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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.
| 2951. |
The highest electrical conductivity of the following solutions is of :A. 0.1 M acetic acidB. 0.1 M chloroacetic acidC. 0.1 M fluoroacetic acidD. 0.1 M difluroacetic acid |
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Answer» Correct Answer - D (d) 0.1 M difluoroacetic acid is maximum dissociated into ions as it is the stronger acid. |
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| 2952. |
Electrolytic reduction of alumina to aluminium by Hall-Heroult process is carried out:A. In the presence of `NaCl`B. In the presence of fluorideC. In the presence of cryolite which forms a melt with lower melting temperatureD. In the presence of cryolite which forms a melt with higher melting temperature |
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Answer» Correct Answer - C Cryolite is used to loweer the m.pt. of alumina as well as to make to good conductor of current |
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| 2953. |
The cell reaction for the given cell is spontaneoous if: `Pt|Cl_(2)(P_1)|Cl^(-)(1M)||Cl^(-)(1M)|Pt|Cl_(2)(P_2)`A. `P_(1) gt P_(2)`B. `P_(1) lt P_(2)`C. `P_(1) = P_(2)`D. `P_(2) = 1` atm |
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Answer» Correct Answer - B `E_("cell") = E_(OP_(Cl))^(@) + E_(RP_(Cl))^(@) + (0.059)/(2)"log"([Cl^(-)]_(LHS)^(2)P_(2))/([Cl^(-)]_(RHS)^(1)P_(2)` `= (0.059)/(2)"log"(P_(2))/(P_(1))` `E_("cell")` is `+ve` when `P_(2)gtP_(1)` |
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| 2954. |
Aluminium oxide may be electrolysed at `100^(@)C` to furnish aluminium metal (atomic mass=27 amu). The cathodic reaction is : `Al^(3+)+3e^(-) to Al` To prepare 5.12 kg of aluminium metal by this reaction would require :A. `549xx10^(7)" C"` of electricityB. `1.83xx10^(7)" C"` of electricityC. `5.49xx10^(7)" C "` of electricityD. `5.49xx10^(10)" C "` of electricity |
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Answer» Correct Answer - C (c ) `Al^(3+)+underset(3xxF)3e^(-) to underset(27g)Al` To deposit 27 g of Al metal, electricity needed `=3xxF` To deposit 5120 g of Al metal, electricity needed `=(3F)/((27 g))xx(5120 g)` `=568.89 F=5.49xx10^(7)" C "` |
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| 2955. |
Element A (atomic mass 112) and element B (atomic mass 27) form chlorides . Solutoins of these chlorides are electrolysed separately and it is found that the same quantity of electricity is passed 5.6 gm of A deposited while only 0.9 gm of B was deposited . if the valency of B is 3 , the valency of A isA. 2B. 3C. 4D. `-2` |
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Answer» Correct Answer - A Equivalent mass of A = `("Atomic mass")/("valency") = (112)/(x)` Equivalent mass of B = `("Atomic mass")/("valency") = (27)/(3) = 9` `("Mass of A")/("Mass of B") = ("Equivalent mass of A")/("Equivalent mass of B") = (5.6)/(0.9)` `therefore (112//x)/(9) = (5.6)/(0.9)` `therefore x = (112 xx 0.9)/(9 xx 5.6) = 2` Valency of A = 2 |
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| 2956. |
How long does it take to deposit 100 g of Al from an electrolytic cell containing `Al_(2)O_(3)` using a current of 125 ampere ?A. 1.54 hB. 1.42 hC. 1.32 hD. 2.15 h |
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Answer» Correct Answer - B `Al^(3+)+ 3e^(-) to Al` , Eq. wt. at `Al=27/3`=9 W = Z x I x t `rArr Z = "Eq. wt. "/96500` `t=(Wxx96500)/("eq.wt. x I")=(50 xx 96500)/(9 xx 105)` = 5105.82 s or 1.42 h |
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| 2957. |
