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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 2701. |
`E^(c-)` of some elements are given as `:` `{:(I_(2)+2e^(-)rarr 2I^(c-),,,,E^(c-)=0.54V),(MnO_(4)^(c-)+8H^(o+)+5e^(-)rarr Mn^(2+)+4H_(2)O,,,,E^(c-)=1.52V),(Fe^(3+)+e^(-)rarr Fe^(2+),,,,E^(c-)=0.77V),(Sn^(4+)+2e^(-)rarr Sn^(2+),,,,E^(c-)=0.1V):}` `a.` Select the stronges reductant and weakes oxidant among these elements. `b.` Select the weakest reductant and strongest oxidant among these elements. |
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Answer» (a) More the `E_(OP)^(@)` more is the tendency for oxidation. Therefore, since maximum `E_(OP)^(@)` stands for: `{:(Sn^(2+)rarrSn^(4+)+2e^(-),,E_(OP)^(@)=-0.1V),( :."Strongest reductant",:,Sn^(2+)),("and weakest oxidant",:,Sn^(4+)):}` (b) More `+ve` is `E_(RP)^(@)` more is the tendency for reduction. Therefore, since maximum `E_(RP)^(@)` stands for: `{:(MnO_(4)^(-)+8H^(+)+5e^(-)rarrMn^(2+)+4H_(2)O,,E_(RP)^(@)=+1.52V),( :."Strongest oxidant",:,MnO_(4)^(-)),("and weakest reductant",:,Mn^(2+)):}` Note-Stronger is oxidant, weaker is its conjugate reducant and vice-verse. (c) For (i) `E_(Ceff) = E_(OP_(Fe^(2+)//Fe^(3+))) +E_(RP_(Sn^(4+)//Sn^(2+))) =- 0.77 +0.1` `Fe^(2+)` oxidizes and `Sn^(4+)` reduces in change. `:. E_(Ceff) = E_(OP_(Fe^(2+)//Fe^(3+))) +E_(RP_(Sn^(4+)//Sn^(2+))) =- 0.77 +0.1 =- 0.67 V` `E_(Cell)` is negative. `:.` (i) Is non-spontaneous change. For (ii) `E_(Cell) = E_(OP_(Fe^(2+)//Fe^(3+)))E_(RP_(I_(2)//I^(-))) =- 0.77 +0.54 =- 0.23 V` `:.` (ii) reaction is non-spontaneous change. For (iii) `:. E_(Ceff) = E_(OP_(I^(-)//I_(2))) +E_(RP_(Sn^(4+)//Sn^(2+))) =- 0.54 +0.1 =- 0.44V` `:.` (iii) Reaction is non-spontaneous change. For (iv) `E_(Cell) = E_(OP_(Sn^(2+)//Sn^(4+)))+E_(RP_(I_(2))//I^(-)) =- 0.1 +0.54 = +0.44V` (iv) Reaction is spontaneous change. |
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| 2702. |
A dilute aqueous solution of sodium fluoride is electrolyzed, the products at the and cathode areA. `O_(2),H_(2)`B. `F_(2),Na`C. `O_(2),Na`D. `F_(2),H_(2)` |
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Answer» Correct Answer - a `NaF:` Cathode `: 2H^(o+)+2e^(-) rarr H_(2)` Anode`: 4 overset(c-)(O)H rarr O_(2) +2H_(2)O+4e^(-)` |
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| 2703. |
Given the standard electrode potentials `K^(+)//K=-2.93V,Ag^(+)//Ag=0.80V`, `Hg_(2)^(2+)//Hg=0.79V,Mg^(2+)//Mg=-2.37V,Cr^(2+)//Cr=-0.74V` Arrange these metals in their increasing order of reducing power. |
| Answer» Higher the oxidation potential, more easily it is oxidized and hence greater is the reducing power. Thus, increasing order of reducing power will be `Ag lt Hg lt Cr lt Mg lt K`. | |
| 2704. |
For a cell reaction involvinig a two electron change, the standard emf of the cell is found to be 0.295 V at `25^(@)` C. The equilibrium constant of the reaction at `25^(@)C` will be:A. `1xx10^(-10)`B. `29.5xx10^(-2)`C. 10D. `1xx10^(10)` |
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Answer» Correct Answer - D `E^(@)=(0.0591)/(n)"log" k( "At" 25^(@)C)` here n=2 and `E^(@)=0.295V` `therefore 0.295=(0.0591)/(2)"log"K` `"log"K=(0.590)/(0.0591)=10` `K =10^(10)` |
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| 2705. |
Given standard electrode potentials `K^(o+)|K=-2.93V, Ag^(o+)|Ag=0.80V`, `Hg^(2+)|Hg=0.79V``Mg^(2+)|Mg=-2.37V,Cr^(3)|Cr=-0.74V` Arrange these metals in their increasing order of reducing power. |
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Answer» It may be noted that lesser the `E^(@)` value for an electrode, more will be reducing power. The increasing order of reducing power is : `Ag^(+)//Ag lt H^(2+)//Hg lt Cr^(3+)//Cr lt Mg^(2+)//Mg lt K^(+)//K` |
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| 2706. |
Stadard reduction electrode potentials of three metals A,B and C are respectively `+0.5V, -3.0V` and `-1.2V`. The reducing powers of these metals are:A. `BgtCgtA`B. `AgtBgtC`C. `CgtBgtA`D. `AgtCgtB`. |
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Answer» Correct Answer - A Higher the R.P. easier to reduce hence b etter O.A. thus species with higher reducing power (better R.A.) has a lower R.P. Therefore, correct order of reducing power is `BgtCgtA` |
