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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.
| 1601. |
Which of the following is displaced by Fe ?A. AgB. ZnC. NaD. All of these |
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Answer» Correct Answer - A |
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| 1602. |
In the reaction : `43Fe(s)+3O_(2)(g)to4Fe^(3+)(aq)+6O^(2-)(aq)` which of the following statement is incorrect ?A. It is a redox reductionB. Fe is reducing agentC. `O_(2)` is an oxidizing agentD. Fe is reducing to `Fe^(3+)` |
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Answer» Correct Answer - D |
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| 1603. |
The S.I. unit of cell constant for conductivity cell is :(a) m-1 (b) S·m-2(c) cm-2 (d) S·dm2·mol-1 |
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Answer» Option : (a) m-1 |
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| 1604. |
The amount of electricity equal to 0.05 F is : (a) 48250 C (b) 3776 C (c) 4825 C (d) 4285 C |
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Answer» Option : (c) 4825 C |
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| 1605. |
The charge of how many coulomb is required to deposit 1.0 g of sodium metal (molar mass 23.0 g mol-1) from sodium ions is :(a) 2098 (b) 96500 (c) 193000 (d) 4196 |
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Answer» Option : (d) 4196 |
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| 1606. |
The number of coulombs required to reduce 12.3g of nitrobenzene to anilineA. 115800CB. 5790CC. 28950CD. 57900C |
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Answer» Correct Answer - D `C_(6)H_(5)NO_(2)+6H^(+)+6e^(-)toC_(6)H_(5)NH_(2)+2H_(2)O` 1 mole=123 gm nitrogen requires 7 mole electron `=6xx96500`coulomb charge `therefore12.3` gm nitrobenzene will require=`(6xx96500xx12.3)/(123)` `=6xx9650=57900C`. |
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| 1607. |
When a lead storage battery is discharged.A. `SO_(2)` is evolvedB. Lead is formedC. Lead sulphate is consumedD. Sulphuric acid is consumed |
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Answer» Correct Answer - d When a lead storage battery is discharged, sulphuric acid is consumed. |
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| 1608. |
The standard reductino potentials `E^(c-)` for the half reactinos are as follows`:` `ZnrarrZn^(2+)+2e^(-)" "E^(c-)=+0.76V` `FerarrFe^(2+)+2e^(-) " "E^(c-)=0.41V` The `EMF` for the cell reaction `Fe^(2+)+Znrarr Zn^(2+)+Fe` isA. `-0.35V`B. `+0.35V`C. `+1.17V`D. `-1.17V` |
| Answer» Correct Answer - b | |
| 1609. |
The standard reduction potential values of three metallic cations, `X,Y,` and `Z` are `0.52,-3.03,` and `-0.18V`, respectively. The order of reducing power of the corresponding metal isA. `YgtZgtX`B. `XgtYgtZ`C. `ZgtYgtX`D. `ZgtXgtY` |
| Answer» Correct Answer - a | |
| 1610. |
The standard reduction potential values of three metallic cations, `X,Y,` and `Z` are `0.52,-3.03,` and `-0.18V`, respectively. The order of reducing power of the corresponding metal isA. `YgtZgtX`B. `XgtYgtZ`C. `ZgtYgtX`D. `ZgtXgtY`. |
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Answer» Correct Answer - A Strongest reducing agent is the one with lowest reduction potential (strongest R.A. `rArr` Easy to oxidise `rArr` highest O.P `rArr` Lowest R.P) |
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| 1611. |
Given`^^^""^(@)=((1)/(3)Al^(3+))=63 cm^(2)//Omega` mol and `^^^ ""^(@)((1)/(2)SO_(4)^(2-))=80 cm^(2)//Omega mol`. The value of `^^^ ""^(@) Al_(2)(SO_(4))_(3)` would beA. `143 Omega^(-1) cm^(2) mol^(-1)`B. `206 Omega^(-1) cm^(2) mol^(-1)`C. `286 Omega^(-1) cm^(2) mol^(-1)`D. `858 Omega^(-1) cm^(2) mol^(-1)` |
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Answer» Correct Answer - D `Lambda_((Al^(3+))^(@) = 3 Lambda_(((1)/(3) Al^(3+)))` =` 3 xx 63 = 189 Omega^(-1) cm^(2) mol^(-1)` . `Lambda^(@) (SO_(4))^(2-) = 2 Lambda^(@) ((1)/(2) SO_(4)^(2-))` `= 2 xx 80 = 160 Omega^(-1) cm^(2) mol^(-1)` . `Lambda^(@) [Al_(2) (SO_(4))_(3)] = 2Lambda^(@) (Al^(3+) ) + 3 Lambda^(@) (SO_(4)^(2-))` `2 xx 189 +3 xx 160 = 858 Omega^(-1) cm^(2) mol^(-1)` . |
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| 1612. |
Why does single electrode potential depend upon the concentration of the electrolyte ? |
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Answer» (i) Effect of concentration on direction of equilibrium (ii) The particles present on metal rod and electrolyte. (iii) The reason for the origin of difference in potential between the electrode and electrolyte. (iv) The factor responsible for the potential difference. (v) Effect of concentration on the above factor. |
