Chapter 17
43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry · 82 exercises
Problem 1
For the given cell; \(\mathrm{Cu}(\mathrm{s})\left|\mathrm{Cu}^{2+}\left(\mathrm{C}_{1} \mathrm{M}\right) \| \mathrm{Cu}^{2+}\left(\mathrm{C}_{2} \mathrm{M}\right)\right| \mathrm{Cu}(\mathrm{s})\) change in Gibbs energy \((\Delta \mathrm{G})\) is negative, if : (a) \(\mathrm{C}_{1}=\mathrm{C}_{2}\) (b) \(\mathrm{C}_{2}=\frac{\mathrm{C}_{1}}{\sqrt{2}}\) (c) \(\mathrm{C}_{1}=2 \mathrm{C}_{2}\) (d) \(\mathrm{C}_{2}=\sqrt{2} \mathrm{C}_{1}\)
5 step solution
Problem 3
Given that the standard potentials \(\left(\mathrm{E}^{0}\right)\) of \(\mathrm{Cu}^{2+} / \mathrm{Cu}\) and \(\mathrm{Cu}^{+} / \mathrm{Cu}\) are \(0.34\) \(\mathrm{V}\) and \(0.522 \mathrm{~V}\) respectively, the \(\mathrm{E}^{0}\) of \(\mathrm{Cu}^{2+} / \mathrm{Cu}^{+}\)is: (a) \(+0.182 \mathrm{~V}\) (b) \(+0.158 \mathrm{~V}\) (c) \(-0.182 \mathrm{~V}\) (d) \(-0.158 \mathrm{~V}\)
4 step solution
Problem 4
The decreasing order of electrical conductivity of the following aqueous solutions is: [Main April 12, 2019 (II)] \(0.1 \mathrm{M}\) Formic acid (A), \(0.1 \mathrm{M}\) Acetic acid (B), \(0.1 \mathrm{M}\) Benzoic acid (C). (a) \(\mathrm{A}>\mathrm{C}>\mathrm{B}\) (b) \(\mathrm{C}>\mathrm{B}>\mathrm{A}\) (c) \(\mathrm{A}>\mathrm{B}>\mathrm{C}\) (d) \(\mathrm{C}>\mathrm{A}>\mathrm{B}\)
5 step solution
Problem 4
The standard Gibbs energy for the given cell reaction in \(\mathrm{kJ} \mathrm{mol}^{-1}\) at \(298 \mathrm{~K}\) is: [Main April 9, 2019 (I)] \(\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})\) \(\mathrm{E}^{\circ}=2 \mathrm{~V}\) at \(298 \mathrm{~K}\) (Faraday's constant, \(\left.\mathrm{F}=96000 \mathrm{C} \mathrm{mol}^{-1}\right)\) (a) \(-384\) (b) 384 (c) 192 (d) \(-192\)
5 step solution
Problem 5
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 : [Main April 10, 2019 (I)] (a) Both S1 and \(\mathrm{S} 2\) are wrong (b) \(\mathrm{S} 1\) is wrong and \(\mathrm{S} 2\) is correct (c) Both S1 and S2 are correct (d) \(\mathrm{S} 1\) is correct and \(\mathrm{S} 2\) is wrong
3 step solution
Problem 6
A solution of \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}\) is electrolysed between platinum electrodes using \(0.1\) Faraday electricity. How many mole of Ni will be deposited at the cathode? (a) \(0.05\) (b) \(0.20\) (c) \(0.15\) (d) \(0.10\)
4 step solution
Problem 6
For the cell \(\mathrm{Zn}(\mathrm{s}) \mid \mathrm{Zn}^{2+}(\mathrm{aq}) \| \mathrm{M}^{\mathrm{x}+}\) (aq) \(\mid \mathrm{M}(\mathrm{s})\), different half cells and their standard electrode potentials are given below: (aq)/ (aq)/ \(\mathrm{Au}^{3+}(\mathrm{aq}) / \mathrm{Ag}^{+}(\mathrm{aq}) / \mathrm{Fe}^{3+}(\mathrm{aq}) / \mathrm{Fe}^{2+}(\mathrm{a}\) \(\mathrm{Au}(\mathrm{s}) \quad \mathrm{Ag}(\mathrm{s}) \quad \mathrm{Fe}^{2+}(\mathrm{aq}) \mathrm{Fe}(\mathrm{s})\) \(\frac{\mathrm{M}^{\mathrm{x}+}(\mathrm{aq}) /}{\mathrm{M}(\mathrm{s})}\) (aq) 1 \(\mathrm{E}_{\mathrm{M}^{\mathrm{x}+} / \mathrm{M}}^{\circ} /(\mathrm{V}) \overline{1.40}\) 1 0 \(\begin{array}{lll}0.80 & 0.77 & -0.44\end{array}\) If \(\mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}=-0.76 \mathrm{~V}\), which cathode will give a maximum value of \(\mathrm{E}_{\mathrm{cell}}^{\circ}\) per electron transferred? [Main Jan. \(11,2019(\mathrm{I})]\) (a) \(\mathrm{Ag}^{+} / \mathrm{Ag}\) (b) \(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\) (c) \(\mathrm{Au}^{3+} / \mathrm{Au}\) (d) \(\mathrm{Fe}^{2+} / \mathrm{Fe}\)