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| 2707. |
Consider the following standard reduction potentials:- `{:(Fe^(+)+2e^(-)hArrFe,,,E^(@) =- 0.41V),(Ag^(+)+e^(-)hArrAg,,,E^(+)=0.80V),(O_(2)+2H_(2)O+4e^(-)hArr4OH^(-),,,E^(@)=0.40V):}` What would happens if a block of silver metal is connected to a buried iron pipe via a wire:-A. The silver metal would corrode, a current would be produced in the wire, and `O_(2)` would be reduced on the surface of the iron pipe.B. The silver metal would corrode, a current would be produced in the wire, and `Fe^(2+)` would be reduced on the surface of the iron pipe.C. The iron pipe would corrode, a current would be produced in the wire, and `Ag^(+)` would be reduced on the surface of the silver metal.D. The iron pipe would corrode, no current would be produced in the wire, and `O_(2)` would be reduced on the surface of the iron pipe. |
| Answer» Correct Answer - D | |
| 2708. |
Serveral blocks of magnesium are fixed to the bottom of a ship toA. keep away the sharksB. make the ship lighterC. prevent action of water and saltD. prevent puncturing by under-sea rocks. |
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Answer» Correct Answer - C Magnesium with lower R.P. than iron. Therefore, when in contact with iron it will protect iron from corroison due to salt and water. |
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| 2709. |
The standard reduction potentials at 298K, for the following half cells are given: `Zn^(2+)(aq)+2e^(-)hArrZn(s):E^(@)=-0.762V` `Cr^(3+)(aq)+3e^(-)hArrCr(s):E^(@)=-0.740V` `2H^(+)(aq)+2e^(-)hArrH_(2)(g), E^(@)=0.000V` `Fe^(3+)(aq)+e^(-)hArrFe^(2+)(aq),E^(@)=0.770V` Which is the stronget reducing agent?A. Zn(s)B. Cr(s)C. `H_(2)(g)`D. `Fe^(2+)(aq)` |
| Answer» Correct Answer - a | |
| 2710. |
The standard reduction potentials at 298K, for the following half cells are given: `Zn^(2+)(aq)+2e^(-)hArrZn(s):E^(@)=-0.762V` `Cr^(3+)(aq)+3e^(-)hArrCr(s):E^(@)=-0.740V` `2H^(+)(aq)+2e^(-)hArrH_(2)(g), E^(@)=0.000V` `Fe^(3+)(aq)+e^(-)hArrFe^(2+)(aq),E^(@)=0.770V` Which is the stronget reducing agent?A. Zn(s)B. Cr(s)C. `H_(2) (s)`D. `Fe^(2+) (aq)` |
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Answer» Correct Answer - A Higher the oxidation potential (or lower the reduction potential ) stronger the reducing agent . |
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| 2711. |
The standard reduction potentials at 298K, for the following half cells are given: `Zn^(2+)(aq)+2e^(-)hArrZn(s):E^(@)=-0.762V` `Cr^(3+)(aq)+3e^(-)hArrCr(s):E^(@)=-0.740V` `2H^(+)(aq)+2e^(-)hArrH_(2)(g), E^(@)=0.000V` `Fe^(3+)(aq)+e^(-)hArrFe^(2+)(aq),E^(@)=0.770V` Which is the stronget reducing agent?A. Zn(s)B. Cr(s)C. `H_(2)(g)`D. `Fe^(2+)(aq)`. |
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Answer» Correct Answer - A Strength reducing agent is the one with highest oxidation potential lowest reduction potential. |
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| 2712. |
Which of the following reaction is reaction is used to make a fuel cell .A. `Cd(s)+2Ni(OH)_(3)(s)rarrCuO(s)+ 2Ni(OH)_(@)+H_(2)O(l)`B. `Pb(s)+PbO_(2)(s)+ 2H_(2)SO_(4)(aq)rarr2PbSO_(4)(s)+2H_(2)O(l)`C. `2H_(2)(g) +O_(2)(g)rarr2H_(2)O(l)`D. `2Fe(s)+O_(2)(g)+4H^(+) (aq)rarr2Fe^(2+)(aq)+2H_(2)O(l)` |
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Answer» Correct Answer - C Chemical reaction in a fuel cell is `2H_(2)(g)+O_(2)(g)rarr2H_(2)O(l)` |
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| 2713. |
The standard reduction potentials of Zn and Ag in water at 298K are. `Zn^(2+)+2e^(-)hArrZn,E^(@)=-0.76V` and `Ag^(+)+e^(-)hArrAg:E^(@)=+0.80V` Which of the following reactions take place?A. ` Zn^(2+)(aq)+Ag^(+)(aq)rarrZn(s)+Ag(s)`B. `Zn(s)+Ag(s)rarrZ^(2+)(aq)+Ag^(+)(aq)`C. `Zn^(2+)(aq)+2Ag(s)rarr2Ag^(+)(aq)+Zn(s)`D. `Zn(s)+2Ag^(+)(aq)rarrZn^(2+)(aq)+2Ag(s)` |
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Answer» Correct Answer - D More the R.P. easier to reduce. Thus `Ag^(+)` will be reduced and Zn will be oxidised. `Zn(s)+2Ag^(+)(aq)rarrZn^(2+)(aq)+2Ag(s)` |