The time required to deposit 90 g of Al from an electrolytic cell containing `Al_(2)O_(3) ` by a current of 965 A (At.mass of Al = 27)A. 20 min . 20 secB. 16 min . 40 secC. 40 min 20 secD. 26 min |
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Answer» Correct Answer - B W = 90 gm , i= 965 A , t = ? W = Zit `therefore t = (W)/(Zi)` `Z = ("Eq. mass")/(96500)` for Al , Eq. mass `= (27)/(3) therefore Z = (27)/(3 xx 96500)` `t = (90 xx 3 xx 96500)/(27 xx 965) = 1000` sec = 16 min . 40 sec |
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| 2958. |
How long does it take to deposit 100 g of Al from an electrolytic cell containing `Al_(2)O_(3)` using a current of 125 ampere ?A. 95.30 minB. 143 minC. 47.65 minD. 10 min |
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Answer» Correct Answer - B Equivalent mass of Al = `("Atomic mass")/("valency") = (27)/(3) = 9` `W = (ixxtxxE)/(96500)` `100 = (125 xx t xx 9)/(96500) implies t = 8577. 77` sec = 143 min . |
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| 2959. |
The number of coulombs required to liberate 0.224 `dm^(3)` of chlorine at `0^(@)C` and 1 atm pressure isA. `2 xx 965`B. 965/2C. 965D. 9650 |
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Answer» Correct Answer - A `22.4 dm^(3) `= 1 mole of `Cl_(2)` `therefore 0.224 dm^(3) = 0.01` mole of `Cl_(2)` As 96,500 coulombs `-= 35.5` gm = 0.5 moles of `Cl_(2)` OR `(1)/(2)` moles of `Cl_(2) = 96500 implies therefore 0.01` moles `-= 96500 xx 0.01 xx 2` `= 965 xx 2` coulombs |
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| 2960. |
Number of moles of oxygen liberated by electrolysis of 90g of waterA. 9 molesB. 4-5 molesC. 2.5molesD. 5 moles. |
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Answer» Correct Answer - c Electrolysis of water `2H_(2)Ooverset("Electrolysis")rarr2H_(2)+O_(2)` `therefore 2` mol `H_(2)O=1` mol of `O_(2)` and 5 mol `H_(2)O=(1)/(2)xx5=2.5` mol of `O_(2)` |
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| 2961. |
The number of moles of oxygen obtained by the electrolytic decomposition of 90g water is :A. 1 moleB. 2.5 moleC. 2 moleD. Data sufficient |
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Answer» Correct Answer - B `H_(2)O to H_(2) + (1)/(2) O_(2)` `because` 18 gms of water gives 0.5 moles oxygen `because` 90 gms of water give `(0.5)/(18) xx 90 = 2.5` moles of oxygen |
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| 2962. |
`50mL` of `0.1M CuSO_(4)` solution is electrolysed with a current of `0.965A` for a period of 200sec. The reactions at electrodes are: Cathode: `Cu^(2+) +2e^(-) rarr Cu(s)` Anode: `2H_(2)O rarr O_(2) +4H^(+) +4e`. Assuming no change in volume during electrolysis, calculate the molar concentration of `Cu^(2+),H^(+)` and `SO_(4)^(2-)` at the end of electrolysis. |
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Answer» Correct Answer - `Cu^(2+) = 0.08M, H^(+) = 0.04M, SO_(4)^(2-) = 0.1M` `n_(e) = (It)/(96500) = (0.965 xx 200)/(96500) = 2xx 10^(-3)` `n_(Cu^(2+)) = 10^(-3)` `n_(Cu^(2+)) Initial = 5 xx 10^(-3)` `n_(Cu^(2+))` left in solution `=4xx 10^(-3)` `m_(Cu^(2+)) = (4 xx 10^(-3))/(50) xx 100 = 0.08M` `2H_(2)O rarr _(2) +4H^(+) +4e^(-)` `n_(E) = 2 xx 10^(-3) = n_(H^(+))` `m_(H^(+)) = (2 xx 10^(-3))/(50) xx 10^(-3) = 0.04M` Molarity of `SO_(4)^(2-)` will not be changed `m_(SO_(4)^(2-)) = 0.1` |
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| 2963. |
A metal is know to form fluoride `MF_(2)`. When `10A` of electricity is passed through a molten sat for `330sec, 1.95g` of metal is deposited. Find the atomic weight of M. what will be the quantity electricity required to deposit the same mass of `Cu` form `CuSO_(4)`? |