5 step solution
Problem 7
The anodic half-cell of lead-acid battery is recharged using electricity of \(0.05\) Faraday. The amount of \(\mathrm{PbSO}_{4}\) electrolyzed in g during the process is: (Molar mass of \(\mathrm{PbSO}_{4}=303 \mathrm{~g} \mathrm{~mol}^{-1}\) ) [Main Jan. 9, 2019 (I)] (a) \(22.8\) (b) \(15.2\) (c) \(7.6\) (d) \(11.4\)
5 step solution
Problem 7
If the standard electrode potential for a cell is \(2 \mathrm{~V}\) at \(300 \mathrm{~K}\), the equilibrium constant \((\mathrm{K})\) for the reaction \(\mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \rightleftharpoons \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s})\) at \(300 \mathrm{~K}\) is approximately \(\left(\mathrm{R}=8 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}, \mathrm{~F}=96000 \mathrm{C} \mathrm{mol}^{-1}\right)\) [Main Jan. 9, 2019 (II)] (a) \(\mathrm{e}^{-80}\) (b) \(\mathrm{e}^{-160}\) (c) \(\mathrm{e}^{320}\) (d) \(\mathrm{e}^{160}\)
6 step solution
Problem 8
What is the standard reduction potential \(\left(\mathrm{E}^{\circ}\right)\) for \(\mathrm{Fe}^{3+} \rightarrow \mathrm{Fe}\) ? Given that : [Main Online April 8, 2017] \(\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Fe} ; \mathrm{E}_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\circ}=-0.47 \mathrm{~V}\) \(\mathrm{Fe}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+} ; \mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}=+0.77 \mathrm{~V}\) (a) \(-0.057 \mathrm{~V}\) (b) \(+0.057 \mathrm{~V}\) (c) \(+0.30 \mathrm{~V}\) (d) \(-0.30 \mathrm{~V}\)
5 step solution
Problem 9
How long (approximate) should water be electrolysed by passing through 100 amperes current so that the oxygen released can completely burn \(27.66 \mathrm{~g}\) of diborane? (Atomic weight of \(\mathrm{B}=10.8 \mathrm{u}\) ) [Main 2018] (a) \(6.4\) hours (b) \(0.8\) hours (c) \(3.2\) hours (d) \(1.6\) hours
6 step solution
Problem 9
For the following cell, \(\mathrm{Zn}(\mathrm{s})\left|\mathrm{ZnSO}_{4}(\mathrm{aq}) \| \mathrm{CuSO}_{4}(\mathrm{aq})\right| \mathrm{Cu}(\mathrm{s})\) when the concentration of \(\mathrm{Zn}^{2+}\) is 10 times the concentration of \(\mathrm{Cu}^{2+}\), the expression for \(\Delta G\) (in \(\mathrm{J} \mathrm{mol}^{-1}\) ) is \([F\) is Faraday constant; \(R\) is gas constant; \(T\) is temperature; \(E^{0}\) (cell) \(\left.=1.1 \mathrm{~V}\right]\) [Adv. 2017] (a) \(1.1 \mathrm{~F}\) (b) \(2.303 R T-2.2 F\) (c) \(2.303 R T+1.1 F\) (d) \(-2.2 F\)
5 step solution
Problem 10
When an electric current is passes through acidified water, \(112 \mathrm{~mL}\) of hydrogen gas at N.T.P. was collected at the cathode in 965 seconds. The current passed, in ampere, is : [Main Online April 15, 2018 (I)] (a) \(2.0\) (b) \(0.1\) (c) \(0.5\) (d) \(1.0\)
6 step solution
Problem 10