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| 2714. |
In the electroysis of an aqueous solution of NaOH, 2.8 litre of oxygen gas at NTP was libreated at anode. How much of hydrogen gas was liberated at cathode ?A. 2.8 litreB. 5.6 litreC. 11.2 litreD. 22.4 litre |
| Answer» Correct Answer - B | |
| 2715. |
Two metals `A` and `B` have `E_(RP)^(@) = +0.76V` and `-0.80V` respectively. Which will liberated `H_(2)` from `H_(2)SO_(4)` ? |
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Answer» Given , For `A, A^(n+)+n e rarr A, E_(RP)^(@) = +0.76V` For `B, B^(n+)+n e rarr B, E_(RP)^(@) = -0.80V` We have, for `H, H^(+)+e rarr 1//2H_(2), E_(RP)^(@) = 0` Now coupling `A` with `H_(2)SO_(4)` : `2A + nH_(2)SO_(4) rarr A_(2)(SO_(4))_(n)+nH_(2)` `E_(Cell)^(@) = E_(OP_(A))^(@)+E_(RP_(H))^(@) = -0.76 + 0.0 = -0.76V` Since `E^(@)` is `+ve`, `2A + nH_(2)SO_(4) rarr A_(2)(SO_(4))_(n)+nH_(2)` Reaction is non-spontaneous. A will not liberate `H_(2)` from `H_(2)SO_(4)`. Now coupling `B` with `H_(2)SO_(4)` `2B+nH_(2)SO_(4) rarr B_(2)(SO_(4)(n)+nH_(2)` `E_(Cell)^(@) = E_(OP_(B))^(@)+E_(RP_(H))^(@) = +0.80 + 0 = +0.80` Since `E^(@)` is `+ve`, `:. 2B+nH_(2)SO_(4) rarr B_(2)(SO_(4))_(n)+nH_(2)` Reaction will be spontaneous. `B` will liberate `H_(2)` from `H_(2)SO_(4)`. |
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| 2716. |
In the electrolysis of an aqueous solution of `NaOH`, `2.8` litres of `O_(2)(g)` is liberated at the anode at `NTP`. Volume of hydrogen gas liberated at the cathode at `NTP` will beA. `5.6`B. `6.5`C. `22.2`D. `11.2` |
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Answer» Correct Answer - A 96500 or 1 F will liberate 1 eq. of `O_(2) = 22.4 dm^(3)` (Lit) or 1/4 mole `O_(2)` or 5.6 litre `O_(2)` at NTP . |
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| 2717. |
The reduction electrode potential E , or 0.1 M solution of `M^(+)` ions (`E_(RP)^(@) = -2.36` V) isA. `-2.41`B. `+2.41`C. `- 4.82`D. None |
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Answer» Correct Answer - A `E = E_(SRP)^(@) + (0.0592)/(2) log_(10) 0.1` `= -2.36 + 0.0592 xx (-1) = -2.41 V` |
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| 2718. |
Assertion: A negative value of standard reduction potential means that reduction takes place on this electrode with reference to standard hydrogen electrode. Reason: The standard electrode potential of a half cell has a fixed value.A. if both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is falseD. If assertion is false but reason is true. |
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Answer» Correct Answer - D A negative value of standard reduction potential means that oxidation takes place ont his electrode with reference to SHE. |
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| 2719. |
A certain electrode has a standard reduction potential of `0.140 V` when measure against a saturaed calomel electrode `(E_(RP)^(@) = + 0.244 V)`. Calculate its standard reduction potential against a standard `H`-electrods. |
| Answer» Correct Answer - `0.384 V ;` | |
| 2720. |
During electrolysis of fused aluminium chloride 0.9gm of aluminium was deposited on the cathode. Thevolume of chlorine liberated at the anode will beA. 2.24 litresB. 11.2 litresC. 1.12 litresD. 5.6 litres |
| Answer» Correct Answer - A | |
| 2721. |
Some quantity of electricity being used to liberate iodine (at anode) and a metal `x` (at cathode). The mass of `x` liberated at cathode is `0.617 g` and the liberated iodine completely reduced by `46.3 mL` of `0.124 M` sodium thio-sulphate solution. what is equivalent weight of metal ? |
| Answer» Correct Answer - `107.47 ;` | |
| 2722. |
`100mL CuSO_(4)(aq)` was electrolyzed using inert electrodes by passing `0.965A `till the `pH` of the resulting solution was 1. The solution after electrollysis was neutralized, treated with excess `KI` and titrated with `0.04M Na_(2)S_(2)O_(3)`. Volume of `Na_(2)S_(2)O_(3)` required was `35mL`. Assuming no volume change during electrolysis, calculate: (a) duration of electrolysis if current efficiency is `80%` (b) initial concentration `(M)` of `CuSO_(4)`. |