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Answer» Correct Answer - `A = 114, Q = 5926, 8C` `M^(+2) + 2e^(-) rarr M rArr n_(e^(-)) = (2xx 1.95)/(M)` `rArr n_(M) = (1.95)/(M)` `=(2 xx 1.95)/(M) = (10 xx 330)/(96500)` `M = 114` `Cu^(+2) +2e^(-) rarr Cu` `n_(e^(-)) = (2 xx 1.95)/(63.5) n_(Cu) = (1.95)/(63.5)` `n_(E) = (Q)/(96500)` `Q = 5926.8C` |
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| 2964. |
In a Daniel cell,A. the chemical energy liberted during the redox reaction is converted to electrical energyB. the electrical energy of the cell is converted to chemical energyC. the energy of the cell is utilised in conduction of the redox reactionD. the potential energy of the cell is converted into electrical energy. |
| Answer» Correct Answer - A | |
| 2965. |
What will happen during the electrolysis of aqueous solution of `CuSO_(4)` by using platinum electrodes ?A. Copper will deposit at cathodeB. Copper will deposit at anodeC. Oxygen will be released at anodeD. Copper will dissolve at anode. |
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Answer» Correct Answer - A::C (a, c) are both correct options. |
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| 2966. |
What will happen during the electrolysis of aqueous solution of `CuSO_(4)` by using platinum electrodes?A. Copper will deposit at cathodeB. Copper will deposit at anodeC. Oxygen will be released at anodeD. Copper will dissolve at anode |
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Answer» For electrolysis of aqueous solution `CuSO_(4)` `CuSO_(4)(aq)rarrCu^(2+)+SO_(4)^(2-)` `H_(@)Orarr2H^(+)+O^(2-)` At anode `2O^(2-)rarrO_(2)+2e^(-)` At cathode `Cu^(2+)+2e^(-)rarrCu(s)` |
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| 2967. |
What will happen during the electrolysis of aqueous solution of `CuSO_(4)` by using platinum electrodes?A. Coper will deposit at cathode.B. Copper will deposti at anode.C. Oxygen will be released at anode.D. Copper will dissolve at anode. |
| Answer» Correct Answer - A::C::D | |
| 2968. |
The electrochemical cell shows below is a concentration cell. `M|M^(2+)` (saturated solution of a sparingly soluble salt, `MX_(2))||M^(2+)(0.001)" mol "dm^(-3))|M`. The emf of the cell Depends on the difference in concentration of `M^(2+)` ions at the two electrodes. the emf of the cell at 298K is 0.059V. Q. The value of `DeltaG(kJ" "mol^(-1))` for the given cell is (take `1F=96500" C "mol^(-1)`)A. `-5.7`B. `5.7`C. `11.4`D. `-11.4` |
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Answer» Correct Answer - D `DeltaG=-nFE_(cell)=2xx96500xx0.059xx10^(-3)kJ//`mole `=-11.4kJ//`mole. |
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| 2969. |
The electrochemical cell shows below is a concentration cell. `M|M^(2+)` (saturated solution of a sparingly soluble salt, `MX_(2))||M^(2+)(0.001)" mol "dm^(-3))|M`. The emf of the cell Depends on the difference in concentration of `M^(2+)` ions at the two electrodes. the emf of the cell at 298K is 0.059V. Q. The solubility product `(K_(sp),mol^(3)dm^(-9))` of `MX_(2)` at 298K based on the information available for the given concentration cell is (take `2.303xxRxx298//F=0.059V`)A. `1xx10^(-15)`B. `5.7`C. `1xx10^(-12)`D. `4xx10^(-12)` |
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Answer» Correct Answer - B `M|M^(2+)underset(0.001M)((aq)||M^(2+))(aq)|M` Anode: `MtoM^(2+)(aq)+2e^(-)` Cathode: `underline(M^(2+)(aq)+2e^(-)toM" ")` `M^(2+)(aq)hArrM^(2+)(aq)` `E_(cell)=0-(0.059)/(2)"log"((M^(2+)(aq))/(10^(-3)))` `0.059=-(0.059)/(2)"log"((M^(2+)(aq))/(10^(-3)))implies-2="log"((M^(2+)(aq))/(10^(-3)))` `10^(-2)xx10^(-3)=M^(2+)(aq)="solubility"=s` `K_(sp)=4S^(3)=4xx(10^(-5))^(3)=4xx10^(-15)`. |