Galvanization is applying a coating of: (a) \(\mathrm{Cu}\) (b) \(\mathrm{Zn}\) (c) \(\mathrm{Pb}\) (d) \(\mathrm{Cr}\)
3 step solution
Problem 11
Given \(\mathrm{E}_{\mathrm{Cl}_{2} / \mathrm{Cl}^{-}}=1.36 \mathrm{~V}, \mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^{\circ}=-0.74 \mathrm{~V}\) \(\mathrm{E}_{\mathrm{Cr}_{2} / \mathrm{O}_{7}^{2-} / \mathrm{Cr}^{3+}}=1.33 \mathrm{~V}, \mathrm{E}_{\mathrm{MnO}_{4}^{\circ} / \mathrm{Mn}^{2+}}=1.51 \mathrm{~V}\) Among the following, the strongest reducing agent is (a) \(\mathrm{Cr}\) (b) \(\mathrm{Mn}^{2+}\) (c) \(\mathrm{Cr}^{3+}\) (d) \(\mathrm{Cl}^{-}\)
5 step solution
Problem 11
Identify the correct statement : [Main Online April 10, 2016] (a) Corrosion of iron can be minimized by forming a contact with another metal with a higher reduction potential (b) Iron corrodes in oxygen free water (c) Corrosion of iron can be minimized by forming an impermeable barrier at its surface (d) Iron corrodes more rapidly in salt water because its electrochemical potential is higher
4 step solution
Problem 12
What will occur if a block of copper metal is dropped into a beaker containing a solution of \(1 \mathrm{M} \mathrm{ZnSO}_{4}\) ? [Main Online April 9, 2016] (a) The copper metal will dissolve with evolution of oxygen gas (b) The copper metal will dissolve with evolution of hydrogen gas (c) No reaction will occur (d) The copper metal will dissolve and zinc metal will be deposited.
3 step solution
Problem 12
For the following electrochemical cell at \(298 \mathrm{~K}\), \(\operatorname{Pt}(\mathrm{s})\left|\mathrm{H}_{2}(\mathrm{~g}, 1 \mathrm{bar})\right| \mathrm{H}^{+}(\mathrm{aq}, 1 \mathrm{M}) \| M^{4+}(\mathrm{aq}), M^{2+}(\mathrm{aq}) \mid \mathrm{Pt}(\mathrm{s})\) \(E_{\mathrm{ccll}}=0.092 \mathrm{~V}\) when \(\frac{\left[M^{2+}(\mathrm{aq})\right]}{\left[M^{4+}(\mathrm{aq})\right]}=10^{x} .\) [Adv. 2016] Given : \(E_{M^{4+} / M^{2+}}^{\circ}=0.151 \mathrm{~V} ; 2.303 \frac{R T}{F}=0.059 \mathrm{~V}\) The value of \(x\) is (a) \(-2\) (b) \(-1\) (c) 1 (d) 2
5 step solution
Problem 13
Two Faraday of electricity is passed through a solution of \(\mathrm{CuSO}_{4}\). The mass of copper deposited at the cathode is (at. mass of \(\mathrm{Cu}=63.5 \mathrm{amu}\) ) [Main 2015] (a) \(2 \mathrm{~g}\) (b) \(127 \mathrm{~g}\) (c) \(0 \mathrm{~g}\) (d) \(63.5 \mathrm{~g}\)
6 step solution
Problem 13
Given below are the half-cell reactions: $$ \begin{aligned} &\mathrm{Mn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn} ; \mathrm{E}^{\circ}=-1.18 \mathrm{~V} \\ &2\left(\mathrm{Mn}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+}\right) ; \mathrm{E}^{\circ}=+1.51 \mathrm{~V} \end{aligned} $$ The \(\mathrm{E}^{\circ}\) for \(3 \mathrm{Mn}^{2+} \rightarrow \mathrm{Mn}+2 \mathrm{Mn}^{3+}\) will be: (a) \(-2.69 \mathrm{~V} ;\) the reaction will not occur (b) \(-2.69 \mathrm{~V}\); the reaction will occur (c) \(-0.33 \mathrm{~V}\); the reaction will not occur (d) \(-0.33 \mathrm{~V}\); the reaction will occur
4 step solution
Problem 14