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Answer» Correct Answer - `1250 S, 0.064M` `2H_(2)O rarr 4H^(+) +O_(2) +4e^(-)` `n_(H^(+)) =10^(-2) n_(E) = 10^(-2)` `n_(E)` actual `= (0.10 xx 100)/(80) rArr 0.0125` `I = (q)/(t)` `0.965 = (0.0125 xx 96500)/(t)` `t =(0.0125 xx 96500)/(0.965) rArr 1250 S` `2CuSO_(4) +4Ki rarr 2CuI +I_(2) +2K_(2)SO_(4)` `I_(2) +2Na_(2)S_(2)O_(3) rarr 2NaI +Na_(2)S_(4)O_(6)` Mili moles of `Na_(2)S_(2)O_(3) = 0.04 xx 35 rArr 1.4` Mili moles of `I_(2) = (1.4)/(2)` Mili moles of `CuSO_(4) = 2xx (1.4)/(2) rarr 1.4` `Cu^(2+) +2e^(-) rarr Cu` `n_(CU^(2+)) = (1)/(2) xx 10^(-2)|n_(E) = 10^(-2)` Total moles of `Cu^(2+) = 5 xx 10^(-3) +14.2 xx 10^(-3)` = total moles of `Cu^(2+) = 5 xx 10^(-3) +1.2 xx 10^(-3)` `= 6.42 xx 10^(-3)` `=(6.42 xx 10^(-3))/(100) = 0.064M` |
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| 2723. |
Certain cells are given below.1. In Daniel cell, if Zn electrode is replaced by Mg what happens? Why?2. Zinc displaces copper from copper sulphate solution while copper doesn’t replace zinc from zinc sulphate solution. Justify your answer. |
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Answer» 1. If Zinc electrode is replaced by Mg, a greater voltage (> 1.1 V) will be produced by the cell. Because Mg lies above Zn in electrochemical series. \(E^\circ\)(Mg2+Mg) = -2.36 V \(E^\circ\)(Zn2+Zn) = -0.76 V i. e., Mg is a stronger reducing agent than Zn. 2. Zinc lying above hydrogen in the electrochemical series has relatively greater tendency to lose electrons. Hence Zn is capable of displacing hydrogen from acids. Moreover, a more active metal Zn, displaces, a less reactive copper from the salt solution. But Cu, being less active than Zn cannot displace Zn from its salt, ZnSO4. |
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| 2724. |
A certain electrode has standard `(` reduction potential `)` of `0.384 V`. The potential when measured against a normal calomel electrode `(` with electrode potential`=0.28V)` isA. `0.104`B. `0.664`C. `0.3322`D. `0.218` |
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Answer» Correct Answer - a `E=0.384-0.28=0.104V` |
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| 2725. |
A saturated calomel electrode is coupled through a salt bridge with a quinhydrone electrode dipping in `0.1 M NH_(4)Cl`. The observed `EMF` at `25^(@)C` is `0.152V`. Find the dissociated constant of `NH_(4)OH`. The oxidation potential of saturated calomel electrode`=-0.699V` at `25^(@)C`. |
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Answer» For the cell`:` Saturated calomel `||0.1 MNH_(4)Cl|H_(2)Q,Q|Pt` `E_(cell)=E_(qui nhydron e el etrode)-E_(saturated calomel)` `E_(cell)=0.152,,,,[{:(E_(saturated calomel(o x i d)),=,-0.242V),(E_(saturated calomel (red)),=,-0.242V):}]` `(E_(qui nhydron e)=0.6994-0.059pH)` `( :. E^(c-)._(o x i de)=-0.6994,so E^(c-)._(Red)=0.6994)` `0.152=0.6994-0.059pH-0.242` `:. pH=(0.457-0.152)/(0.059)=5.17` For `0.1 M NH_(4)Cl,` salt of `S_(A)//W_(B)` which on hydrolysis gives `NH_(4)OH` `pH=(1)/(2)(pK_(w)=pK_(b)-logc)` `5.17=(1)/(2)(14-pK_(b)-log0.1) Solving , we get `K_(b)=2.18xx10^(-5)` |
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| 2726. |
The quinhydrone electrode `(Q,H^(o+)|H^(2)Q)` is used in conjunction with a saturated calomel electrode, as represented below`:` `EMF_(cell)=0.264V` at `30^(@)C`. Calculate the `pH` of unknown solution at this temperature. Given `:E_(calomel)=0.24V` and `E^(c-)._(2H^(o+),Q|H_(2)Q)=0.7V` |
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Answer» `(` Note that the value `2.303(RT)/(F)`at`30^(@)C=0.06)` `E_(S C E )=0.242V` `E_(qui nhydron e)=E^(c-)._(qui n hydron e )-0.06pH(`at`30^(@)C)` `:. E_(cell)=E_(RHS)-E_(LHS)=E_(qui n hydron e)-E_(SCE)` `0.264V=0.7-0.06pH-0.24(` at` 30^(@)C)` `:. pH=(0.7-0.24-0.264)/(0.06)=3.266~~3.27` |
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| 2727. |
Same quantity of charge is being used to liberate iodine (at anode) and a metal `M` (at cathode). The mass of metal `M` liberated is `0.617 g` and the liberated iodine is completely reduced by `46.3 mL` of `0.124 M` sodium thio-sulphate. Calculate equivalent weight of metal. Also calculate the total time to bring this change if `10` ampere current passed through solution of metal iodide. |
