode. The amount of products formed can be calculated by using mole concept.
The following questions are multiple choice questions. Choose the most appropriate answer :
(i) The total number of moles of chlorine gas evolved is(a) 0.5 (b) 1.0 (c) 1.5 (d) 1.9 (ii) If cathode is a Hg electrode, then the maximum weight of amalgam formed from this solution is
(a) 300 g (b) 446 g (c) 396 g (d) 296 g (iii) In the electrolysis, the number of moles of electrons involved are
(a) 2 (b) 1 (c) 3 (d) 4 (iv) In electrolysis of aqueous NaCl solution when Pt electrode is taken, then which gas is liberated at cathode?
(a) H2 gas (b) C2 gas (c) O2 gas (d) None of these (a) -
Read the passage given below and answer the following questions:
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).
The following questions are multiple choice questions. Choose the most appropriate answer:
(i) For the above cell,(a) \(E_{\text {cell }}<0 ; \Delta G>0\) (b) \(E_{\text {cell }}>0 ; \Delta G<0\) (c) \(E_{\text {cell }}<0 ; \Delta G^{\circ}>0\) (d) \(E_{\text {cell }}>0 ; \Delta G^{\circ}<0\) (ii) The value of equilibrium constant for a feasible cell reaction is
(a) < 1 (b) = 1 (c) > 1 (d) zero (iii) What is the emf ofthe cell when the cell reaction attains equilibrium?
(a) 1 (b) 0 (c) > 1 (d) < 1 (iv) The potential of an electrode change with change in
(a) concentration of ions in solution (b) position of electrodes (c) voltage of the cell (d) all of these (a) -
Read the passage given below and answer the following questions:
The electrochemical cell shown below is 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 following questions are multiple choice questions. Choose the most appropriate answer:
(i) The solubility product (Ksp, mol3 dm-9) of MX2 at 298 K based on the information available for the given concentration cell is (take 2.303 x R x 298/P = 0.059)(a) 2 x 10-15 (b) 4 x 10-15 (c) 3 x 10-12 (d) 1 x 1012 (ii) The value of \(\Delta G\) (in kJ mol-1) for the given cell is (take 1F = 96500 C mol-1)
(a) 3.7 (b) -3.7 (c) 10.5 (d) -11.4 (iii) The equilibrium constant for the following reaction is
\(\mathrm{Fe}^{2+}+\mathrm{Ce}^{4+} \rightleftharpoons \mathrm{Ce}^{3+}+\mathrm{Fe}^{3+}\)
(Given, \(E^{0} \mathrm{Ce}^{4+} / \mathrm{Ce}^{3+}=1.44\) and Eo \(E_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}=0.68 \mathrm{~V}\))(a) 7.6 x 1012 (b) 6.5 x 1010 (c) 5.2 x 109 (d) 3.4 x 1012 (iv) To calculate the emf of the cell, which of the following options is correct?
(a) emf = Ecathode - Eanode (b) emf = Eanode - Ecathode (c) emf = Eanode + Ecathode (d) None of these (a)
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CBSE 12th Standard Chemistry Subject Electrochemistry Case Study Questions With Solution 2021 Answer Keys
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(i) (b) : Equivalent conductance of CaSO4 :
\(\Lambda_{\mathrm{CaSO}_{4}}^{\infty}=\lambda_{\mathrm{Ca}^{2+}}^{\infty}+\lambda_{\mathrm{SO}_{4}^{2-}}^{\infty}\)
\(\lambda_{\mathrm{Ca}^{2+}}^{\infty}=\left(\mu_{\mathrm{Ca}^{2+}}\right) F ; \lambda_{\mathrm{SO}_{4}^{2-}}^{\infty}=\left(\mu_{\mathrm{SO}_{4}^{2-}}\right) F\)
\(\mu_{\mathrm{Ca}^{2+}}\) and \(\mu_{\mathrm{SO}_{4}^{2-}}\) - are ionic mobilities.