A variable, opposite external potential \(\left(\mathrm{E}_{\text {ex }}\right.\) ) is applied to the cell \(\mathrm{Zn} \mathrm{Zn}^{2+}(1 \mathrm{M}) \| \mathrm{Cu}^{2+}(1 \mathrm{M}) \mid \mathrm{Cu}\), of potential \(1.1 \mathrm{~V}\). When \(\mathrm{E}_{\mathrm{ext}}<1.1 \mathrm{~V}\) and \(\mathrm{E}_{\text {ext }}>1.1 \mathrm{~V}\), respectively electrons flow from : [Main Online April 10, 2015] (a) anode to cathode in both cases (b) cathode to anode and anode to cathode (c) anode to cathode and cathode to anode (d) cathode to anode in both cases
4 step solution
Problem 14
Given : \(E_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^{\circ}=-0.74 \mathrm{~V} ; \mathrm{E}_{\mathrm{MnO}_{4}^{-} / \mathrm{Mn}^{2+}}^{\circ}=1.51 \mathrm{~V}\) \(\mathrm{E}_{\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} / \mathrm{Cr}^{3+}}=1.33 \mathrm{~V} ; \mathrm{E}_{\mathrm{Cl} / \mathrm{Cl}^{-}}^{\circ}=1.36 \mathrm{~V}\) Based on the data given above, strongest oxidising agent will be : [Main 2013] (a) \(\mathrm{Cl}\) (b) \(\mathrm{Cr}^{3+}\) (c) \(\mathrm{Mn}^{2+}\) (d) \(\mathrm{Mn} \mathrm{O}_{4}\)
4 step solution
Problem 15
Resistance of \(0.2 \mathrm{M}\) solution of an electrolyte is \(50 \Omega\). The specific conductance of the solution is \(1.4 \mathrm{~S} \mathrm{~m}^{-1}\). The resistance of \(0.5 \mathrm{M}\) solution of the same electrolyte is \(280 \Omega\). The molar conductivity of \(0.5 \mathrm{M}\) solution of the electrolyte in \(\mathrm{S} \mathrm{m}^{2} \mathrm{~mol}^{-1}\) is: [Main 2014] (a) \(5 \times 10^{-4}\) (b) \(5 \times 10^{-3}\) (c) \(5 \times 10^{3}\) (d) \(5 \times 10^{2}\)
3 step solution
Problem 15
Consider the following cell reaction: [2011] \(2 \mathrm{Fe}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow 2 \mathrm{Fe}^{2+}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) ; E^{\circ}=1.67 \mathrm{~V}\) At \(\left[\mathrm{Fe}^{2+}\right]=10^{-3} \mathrm{M}, \mathrm{P}\left(\mathrm{O}_{2}\right)=0.1\) atm and \(\mathrm{pH}=3\), the cell potential at \(25^{\circ} \mathrm{C}\) is (a) \(1.47 \mathrm{~V}\) (b) \(1.77 \mathrm{~V}\) (c) \(1.87 \mathrm{~V}\) (d) \(1.57 \mathrm{~V}\)
4 step solution
Problem 16
The rusting of iron takes place as follows \(2 \mathrm{H}^{+}+2 \mathrm{e}^{-}+1 / 2 \mathrm{O}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O}(1) ; E^{\circ}=+1.23 \mathrm{~V}\) \(\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(\mathrm{s}) ; E^{\circ}=-0.44 \mathrm{~V}\) Calculate \(\Delta G^{\circ}\) for the net process (a) \(-322 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (b) \(-161 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (c) \(-152 \mathrm{~kJ} \mathrm{~mol}^{-1}\) (d) \(-76 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
6 step solution
Problem 17
The standard electrode potentials \(\left(\mathrm{E}^{\circ} \mathrm{M}^{+} / \mathrm{M}\right)\) of four metals \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) and \(\mathrm{D}\) are \(-1.2 \mathrm{~V}, 0.6 \mathrm{~V}, 0.85 \mathrm{~V}\) and \(-0.76 \mathrm{~V}\), respectively. The sequence of deposition of metals on applying potential is: [Main Online April 9, 2014] (a) \(A, C, B, D\) (b) \(\mathrm{B}, \mathrm{D}, \mathrm{C}, \mathrm{A}\) (c) \(\mathrm{C}, \mathrm{B}, \mathrm{D}, \mathrm{A}\) (d) \(\mathrm{D}, \mathrm{A}, \mathrm{B}, \mathrm{C}\)
5 step solution
Problem 18