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| 2970. |
Dal lake has water `8.2 xx 10^(12)`litre approximately. A power reactor produces electricity at the rate of `1.5 xx 10^(6)` colulomb per second at an appropriate voltage. How many years would it take to electrolyse the lake? |
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Answer» Correct Answer - `1.9` million year At anode: `2H_(2)O rarr 4H^(+) +O_(2) +4e^(-)` At anode: `2H_(2)O +2e^(-) rarr H_(2) +2OH^(-)` Not reaction: `2H_(2)O rarr 2H_(2) +O_(2)` 4 Faraday `rarr 2` mole `H_(2)O = 36g = 36 mL` `1mL H_(2)O rarr (4 xx 96500)/(36)` Faraday `8.2 xx 10^(15) mL H_(2)O.. (4 xx 96500)/(36) xx8.2 xx 10^(15)` Faraday `t = (4 xx 96500 xx 8.2 xx 10^(15))/(36 xx 1.5 xx 10^(-6)) sec =- 5.861 xx 10^(13)sec`. `=1.858 xx10^(6)` years. |
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| 2971. |
Hydrogen peroxide can be prepared by successive reaction: `2NH_(4)HSO_(4) rarr H_(2)+(NH_(4))_(2)S_(2)O_(8)` `(NH_(4))_(2)S_(2)O_(8)+2H_(2)O rarr 2NH_(4)HSO_(4)+H_(2)O_(2)` The first reaction is an electrolytic reaction the second is steam distillation. what amount ofcurrent would have to be used in first reaction to produce enough intermediate to yield 100g pure `H_(2)O_(2)` per hour ? Assume `50%` anode current efficiency. |
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Answer» Correct Answer - `315.36A` moles of `H_(2)O_(2) =` moles of `H_(2) = (100)/(34)` wt. of `H_(2) = (100)/(17) = Z` it `(100)/(17) = (1 xx1 xx 0.5 xx 3600)/(96500)` `i = 315.36 Amp`. |
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| 2972. |
The depolariser used in dry cell isA. `MnO_(2)`B. `NH_(4)Cl`C. CarbonD. `ZnCl_(2)` |
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Answer» Correct Answer - A In dry cell , `MnO_(2)` is used as depolarizer . |
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| 2973. |
Specific conductivity of a solutionA. increases with dilutionB. decreases with dilutionC. remains unchanged with dilutionD. depends on mass of electrolyte. |
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Answer» Correct Answer - B Specific conductivity of a solution decreases with dilution. |
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| 2974. |
Which substance is the best oxidizing agent in Q. no. 69?A. `IO_(3)^(-)`B. `IO^(-)`C. PbOD. `PO_(4)^(3-)` |
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Answer» Correct Answer - A |
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| 2975. |
From the fllowing half-cell reactions and their standard potentials ,what is the smallest possible standard e.m.f for spontaneous reactions? `PO_4^(3-)(aq)+2H_2O(l)+2e^(-)toHPO_3^(2-)+3OH^(-)(aq), E^(@)=-1.05V` `PbO_2(s)+H_2O(l)+2e^(-)toPbO(s)+2OH^(-)(aq),E^(@)=+0.28 V` |
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Answer» Correct Answer - D |
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| 2976. |
Determine which substance is the best reducing agent in Q. no. 69 :A. `HPO_(3)^(2-)`B. `PO_(4)^(3-)`C. `IO^(-)`D. `IO_(3)^(-)` |
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Answer» Correct Answer - A |
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| 2977. |
The `E^(@)` for the following cell is +0.34 V. In(s)|In `(OH)_3(aq)"||"Sb_2^(-)(AQ)"|"Sb(s)`. Using `E^(@)=-1.0 V` for the In `(OH)_3`| In, couple, calculate `E^(@)` for the `Sb_2^(-)`|Sb half-reaction:A. -1.34B. 0.66C. 0.82D. -0.66 |
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Answer» Correct Answer - D |