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Answer» Correct Answer - Eq.wt `=107.468` `I_(2) +2Na_(2)S_(2)O_(3) rarr 2NaI +Na_(2)S_(4)O_(6)` Mili moles of `Na_(2)S_(2)O_(3) = 46.3 xx 0.124 = 5.7412` Mili mole of `I_(2) = 2.8706` `2I^(-) rarr I_(2) +2e^(-)` `n_(E) = 2n_(I_(2)) = 5.7412 xx 10^(-3)` `5.7412 xx 10^(-3)` moles `e^(-)` deposited `= 0.617g` So 1 mole `e^(-)` will deposite `= (0.617)/(5.7412 xx 10^(-3))` `= 107.468g` Eq.wt `=107.468g` |
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| 2728. |
During the discharge of a lead storage battery, the density of sulphuric acid fell from `1.294 g mL^(-1)` to `1.139 g mL^(-)`. Sulphuric acid of density `1.294 g mL^(-1)` is `39%` by weight and that of density `1.139 g mL^(-1)` is `20%` by weight. The battery hold `3.5` litre of acied and discharge. Calculate the no. of ampere hour for which the battery must have been used. The charging and discharging reactions are: `Pb+SO_(4)^(2-) rarr PbSO_(4)+2e` (charging) `PbO_(2)+4H^(+)+SO_(4)^(2-)+2e rarr PbSO_(4)+2H_(2)O` (discharging) |
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Answer» Correct Answer - `265 Amp.hr` The overall battery reaction is `Pb +PbO_(2) +2H_(2)SO_(4) = 2PbSO_(4) +2H_(2)O` `:.` two moles of electrons are involved for the reaction of two moles of `H_(2)SO_(4)`. `:.` eq.wt.of `H_(2)SO_(4) =mol.wt.of H_(2)SO_(4) = 98` no.of eq.of `H_(2)SO_(4)` present in `3.5 L` of solution of a charged battery `= (39)/(98) xx (1.294)/(100) xx 3500 = 18.0235` No. of equivalents of `H_(2)SO_(4)` present in `3.5L` of solution after getting discharged `=(20)/(98) xx (1.139)/(100) xx 3500 = 8.1357` Number of eq. of `H_(2)SO_(4)` lost `(H_(2)SO_(4) =18.0235 -8.1357 = 9.8878)` `:.` moles of electric charge produced by the battery `=9.8878F` `= 9.8878 xx 96500` coulombs `=9.8878 xx 96500` amp-seconds `= (9.8878 xx 96500)/(60 xx 60)` amp-hours `= 265` amp-hours. |
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| 2729. |
The standard EMF of a Daniell cell is `1.10 ` volt. The maximum electrical work obtained from the Daniell cell is .A. 212.3kJB. 175.4kJC. 106.15kJD. 53.07kJ |
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Answer» Correct Answer - A Electrical work obtained `=nFE_("cell")^(@)` `=2xx96500xx1.1J=212.3KJ`. |
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| 2730. |
The standard reduction potential of normal calomel electrode and reduction potential of saturated calomel electrodes are 0.27 and 0.33 volt respectively. What is the concentration of `Cl^-` in saturated solution of KCL?A. 0.1MB. 0.01MC. 0.001MD. None |
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Answer» Correct Answer - A |
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| 2731. |
The e.m.f. of the cell obtained by combining `Zn` and `Cu` electrode of a Daniel cell with N calomel electrode in two different arrangements are `1.083V` and `0.018V` respectively at `25^(@)C`. If the standard reduction potential of N calomel electrode is `0.28V` find the emf og Daniel cell. |
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Answer» Correct Answer - `E = 1.1V` `E_(cell) = E_(ZN//Zn^(2+)) +E_(calomel(Red))` `E_(ZN//Zn^(2+)) = 0.803` `E_(cell) = E_(Cu^(2+)//Cu)` `E_(cell) = 1.1V` |
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| 2732. |
The EMF of a cell formed by combining a particular electrode with standard calomel electrode is found to be 0.344 V and calomel electrode is found to act as cathode. If the same electrode is combined with standard hydrogen electrode, the EMF of the cell will be (Given standard reduction potential, `E_("calomel")^(@)=+0.244V`)A. 0.344VB. 0.244 VC. 0.588 VD. 0.100 V |
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Answer» Correct Answer - D `E_(cell)=E_("cathode")^(@)-E_("anode")^(@)` `0.344=E_("calomel")^(@)-E_(X)^(@)` `0.344=0.244-E_(X)^(@)` or `E_(X)^(@)=-0.100V` As `E_(H^(+)//H_(2))^(@)=0,E_(cell)^(@)` with SHE=0.100V. |
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| 2733. |