Given \(\mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+}(\mathrm{aq}) ; \mathrm{E}^{\circ}=+0.77 \mathrm{~V}\) \(\mathrm{Al}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Al}(\mathrm{s}) ; \mathrm{E}^{\circ}=-1.66 \mathrm{~V}\) \(\mathrm{Br}_{2}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br} ; \mathrm{E}^{\circ}=+1.09 \mathrm{~V}\) Considering the electrode potentials, which of the following represents the correct order of reducing power? [Main Online April 11, 2014] (a) \(\mathrm{Fe}^{2+}<\mathrm{Al}<\mathrm{Br}\) (b) \(\mathrm{Br}^{-<\mathrm{Fe}^{2+}<\mathrm{Al}}\) (c) \(\mathrm{Al}<\mathrm{Br}^{-}<\mathrm{Fe}^{2+}\) (d) \(\mathrm{Al}<\mathrm{Fe}^{2+}<\mathrm{Br}\)
4 step solution
Problem 18
In the electrolytic cell, flow of electrons is from (a) Cathode to anode in solution (b) Cathode to anode through external supply (c) Cathode to anode through internal supply (d) Anode to cathode through internal supply
4 step solution
Problem 19
How many electrons would be required to deposit \(6.35 \mathrm{~g}\) of copper at the cathode during the electrolysis of an aqueous solution of copper sulphate? (Atomic mass of copper \(=63.5 \mathrm{u}, \mathrm{N}_{\mathrm{A}}=\) Avogadro's constant): (a) \(\frac{\mathrm{N}_{\mathrm{A}}}{20}\) (b) \(\frac{\mathrm{N}_{\mathrm{A}}}{10}\) (c) \(\frac{\mathrm{N}_{\mathrm{A}}}{5}\) (d) \(\frac{\mathrm{N}_{\mathrm{A}}}{2}\)
5 step solution
Problem 20
A solution of copper sulphate \(\left(\mathrm{CuSO}_{4}\right)\) is electrolysed for 10 minutes with a current of \(1.5\) amperes. The mass of copper deposited at the cathode (at. mass of \(\mathrm{Cu}=63 \mathrm{u}\) ) is: [Main Online April 25, 2013] (a) \(0.3892 \mathrm{~g}\) (b) \(0.2938 \mathrm{~g}\) (c) \(0.2398 \mathrm{~g}\) (d) \(0.3928 \mathrm{~g}\)
6 step solution
Problem 21
For the electrochemical cell, \(M\left|M^{+} \| X^{-}\right| X, E^{\circ}\left(M^{+} / M\right)=0.44 \mathrm{~V}\) and \(E^{\circ}\left(X / X^{-}\right)=0.33 \mathrm{~V}\) From this data one can deduce that [2000S] (a) \(M+X \rightarrow M^{+}+X^{-}\)is the spontaneous reaction (b) \(M^{+}+X^{-} \rightarrow M+X\) is the spontaneous reaction (c) \(\mathrm{E}_{\text {cell }}=0.77 \mathrm{~V}\) (d) \(\mathrm{E}_{\mathrm{ccll}}=-0.77 \mathrm{~V}\)
5 step solution
Problem 22
Electrolysis of dilute aqueous \(\mathrm{NaCl}\) solution was carried out by passing 10 milli ampere current. The time required to liberate \(0.01\) mol of \(\mathrm{H}_{2}\) gas at the cathode is \(\left(1\right.\) Faraday \(\left.=96500 \mathrm{C} \mathrm{mol}^{-1}\right)\) [2008S] (a) \(9.65 \times 10^{4} \mathrm{sec}\) (b) \(19.3 \times 10^{4} \mathrm{sec}\) (c) \(28.95 \times 10^{4} \mathrm{sec}\) (d) \(38.6 \times 10^{4} \mathrm{sec}\)
4 step solution
Problem 22
A gas \(X\) at 1 atm is bubbled through a solution containing a mixture of \(1 \mathrm{M} Y^{-}\)and \(1 \mathrm{M} Z^{-}\)at \(25^{\circ} \mathrm{C}\). If the reduction potential of \(Z>Y>X\) then, [1999 - 2 Marks] (a) \(Y\) will oxidize \(X\) and \(\operatorname{not} Z\) (b) \(Y\) will oxidize \(Z\) and \(\operatorname{not} X\) (c) \(Y\) will oxidize both \(X\) and \(Z\) (d) \(Y\) will reduce both \(X\) and \(Z\)