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| 2978. |
Consider the following half-cell reaction and associated standerd half-cell potentials and determine the maximum voltage thatr can be obtained by combination resulting in spontenous process : `AuBr_(4)^(-)(aq)+3e^(-)toAu(s)+4BR^(-)(aq), E^(@)=-086V` `Eu^(3+)(aq)+e^(-)toEu^(2+)(aq), E^(@)=-043V` `Sn^(2+)(aq)+2e^(-)toSn(s), E^(@)=-0.14V` `IO^(-)(aq)+H_(2)O(l)+2e^(-)toI^(-)(aq)+2OH^(-), E^(@)=+0.49V`A. `+0.72`B. `+1.54`C. `+1.00`D. `+1.35` |
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Answer» Correct Answer - D |
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| 2979. |
Using the standerd half-cell potential listed, calculate the equilibrium constant for the reaction : `Co(s)+2H^(+)(aq)toCo^(2+)(aq)+H_(2)(g)" at 298 K"` `Co^(2+)(aq)+2e^(-)toCo(s) E^(@)=-0.277 V`A. `2.3xx10^(9)`B. `4.8xx10^(4)`C. `4.8xx10^(7)`D. `4.8xx10^(11)` |
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Answer» Correct Answer - A |
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| 2980. |
The `E^(@)` at `25^(@)`C for the following reaction is 0.22 V. Calculate the equilibrium constant at `25^(@)`C : `H_(2)(g)+2AgCl(s)to2Ag(s)+2HCl(aq)`A. `2.8xx10^(7)`B. `5.2xx10^(8)`C. `5.2xx10^(6)`D. `5.2xx10^(3)` |
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Answer» Correct Answer - A |
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| 2981. |
A cell, as shown below, consists of there compartments separated by porous pots. The first contains a cobalt eletrode in `5.0L` of `0.10M Co(NO_(3)),` the second contains `5.0L` of `0.10M KNO_(3),` the third contains an `Ag` electrode in `5.0L` of `0.10M AgNO_(3)`. Assuming that current with in the cell iis carried equally by the negative and positive ions by passign `0.1F` of electricity. Given `: Co^(2+)+2e^(-) rarr Co" "E^(c-)=-0.28V` Ag^(o+)+e^(-) rarr Ag" "E^(c-)=0.80V` `2Ag^(o+)+Corarr Co^(2+)+2Ag" "E^(c-)._(cell)=1.08V` Finally, the `II` compartment containsA. `I`B. `I,II,`C. `I,II,III`D. `II,III,IV` |
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Answer» Correct Answer - c `II`. Compartment contains some concentration of `K^(o+)` ions and some concentration of `NO_(3)^(c-)` left due to the movement . It also contains some concentration of `Co^(2+)` due to the movement of `Co^(2+)` ion from `I` compartment. Hence, ions present in `II` compartment are `:Co^(2+),NO_(3)^(c-), `and `K^(o+)` |
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| 2982. |
A solution containing `H^(+)` and `D^(+)` ions is in equilibrium with a mioxture of `H_(2)` and `D_(2)` gases at `25^(@)`C. If the partial pressures of both gases are 1.0 atm, find the ratio of [`D^(+)`]`//`[H^(+)]` : (Given : `E_(D^(+)//(D_(2)))^(@)=-0.003V`]A. 1.23B. 1.12C. 0.11D. 1 |
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Answer» Correct Answer - B |
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| 2983. |
A solution containing `H^(+)` and `D^(+)` ions is in equilibrium with a mixture of `H_(2)` and `D_(2)` gases at `25^(@)C` . If the partial pressure of both the gases are `1.0` atm. The ration `[H^(+)]/[D^(+)]`. Given, `E_(D^(+)//D)^(@) = -0.003 C, E_("cell") = 0.006 V` :A. `1.2`B. `1.1`C. `0.11`D. `1.0` |
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Answer» Correct Answer - B `D_(2) + 2H^(+) rarr H_(2) + 2D^(+)` `E = E_(OP_(D))^(@) + E_(RP_(H))^(@) + (0.0591)/(2)"log"([H^(+)]^(2)xxP_(D_(2)))/(P_(H_(2)) xx [D^(+)]^(2))` `0.006 = 0.003 + 0 + (0.0591)/(1)"log"([H^(+)])/([D^(+)])` `(0.0591)/(1)"log"([H^(+)])/([D^(+)]) = 0.0006 - 0.003` `:. "log"([H^(+)])/([D^(+)]) = 0.0507` `:. ([H^(+)])/([D^(+)]) = 1.12` |