The emf of the cells obtained by combining zinc and copper electrodes of the Daniell cell with calomel electrodes are 1.083 volt and -0.018 volt respectively at `25^(@)C`. If the reduction potential of normal calomel electrode is + 0.28 volt, find the emf of the Daniell cell. |
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Answer» For the cell Zn electrode || calomel electrode `E_(cell)=` Oxid. Pot. Of Zn electrode + Red. Pot. Of calomel electrode So, oxid. Pot. Of Zn electrode `=1.083-0.28=0.803` volt For the cell, Cu electrode || Calomel electrode `E_(cell)=` Oxid. pot. of Cu electrode + Red. pot. of calomel electrode So, oxid. pot. of Cu electrode=- 0.018-0.28=-0.298 volt for the daniell cell, Zn electrode||Cu electrode `E_(cell)=` Oxid. pot. of Zn electrode + Red. pot. of copper electrode `=0.803+0.298=1.101` volt |
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| 2734. |
Which of the following cells has a constant voltage through-out its life?A. Leclanche cellB. Daniell cellC. Mercury cellD. Electrolytic cell |
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Answer» Correct Answer - C Because there is no change in electrolyte composi-tion during operation (the overall cell reaction involves only solid substances) the mercury cell provides a more constant voltage `(1.35 V)` than other cells. |
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| 2735. |
The specific conductance of a `0.1 N KCl` solution at `23^(@)C` is `0.012 ohm^(-1) cm^(-1)`. The resistance of cell containing the solution at the same tempreature was found to be `55 ohm`. The cell constant will beA. `0.142 cm^(-1)`B. `0.66 cm^(-1)`C. `0.918 cm^(-1)`D. `1.12 cm^(-1)` |
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Answer» Correct Answer - B The specific conductance (or conductivity) of the solution is given by the equation : `kappa = (1)/(c) xx` cell constant Thus, the cell constant `(G^(**))` is given by the equation: `(G^(**)) = R kappa` `= (55 "ohm")(0.012 "ohm"^(-1)cm^(-1))` `= 0.66 cm^(-1)` |
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| 2736. |
The specific conductance of a `0.1 N KCl` solution at `23^(@)C` is `0.012 ohm^(-1) cm^(-1)`. The resistance of cell containing the solution at the same tempreature was found to be `55 ohm`. The cell constant will beA. `0.142cm^(-1)`B. `0.66cm^(-1)`C. `0.918cm^(-1)`D. `1.12cm^(-1)` |
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Answer» Correct Answer - B Specific conductance =Conductance xx Cell constant 0.012`ohm^(-1) cm^(-1)=(1)/(55)ohmxx` Cell constant ohm xx Cell constant Cell constant `=53xx0.012cm^(-1)=0.66cm^(-1)` |
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| 2737. |
The specific conductance of a `0.1 N CKI` solutin at `23^@ C ` is `0.012 "ohm"^(-1) cm^(-1)`. The resistance of cell containing the solution at the same trmprature was contaning the solution at the same temperature was found to be `55` ohm. The cell constant will be .A. ` 0. 142 cm^1`B. ` 0.918 cm^1`C. ` 0. 66 cm^1`D. ` 1.12 cm^1` |
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Answer» Correct Answer - C ` K=1/K xx` Cell constant. Cell constant `= K R, 0.012 xx5 = 0. 66 cm^(-1)` . |
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| 2738. |
The specific conductance of a `0.1 N KCl` solution at `23^(@)C` is `0.012 ohm^(-1) cm^(-1)`. The resistance of cell containing the solution at the same tempreature was found to be `55 ohm`. The cell constant will beA. `0.142 cm^(-1)`B. `0.918 cm^(-1)`C. `1.12 cm^(-1)`D. `0.616 cm^(-1)` |
| Answer» Correct Answer - D | |
| 2739. |
At `25^(@)C`, the standard emf of a cell having reaction involving two electrons change is found to be 0.295 V. The equilibrium constant of the reaction is :A. `29.5 xx 10^(-2)`B. 10C. `1 xx 10^(10)`D. `29.5 xx 10^(10)` |
| Answer» Correct Answer - C | |
| 2740. |
Two half-reactions of an electrochemical cell are given below : `MnO_(4)^(-)(aq)+8H^(+)(aq)+5e^(-) toMn^(2+)(aq)+4H_(2)O(l), E^(@)=+1.51" V"` `Sn^(2+)(aq) to Sn^(4+)(aq)+2e^(-),E^(@)=-0.15" V"` |