4 step solution
Problem 23
The correct order of equivalent conductance at infinite dilution of \(\mathrm{LiCl}, \mathrm{NaCl}\) and \(\mathrm{KCl}\) is [2001S] (a) \(\mathrm{LiCl}>\mathrm{NaCl}>\mathrm{KCl}\) (b) \(\mathrm{KCl}>\mathrm{NaCl}>\mathrm{LiCl}\) (c) \(\mathrm{NaCl}>\mathrm{KCl}>\mathrm{LiCl}\) (d) \(\mathrm{LiCl}>\mathrm{KCl}>\mathrm{NaCl}\)
5 step solution
Problem 23
The standard reduction potentials of \(\mathrm{Cu}^{2+} \mid \mathrm{Cu}\) and \(\mathrm{Cu}^{2+} \mid \mathrm{Cu}^{+}\)are \(0.337\) \(\mathrm{V}\) and \(0.153\) respectively. The standard electrode potential of \(\mathrm{Cu}^{+} \mid \mathrm{Cu}\) half cell is [1997-1 Mark] (a) \(0.184 \mathrm{~V}\) (b) \(0.827 \mathrm{~V}\) (c) \(0.521 \mathrm{~V}\) (d) \(0.490 \mathrm{~V}\)
4 step solution
Problem 24
The electric charge for electrode deposition of one gram equivalent of a substance is : [1984-1 Mark] (a) one ampere per second. (b) 96,500 coloumbs per second. (c) one ampere for one hour. (d) charge on one mole of electrons.
4 step solution
Problem 24
A dilute aqueous solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is electrolyzed using platinum electrodes. The products at the anode and cathode are: [1996-1 Mark] (a) \(\mathrm{O}_{2}, \mathrm{H}_{2}\) (b) \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}, \mathrm{Na}\) (c) \(\mathrm{O}_{2}, \mathrm{Na}\) (d) \(\mathrm{S}_{2} \mathrm{O}_{8}^{2-}, \mathrm{H}_{2}\)
3 step solution
Problem 25
Faraday's laws of electrolysis are related to the (a) atomic number of the reactants. (b) atomic number of the anion. (c) equivalent weight of the electrolyte. (d) speed of the cation.
4 step solution
Problem 26
A solution of sodium sulphate in water is electrolysed using inert electrodes. The products at the cathode and anode are respectively [1987 - 1 Mark] (a) \(\mathrm{H}_{2}, \mathrm{O}_{2}\) (b) \(\mathrm{O}_{2}, \mathrm{H}_{2}\) (c) \(\mathrm{O}_{2}, \mathrm{Na}\) (d) \(\mathrm{O}_{2}, \mathrm{SO}_{2}\)
4 step solution
Problem 27
The molar conductivity of a solution of a weak acid \(\mathrm{H} X(0.01 \mathrm{M})\) is 10 times smaller than the molar conductivity of a solution of a weak acid \(\mathrm{H} Y(0.10 \mathrm{M})\). If \(\lambda_{\mathrm{X}^{-}}^{0} \approx \lambda_{\mathrm{Y}}^{0}\) the difference in their \(\mathrm{p} K_{a}\) values, \(\mathrm{p} K_{a}(\mathrm{H} X)\) \(-\mathrm{p} K_{a}(\mathrm{H} Y)\), is (consider degree of ionization of both acids to be \(<<1\) )
5 step solution
Problem 27
The reaction : \(1 / 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{AgCl}(\mathrm{s}) \rightarrow \mathrm{H}^{+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})+\mathrm{Ag}(\mathrm{s})\) occurs in the galvanic cell (a) \(\mathrm{Ag}|\mathrm{AgCl}(\mathrm{s})| \mathrm{KCl}(\mathrm{soln})\left|\mathrm{AgNO}_{3}(\mathrm{soln})\right| \mathrm{Ag}\) (b) \(\mathrm{Pt}\left|\mathrm{H}_{2}(\mathrm{~g})\right| \mathrm{HCl}(\mathrm{soln})\left|\mathrm{AgNO}_{3}(\mathrm{soln})\right| \mathrm{Ag}\) (c) \(\mathrm{Pt}\left|\mathrm{H}_{2}(\mathrm{~g})\right| \mathrm{HCl}(\mathrm{soln})|\mathrm{AgCl}(\mathrm{s})| \mathrm{Ag}\) (d) \(\mathrm{Pt}\left|\mathrm{H}_{2}(\mathrm{~g})\right| \mathrm{KCl}(\mathrm{soln})|\mathrm{AgCl}(\mathrm{s})| \mathrm{Ag}\)