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| 2984. |
How much will the reduction potential of a hydrogen electrode change when its solution initially at `pH=0` is neutralized to `pH=7` ?A. `-0.059`B. `0.059`C. `-0.59`D. `0.59` |
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Answer» Correct Answer - c `pOH=4,pH=14-4=10` `E=-0.059xx10=-0.59V` |
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| 2985. |
Rusting of iron is catalyzed by which of the following?A. FeB. ZnC. `O_2`D. `H^+` |
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Answer» Correct Answer - D |
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| 2986. |
How much will the reduction potential of a hydrogen electrode change when its solution initially at `pH=0` is neutralized to `pH=7` ?A. Increase by `0.059 V`B. Decrease by `0.058 V`C. Increase by `0.41 V`D. Decrease by `0.41 V` |
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Answer» Correct Answer - D `E ^(@) = E + (0.059)/(1)log 10^(-7)` `= E^(@) + (0.059 xx (-7))/(1) = E^(@) -0.41 V` |
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| 2987. |
Rusting of iron is catalyzed byA. `H^(o+)`B. Dissolved `CO_(2)` in waterC. `O_(2)`D. Impurities present in `Fe` |
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Answer» Correct Answer - a,b,c,d Refer Section |
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| 2988. |
In which of the following cells, `EMF` is greater than `E^(c-)._(cell)`?A. `Pt,H_(2)(g)|H^(o+)(pH=5)||H^(o+)(pH=3)||H_(2)(g),Pt`B. `Zn(s)|Zn^(2+)(0.2M)||Cu^(2+)(0.1M)|Cu(s)`C. `Cr(s)|Cr^(3+)(0.1M)||Cu^(2+)(0.2M)|Cu(s)`D. `Pt,H_(2)(g)|H^(o+)(pH=4)||H^(o+)(pH=6)|H_(2)(g)|Pt` |
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Answer» Correct Answer - a `a.` Use `: E_(cell)=E^(c-)._(cell)-(0.059)/(n_(cell))log Q_(cell)` For `E_(cell)gt E^(c-)._(cell),Q_(cell)lt1(i.e., log Q_(cell)lt0)` `a.` `H_(2)(1 atm)+2H^(o+)(10^(-3)M) rarr 2H^(o+)(10^(-5)M)+H_(2)(1atm)` `Q_(cell)=((10^(-5))^(2)xx1)/(1xx(10^(-3))^(2))=10^(-4)lt1` `b.Zn(s)+Cu^(2+)(0.1M) rarr Zn^(2+)(0.2M)+Cu(s)` `Q_(cell)([Zn^(2+)])/([Cu^(2+)])=(0.2)/(0.1)=2gt1` `c.` `2Cr(s)+3Cu^(2+)(0.2M) rarr 2Cr^(3+)(0.1M) +3Cu(s)` `Q_(cell) ([Cr^(3+)]^(2))/([Cu^(2+)]^(3))=((0.1)^(2))/((0.2)^(3))gt1` `d. H_(2)(1atm)+2H^(o+)(10^(-6)M)rarr 2H^(o+)(10^(-4)M)+H_(2)(1atm)` `Q_(cell)=((10^(-4))^(2)xx1)/(1xx(10^(-6))^(2))=10^(4)gt1` |
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| 2989. |
In the atmosphere of industrial smog, copper corrodes to formA. Basic copper carbonateB. Copper sulphideC. Basic copper sulphateD. Copper oxide |
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Answer» Correct Answer - a,c `8Cu+6H_(2)O+2SO_(2)+5O_(2) rarr 2[CuSO_(4).3Cu(OH)_(2)]` `2Cu+H_(2)O+CO_(2)+O_(2) rarr [CuCO_(3).Cu(OH)_(2)]` |
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| 2990. |
If thenA. `a=b`B. `K_(2)=K_(1)^(2)`C. `a=2b`D. `b=a^(2)` |
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Answer» Correct Answer - a,b `K_(1)=([C][D])/([A][B])` and `K_(2)=([C]^(2)[D]^(2))/([A][B])` Also, `E^(c-)` is independent of stoichiometry. |
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| 2991. |
Some statement are given about corrosion . The incorrect statement isA. corrosion can not occur in vacuumB. pure metals can undergo corrosionC. corrosion is prevented by formation of oxide layer on metal .D. Fe is more easily corroded than Cu . |
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Answer» Correct Answer - B Pure metals can not undergo corrosion . |
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| 2992. |