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Answer» Multiply the first equation by 2 and second equation by 5. Add up these equations. `2MnO_(4)^(-)(aq)+16H^(+)(aq)+10e^(-) to 2Mn^(2+)(aq)+8H_(2)O(l),E_(cell)^(@)=+1.51" V"` `(5Sn^(2+)(aq) to 5Sn^(4)(aq)+10e^(-),E^(@)=-0.15" V")/(2MnO_(4)^(-)(aq)+16H^(+)(aq)+5Sn^(2+)(aq) to 2Mn^(2+)(aq)+5Sn^(4+)(aq)+8H_(2)O(l))` `E^(@)=(1.51-0.15)=1.36" V"` Since `E_(cell)^(@)` comes out to be positive, the reaction is reactant favoured. It is likely to proceed in the backward direction. |
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| 2741. |
The molar conductance of NaCl, HCl and `CH_(3)COONa` at infinite dilution are 126.45, 426.16 and 91 `ohm^(-1) cm^(2) mol^(-1)` respectively. The molar conductance of `CH_(3)COOH` at infinite dilution is :A. `201.28 ohm^(-1) cm^(2) mol^(-1)`B. `390.71 ohm^(-1) cm^(2) mol^(-1)`C. `698.28 ohm^(-1) cm^(2) mol^(-1)`D. `540.48 ohm^(-1) cm^(2) mol^(-1)` |
| Answer» Correct Answer - B | |
| 2742. |
Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox titration. Some half cell reaction and their standard potentials are given below: `MnO_(4)^(-)(aq) +8H^(+)(aq) +5e^(-) rarr Mn^(2+)(aq) +4H_(2)O(l) E^(@) = 1.51V` `Cr_(2)O_(7)^(2-)(aq) +14H^(+) (aq) +6e^(-) rarr 2Cr^(3+)(aq) +7H_(2)O(l), E^(@) = 1.38V` `Fe^(3+) (aq) +e^(-) rarr Fe^(2+) (aq), E^(@) = 0.77V` `CI_(2)(g) +2e^(-) rarr 2CI^(-)(aq), E^(@) = 1.40V` Identify the only incorrect statement regarding quantitative estimation of aqueous `Fe(NO_(3))_(2)`A. `MnO_(4)^(-)` can be used in aqueous `HCI`B. `Cr_(2)O_(7)^(2-)` can be used in aqueous `HCI`C. `MnO_(4)^(-)` can be used in aqueous `H_(2)SO_(4)`D. `Cr_(2)O_(7)^(2-)` can be used in aqueous `H_(2)SO_(4)` |
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Answer» Correct Answer - A `2MnO_(4)^(Theta) +16H^(o+) +10CI^(Theta) rarr 5CI_(2)(g) +2Mn^(2+) 1+8H_(2)O` Cell constituted is `Pt|CI_(2)|CI^(Theta) ||MnO_(4)^(-) |Mn^(2+), H^(+)|Pt` `E_(cell) = E_(MnO_(4)^(Theta)//Mn^(2+))^(Theta) -E_(CI_(2)//CI^(Theta))` `= 1.51-1.40 = 0.11V` Since the cell voltage is +ve, hence reaction will take place. `MnO_(4)^(Theta)` will oxidise both `Fe^(+2)` and `HCI`. |
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| 2743. |
Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox titration. Some half cell reaction and their standard potentials are given below `MnO_(4)^(-)(aq)+8H^(+)(aq)+5e^(-)toMn^(2+)(aq)+4H_(2)O(l)" "E^(@)=1.51V` `Cr_(2)O_(7)^(2-)(aq)+14H^(+)(aq)+6e^(-)to2Cr^(3+)(aq)+7H_(2)O(l)" "E^(@)=1.38V` `Fe^(3+)(aq)+e^(-)toFe^(2+)(aq)" "E^(@)=0.77V` `Cl_(2)(g)+2e^(-)to2Cl^(-)(aq)" "E^(@)=1.40V` Identify the only incorrect statement regarding the quantitative estimation of aqueous `Fe(NO_(3))_(2)`A. `MnO_(4)^(-)` can be used in aqueous HClB. `Cr_(2)O_(7)^(2-)` can be used in aqueous HClC. `MnO_(4)^(-)` can be used in aqueous `H_(2)SO_(4)`D. `Cr_(2)O_(7)^(2-)` can be used in aqueous `H_(2)SO_(4)` |
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Answer» Correct Answer - A `MnO_(4)^(-)` will oxidize `Cl^-` ion according to the equation, `2MnO_(4)^(-)+16H^(+)+10Cl^(-)to2Mn^(2+)+8H_(2)O+5Cl_(2)uarr` The cell corresponding to this reaction is as follows: `Pt,Cl_(2)("1 atm")|Cl^(-)||MnO_(4)^(-), Mn^(2+),H^(+)|Pt,` `E_(cell)^(o)=1.51-1.40=0.11V` `E_(cell)^(o)` bring +ve, `DeltaG^(o)` will be -ve and hence the above reaction is feasible. `MnO_(4)^(-)` will not only oxidize `Fe^(2+)` ion but also `Cl^(-)` ion simultaneously. |
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| 2744. |
Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox titration. Some half cell reaction and their standard potentials are given below: `MnO_(4)^(-)(aq) +8H^(+)(aq) +5e^(-) rarr Mn^(2+)(aq) +4H_(2)O(l) E^(@) = 1.51V` `Cr_(2)O_(7)^(2-)(aq) +14H^(+) (aq) +6e^(-) rarr 2Cr^(3+)(aq) +7H_(2)O(l), E^(@) = 1.38V` `Fe^(3+) (aq) +e^(-) rarr Fe^(2+) (aq), E^(@) = 0.77V` `CI_(2)(g) +2e^(-) rarr 2CI^(-)(aq), E^(@) = 1.40V` Identify the only correct statement regarding quantitative estimation of aqueous `Fe(NO_(3))_(2)`A. `MnO_(4)^(-)` can be used in aqueous HCLB. `Cr_(2)O_(7)^(2-)` can be used in aqueous HClC. `MnO_(4)^(-)` can be used in aqueous `H_(2)SO_(4)`D. `Cr_(2)O_(7)^(2-)` can be used in aqueous `H_(2)SO_(4)` |
| Answer» Correct Answer - a | |
| 2745. |
Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox titration. Some half cell reaction and their standard potentials are given below: `MnO_(4)^(-)(aq) +8H^(+)(aq) +5e^(-) rarr Mn^(2+)(aq) +4H_(2)O(l) E^(@) = 1.51V` `Cr_(2)O_(7)^(2-)(aq) +14H^(+) (aq) +6e^(-) rarr 2Cr^(3+)(aq) +7H_(2)O(l), E^(@) = 1.38V` `Fe^(3+) (aq) +e^(-) rarr Fe^(2+) (aq), E^(@) = 0.77V` `CI_(2)(g) +2e^(-) rarr 2CI^(-)(aq), E^(@) = 1.40V` Identify the only correct statement regarding quantitative estimation of aqueous `Fe(NO_(3))_(2)`A. `MnO_(4)^(c-)` can be used in aqueous `HCl`B. `CrO_(4)^(2-)` can be used in aqueous `HCl`C. `MnO_(4)^(c-)` can be used in aqueous `H_(2)SO_(4)`D. `Cr_(2)O_(7)^(2-)` can be used in aqueous `H_(2)SO_(4)` |
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Answer» Correct Answer - a The reaction between `MnO_(4)^(c-)` and `HCl` may be represented as follows`:` `2MnO_(4)^(c-)(aq)+16H^(o+)+10Cl^(c-) rarr 2Mn^(2+)(aq)+8H_(2)O(l) +5Cl_(2)(g)` Thus, on the basis of this reaction following electrochemical cell will be represented `:` `Pt, Cl_(2)(g)(1atm)|Cl^(c-)(aq)||MnO_(4)^(c-)(g)|Mn^(2+)(aq)` Hence, `E^(c-)._(cell)=E^(c-)._(cathode)-E^(c-).(anode)` From the given data , `E^(c-)._(cell)=1.51-1.40=0.11V` `E^(c-)` cell is positive, hence `DeltaG^(c-)` is negative. Thus, the above cell reaction is feasible but `MnO_(4)^(c-)` ion can oxidize, `Fe^(2+)` to `Fe^(3+)` and `Cl^(c-)` to `Cl_(2)` in aqueous medium also . Therefore, for quantitative estimation of aqueous `Fe(NO_(3))_(2)` it is not a suitable reagent. |
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| 2746. |
Assertion: Molar conductivity of a weak electrolyte at infinite dilution cannot be determined experimentally. Reason: Kohlrausch law helps to find molar conductivity of a weak electrolyte at infinite dilution.A. If both assertion and reason are true, and reason is the true explanation of the assertionB. if both assertion and reason are true, but reason is not the true explanation of the assertion.C. if assertion is true, but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - B Correct explanation. The dissociation of weak electrolyte keeps increasing with dilution and is not complete even infinite dilution. |
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| 2747. |
The molar conductance of NaCl, HCl and `CH_(3)COONa` at infinite dilution are 126.45, 426.16 and 91 `ohm^(-1) cm^(2) mol^(-1)` respectively. The molar conductance of `CH_(3)COOH` at infinite dilution is :A. `201. 28 ohm^(-1) cm^(2)`B. `390 . 71 ohm ^(-1) cm^(2)`C. `698.28 ohm^(-1) cm^(2)`D. `540 ohm^(-1) cm^(2)`. |
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Answer» Correct Answer - B `Lambda_(C_(2) H_(5) COOH)^(@) = Lambda_(C_(2) H_(5) COONa)^(@) + Lambda_(HCl)^(@) - Lambda_(NaCl)^(@)` `= 91 + 426.16 - 126. 45` `= 390. 71 ohm^(-1) cm^(2)` . |
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| 2748. |
Molar conductivity of a 1.5 m solustion of an electrolyte is found to be 138.9 S `cm^(2) mol^(-1)`. Calculate the conductivity of this solution. |
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Answer» Correct Answer - `2.083xx10^(-1) S cm^(-1)` `Lambda_(m)=(1000xxk)/(M),k=(Lambda_(m)xxM)/(1000)=((138.95" cm"^(2)mol^(-1))xx(1.5" mol" " cm"^(-3)))/(1000)=2.083xx10^(-1)" S "cm^(-1)`. |
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| 2749. |
The molar conductance of `CH_(3)COOHNa` HCI and NaCI at infinite dilutions are 91.0, 426.0 and 126.0 `ohm^(-1) cm^(-1)` respectively. Calculate the molar conductence of acetic acid at infinie dilution. |
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Answer» Correct Answer - `391 ohm^(-1)cm^(2)mol^(-1)` `Lambda_(m(CH_(3)COOH))^(oo)=Lambda_(m(CH_(3)COONa))^(oo)+Lambda_(m(HCl))^(oo)-Lambda_(m(NaCl))^(oo)` `=(91.0+426.0-126.0)" ohm"^(-1)cm^(2)mol^(-1)=391" ohm"^(-1)cm^(2)mol^(-1)`. |
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| 2750. |
Two students use same stock solution of `ZnSO_(4)` and a solution of `CuSO_(4)`. The `EMF` of one cell is `0.03` higher than the other. The concentration of `CuSO_(4)` in the cell with higher `EMF` value is `0.5M`. Find the concentration of `CuSO_(4)` in the other cell. `(` Take `2.303 RT//F=0.06)` |
| Answer» Correct Answer - 0.05 | |