3 step solution
Problem 28
Potassium chlorate is prepared by the electrolysis of \(\mathrm{KCl}\) in basic solution \(6 \mathrm{OH}^{-}+\mathrm{Cl}^{-} \rightarrow \mathrm{ClO}_{3}+3 \mathrm{H}_{2} \mathrm{O}+6 \mathrm{e}^{-}\) If only \(60 \%\) of the current is utilized in the reaction, the time (rounded to the nearest hour) required to produce \(10 \mathrm{~g}\) of \(\mathrm{KClO}_{3}\) using a current of \(2 \mathrm{~A}\) is (Given : \(\mathrm{F}=\overline{96,500 \mathrm{C}} \mathrm{mol}^{-1} ;\) molar mass of \(\mathrm{KClO}_{3}=122 \mathrm{~g} \mathrm{~mol}^{-1}\) )
5 step solution
Problem 28
A solution containing one mole per litre of each \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2} ; \mathrm{AgNO}_{3}\); \(\mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}\); is being electrolysed by using inert electrodes. The values of standard electrode potentials in volts (reduction potentials) are : [1984-1 Mark] \(\mathrm{Ag} / \mathrm{Ag}^{+}=+0.80,2 \mathrm{Hg} / \mathrm{Hg}_{2}^{++}=+0.79\) \(\mathrm{Cu} / \mathrm{Cu}^{++}=+0.34, \mathrm{Mg} / \mathrm{Mg}^{++}=-2.37\) With increasing voltage, the sequence of deposition of metals on the cathode will be : (a) \(\mathrm{Ag}, \mathrm{Hg}, \mathrm{Cu}, \mathrm{Mg}\) (b) \(\mathrm{Mg}, \mathrm{Cu}, \mathrm{Hg}, \mathrm{Ag}\) (c) \(\mathrm{Ag}, \mathrm{Hg}, \mathrm{Cu}\) (d) \(\mathrm{Cu}, \mathrm{Hg}, \mathrm{Ag}\)
4 step solution
Problem 29
The standard reduction potentials at \(298 \mathrm{~K}\) for the following half reactions are given against each [1981-1 Mark] \(\mathrm{Zn}^{2+}(\mathrm{aq})+2 \mathrm{e} \rightleftharpoons \mathrm{Zn}(\mathrm{s}) \quad-0.762\) \(\mathrm{Cr}^{3+}(\mathrm{aq})+2 \mathrm{e} \rightleftharpoons \mathrm{Cr}(\mathrm{s}) \quad-0.740\) \(2 \mathrm{H}^{+}(\mathrm{aq})+2 \mathrm{e} \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g}) \quad 0.000\) \(\mathrm{Fe}^{3+}(\mathrm{aq})+2 \mathrm{e} \rightleftharpoons \mathrm{Fe}^{2+}(\) aq \() \quad 0.770\) which is the strongest reducing agent? (a) \(\mathrm{Zn}(\mathrm{s})\) (b) \(\mathrm{Cr}(\mathrm{s})\) (c) \(\mathrm{H}_{2}(\mathrm{~g})\) (d) \(\mathrm{Fe}^{2+}(\mathrm{aq})\)
3 step solution
Problem 30
Consider a \(70 \%\) efficient hydrogen-oxygen fuel cell working under standard conditions at 1 bar and \(298 \mathrm{~K}\). Its cell reaction is $$ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(l) $$ The work derived from the cell on the consumption of \(1.0 \times 10^{-3} \mathrm{~mol}\) of \(\mathrm{H}_{2}(g)\) is used to compress \(1.00 \mathrm{~mol}\) of a monoatomic ideal gas in a thermally insulated container. What is the change in the temperature (in \(\mathrm{K}\) ) of the ideal gas? The standard reduction potentials for the two half-cells are given below. $$ \begin{aligned} &\mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 e^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l), E^{0}=1.23 \mathrm{~V}, \\ &2 \mathrm{H}^{+}(a q)+2 e^{-} \rightarrow \mathrm{H}_{2}(g), E^{0}=0.00 \mathrm{~V} \end{aligned} $$ Use \(F=96500 \mathrm{C} \mathrm{mol}^{-1}, R=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\).