The tarnishing of silver ornaments in atmosphere is due toA. `Ag_(2)O`B. `Ag_(2)S`C. `Ag_(2)CO_(3)`D. `Ag_(2)SO_(4)` |
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Answer» Correct Answer - a,b `2aG+(1)/(2)O_(2)rarr Ag_(2)O` `2Ag+H_(2)S rarr Ag_(2)S+H_(2)` |
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| 2993. |
The tarnishing of silver ornaments in atmosphere is due toA. `Ag_(2)O, Ag_(2)S`B. `AgNO_(3), Ag_(2)S`C. `Ag(OH), Ag_(2)CO_(3)`D. `Ag` |
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Answer» Correct Answer - A `Ag` reacts with `O_(2)` and `H_(2)S`. |
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| 2994. |
Tarnishing of silver is an example ofA. rustingB. corrosionC. oxidationD. reduction |
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Answer» Correct Answer - B Tarnishing of silver is called corrosion . |
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| 2995. |
When electrolysis of silver nitrate solution is carried out using silver electrodes, wich of the following reaction occurs at the anode?A. `AgrarrAg^(+)+e^(-)`B. `Ag^(+)+e^(-)rarrAg`C. `2H_(2)Orarr4H^(+)+O_(2)+4e^(-)`D. `4OH^(-)rarr2H_(2)+O_(2)+4e^(-)` |
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Answer» Correct Answer - A Silver electrode will dissolve and go into solution. |
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| 2996. |
In passing `3F` of electricity through three electrolytic cells connect in series containing `Ag^(o+),Ca^(2+),` and `Al^(3+)` ions, respectively. The molar ratio in which the three metal ions are liberated at the electrodes isA. `1:2:3`B. `3:2:1`C. `6:3:2`D. `3:4:2` |
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Answer» Correct Answer - C `Ag^(+)+e^(-)rarrAg` `Ca^(2+)+2e^(-)rarrCa` `Al^(3+)+3e^(-)rarrAl` 3 farad aya li berate 1 mol of Al, 3 moles of Ag and 3/2 moles of Cu. Thus, molar ratio of Ag:Cu: Al is 3:3/2 :1 or 6:3:2. |
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| 2997. |
`0.5F` of electricity is passed through `500mL` of copper sulphate solution. The amount of copper which can be deposited will beA. 63.5gB. 31.75gC. 15.8gD. unpredictable. |
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Answer» Correct Answer - C 2F will deposit `63.5g (Cu^(2+)+2e^(-)rarrCu)` |
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| 2998. |
During the electrolysis of the aqueous solution of copper sulphate using `Pt` electrode, the reaction taking place at anode electrode isA. `Cu^(+2)+2e^(-)rarrCu`B. `CurarrCu^(2+)+2e^(-)`C. `2H_(2)Orarr4H^(+)+O_(2)+4e^(-)`D. `H_(2)O+e^(-)rarrOH^(-)+(1)/(2)H_(2)` |
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Answer» Correct Answer - C Oxidation i.e., loss of electrons occurs at anode. |
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| 2999. |
During the electrolysis of the aqueous solution of copper sulphate using `Pt` electrode, the reaction taking place at anode electrode isA. ` Curarr Cu^(2+) + 2e`B. ` 2SO_4^(2-) + 2 H_2 Orarr 2H_2 SO_4 +O_2+4e`C. ` 2H_2OrarrO_2 +4H^+ +4e`D. ` 2Cl^(-) rarr Cl_2+ 2e` |
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Answer» Correct Answer - C Anodic reaction is ` 2H_2O rarrO_2+4H^(+) +4e`. |
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| 3000. |
`0.5F` of electricity is passed through `500mL` of copper sulphate solution. The amount of copper which can be deposited will beA. `63.5g`B. `31.75g`C. `15.8g`D. Unpredictable |
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Answer» Correct Answer - c `Cu^(2+)+2e^(-) rarr Cu` `2F=1 mol of Cu=63.5g of Cu` `0.5 F=(63.5)/(2)xx0.5=15.8g` |
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