\(\Lambda_{\mathrm{CaSO}_{4}}^{\infty}=F(12.33+16.58) \times 10^{-4}\)
= 96500 x 10- 4 x 28.91 = 279
(ii) (c) : \(\alpha=\frac{\Lambda_{C}}{\Lambda^{\infty}} \Rightarrow 0.1=\frac{\Lambda_{C}}{279} \Rightarrow \Lambda_{C}=27.9\)
(iii) (a)
(iv) (c) : \(\Lambda_{m\left(\mathrm{CaCl}_{2}\right)}^{\circ}=\lambda_{\mathrm{Ca}^{2+}}^{\circ}+2 \lambda_{\mathrm{Cl}^{-}}^{\circ}\)
= (118.88 x 10- 4 ) + 2(77.33 x 10- 4 )
= 273.54 x 10-4 m2 mho mol-1 -
(i) (c) : The half cell having the higher reduction potential will undergo reduction process.
(ii) (d) : Electrode potential of Mn3+ is higher than O2.
(iii) (a) : Due to least electrode potential value.
(iv) (a) : Half reactions :\(\mathrm{I}_{2}+2 e^{-} \rightarrow 2 \mathrm{I}^{-}\) Reduction Eo = 0.54 V \(2 \mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_{2}+2 e^{-}\) Oxidation Eo = -1.36 V ------------------------------- e.m.f = -0.82 V -
(i) (b) : \(n_{\mathrm{NaCl}}=\frac{4 \times 500}{1000}=2 \mathrm{~mol}\)
\(\therefore\) \(n_{\mathrm{Cl}_{2}}=1 \mathrm{~mol}\)
(ii) (b) : nNa deposited = 2 mol
\(\therefore\) nNa _ Hg formed = 2 mol
\(\therefore\) Mass of amalgam formed = 2 x 223 = 446 g
(iii) (a)
(iv) (a) -
(i) (b) : \(\begin{array}{l} M \longrightarrow M^{+}+e^{-} \\ (1 \cdot M)(0.05 M) \end{array}\)
For concentration cell, \(E_{\text {cell }}=-\frac{0.059}{1} \log \frac{0.05}{1}\)
\(E_{\text {cell }}=-\frac{0.059}{1} \log \left(5 \times 10^{-2}\right)\)
\(E_{\text {cell }}=-\frac{0.059}{1}[(-2)+\log 5]-0.059(-2+0.698)\)
= -0.059(-1.302) = 0.0768
\(\Delta G=-n F E_{\text {cell }}\)
If Ecell is positive, \(\Delta G\) is negative.
(ii) (c) : \(K=\operatorname{antilog}\left(\frac{n E^{\circ}}{0.0591}\right)\)
For feasible cell, Eo is positive, hence from the above equation K > 1 for a feasible cell reaction.
(iii) (b)
(iv) (a) -
(i) (b) : \(0.059=\frac{+0.059}{2} \log \frac{0.001}{\left[M^{2+}\right]}\)
\(\log \frac{0.001}{\left[M^{2+}\right]}=2 \text { or }\left[M^{2+}\right]=10^{-5}\)
Let solubility of salt be S mol/litre
\(\begin{array}{cc} \text {Thus,} \ M X_{2} & \rightarrow M^{2+}+2 X^- \\ S & S& 2 S \end{array}\)
\(\therefore\) Ksp = 4S3 = 4 x (10-5 )3 = 4 x 10-15
(ii) (d) : \(\Delta G=-n F E=-2 \times 96500 \times 0.059\)
= -11387 J mol-1 = -11.4 kJ mol-1
(iii) (a) : \(E_{\mathrm{cell}}^{\circ}=\frac{0.059}{1} \log K_{\mathrm{C}}\)
\(E_{\mathrm{cell}}^{\circ}=E_{\mathrm{Fe}^{2+} / \mathrm{Fe}^{3+}}^{\circ}+E_{\mathrm{Ce}^{4+} / \mathrm{Ce}^{3+}}^{\circ}\)
= -0.68 + 1.44 = 0.76 V
\(\log _{10} K_{C}=\frac{0.76}{0.059}=12.88\)
KC = 7.6 x 1012
(iv) (a)