5 step solution
Problem 30
For the electrochemical cell, [Adv. 2018] \(\mathrm{Mg}(\mathrm{s})\left|\mathrm{Mg}^{2+}(\mathrm{aq}, 1 \mathrm{M}) \| \mathrm{Cu}^{2+}(\mathrm{aq}, 1 \mathrm{M})\right| \mathrm{Cu}(\mathrm{s})\) the standard emf of the cell is \(2.70 \mathrm{~V}\) at \(300 \mathrm{~K}\). When the concentration of \(\mathrm{Mg}^{2+}\) is changed to \(x \mathrm{M}\), the cell potential changes to \(2.67 \mathrm{~V}\) at \(300 \mathrm{~K}\). The value of \(x\) is (given, \(\frac{F}{R}=11500 \mathrm{~K} \mathrm{~V}^{-1}\), where \(F\) is the Faraday constant and \(R\) is the gas constant, \(\ln 10=2.30\) )
6 step solution
Problem 31
A current of \(1.70 \mathrm{~A}\) is passed through \(300.0 \mathrm{~mL}\) of \(0.160 \mathrm{M}\) solution of a \(\mathrm{ZnSO}_{4}\) for \(230 \mathrm{sec}\). with a current efficiency of \(90 \%\). Find out the molarity of \(\mathrm{Zn}^{2+}\) after the deposition of \(\mathrm{Zn}\). Assume the volume of the solution to remain constant during the electrolysis.
7 step solution
Problem 31
All the energy released from the reaction \(X \rightarrow Y, \Delta_{1} G^{\circ}=-193 \mathrm{~kJ} \mathrm{~mol}^{-1}\) is used for oxidizing \(M^{*}\) as \(M^{+} \rightarrow M^{3+}+2 \mathrm{e}^{-}, E^{\circ}=-0.25 \mathrm{~V}\) Under standard conditions, the number of moles of \(M^{+}\)oxidized when one mole of \(X\) is converted to \(Y\) is \(\left[\mathrm{F}=96500 \mathrm{C} \mathrm{mol}^{-1}\right]\)
4 step solution
Problem 32
An oxidation-reduction reaction in which 3 electrons are transferred has a \(\Delta \mathrm{G}^{0}\) of \(17.37 \mathrm{~kJ} \mathrm{~mol}^{-1}\) at \(25^{\circ} \mathrm{C}\). The value of \(\mathrm{E}_{\text {cell }}^{0}\) (in \(\mathrm{V}\) ) is \(\times 10^{-2}\) \(\left(1 \mathrm{~F}=96,500 \mathrm{C} \mathrm{mol}^{-1}\right)\)
6 step solution
Problem 33
The Gibbs energy change (in \(\mathrm{J}\) ) for the given reaction at \(\left[\mathrm{Cu}^{2+}\right]=\) \(\left[\mathrm{Sn}^{2+}\right]=1 \mathrm{M}\) and \(298 \mathrm{~K}\) is : \(\mathrm{Cu}(\mathrm{s})+\mathrm{Sn}^{2+}(\mathrm{aq}) \rightarrow \mathrm{Cu}^{2+}(\mathrm{aq})+\mathrm{Sn}(\mathrm{s})\) \(\left(\mathrm{E}_{\mathrm{Sa}^{2+} \mid \mathrm{Su}}^{0}=-0.16 \mathrm{~V}, \mathrm{E}_{\mathrm{Cu}^{2+} \mid \mathrm{Cu}}^{0}=0.34 \mathrm{~V}\right.\) Take \(\mathrm{F}=96500 \mathrm{C} \mathrm{mol}^{-1}\) )
5 step solution