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| 1613. |
An inaccurate ammeter and silver coulometer connnected in series in an electric circuit through which a constant direct current flows. If ammeter reads `0.6` ampere throughout one hour, the silver deposited on coulometer was found to be `2.16 g`. What `%` error is in the reading of ammeter? Assume `100%` current efficiency. |
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Answer» `(w)/(E) = (i.t)/(96500)` `(2.16)/(108) = (i xx 60 xx 60)/(96500)` `:. i = 0.54` ampere error in current `= 0.60-0.54 = 0.06` `:. %` error `= (0.06)/(0.6) xx 100 = 1%` |
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| 1614. |
The products of electrolysis of dilute aqueous solutio of sodium hydride areA. Na at cathode and `H_(2)` at anodeB. `H_(2)` at cathode and `O_(2)` at cathodeC. Na at cathode and `O_(2)` at anodeD. `H_(2)` at both cathode and anode |
| Answer» Correct Answer - D | |
| 1615. |
Electrode potential for the electrode `M^(n+)|M` with concerntration is given by the expression under `STP` conditions `:` `E=E^(c-)+(0.059)/(n)log[M^(n+)]` |
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Answer» Correct Answer - F This expression is valid at `298K(` standard condition `)`, not at `273 K (` STP condition `)`. |
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| 1616. |
The conductance of electrolyte solution increases wih temperature. |
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Answer» Correct Answer - T With increase of temperature the ionic mobility increases. |
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| 1617. |
In a Daniel cells, if `A (E_(0) = -0.76 ) and B(= - 2.36 V)` half - cells are taken thenA. B acts as an anodeB. A acts as an anodeC. B acts as a cathodeD. cannot be predicted |
| Answer» Correct Answer - A | |
| 1618. |
In the electroplating of silver,`AgNO_(3)` solution is usually used as an electrolyte. |
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Answer» Correct Answer - F The solution of sodium argento cyanide is used as an electrolyte,` i.e., Na(Ag(CN)_(2)]`. |
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| 1619. |
The more `…………………………` the standard reduction potential, the `……………………….` its ability to displace hydrogen from acids. |
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Answer» Correct Answer - Negative, more The more negative the standard reduction potential, the more is its ability to displace hydrogen from acids. |
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| 1620. |
The electrical conductivity of a solution of acetic acid will be `…………………` if a solution of sodium hydroxide is added. |
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Answer» Correct Answer - Increased The electrical conductivity of solution of acetic acid will be increased if a solution hydroxide is added because acetate salt is formed which is a strong electrolyte. |
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| 1621. |
The `pK_(SP)` of `Agl` is `16.07`. If the `E^(@)` value for `Ag^(+)//Ag` is `0.7991 V`, find the `E^(@)` for hlaf cell reaction :`AgI_((s)) + e rarr Ag + I^(-)` |
| Answer» Correct Answer - `-0.1490 V ;` | |
| 1622. |
The `e.m.f` of cell `Ag|AgI_((s)),0.05MKI|| 0.05 M AgNO_(3)|Ag` is `0.788 V`. Calculate solubility product of `AgI`. |
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Answer» `E_(cell)=E^(@)_(cell)=(0.591)/(1)log[Al^(+)]_(LHS)/[(Ag^(+))]_(RHS)` `0.8=0-(0.0591)/(1)log(C_(1))/(0.2)` `log(0.2)/(C_(1))=(0.8)/(0.0591)=(13.54)` `(0.2)/(C_(1))="Antilog"(13.54)` `(0.2)/(C_(1))=3.467xx10^(13)` `C_(1)=(0.2)/(3.467xx10^(13))` `C_(1)=5.77xx10^(-15)` `K_(sp)=[Ag^(+)][I^(-)]=5.77xx10^(-15)xx0.19` `k_(sp)=1.1xx10^(-16)` |
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| 1623. |
At `pH = 2, E_(("Quinhydrone"))^(@) = 1.30 V, E_("Quinhydrone")` will be : A. `1.36" V"`B. `1.30" V"`C. `1.42" V"`D. `1.20" V"` |
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Answer» Correct Answer - C (c ) `E=E^(@)-(0.0591)/(2)log [H^(+)]^(2)` `=1.30-(0.0591)/(2)log(10^(-2))^(2)` `=E^(@)-(0.0591)/(2)(-4" log "10)` `=1.30+(0.236)/(2)=1.418" V"` |
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| 1624. |
`DeltaG^(c-)` or the reaction is `,` `4Al+3O_(2)+6H_(2)O+4overset(c-)(O)H rarr 4 Al(OH)_(4)^(c-)` `E^(c-)._(cell)=2.73V` `Delta_(f)G^(c-)._((overset(c)(O)H))=-157k J mol^(-1)` `Delta_(f)G^(c-)._((overset(c-)(O)H))=-237k J mol^(-1)`A. `-3.16xx10^(3)kJ mol^(-1)`B. `-0.79xx10^(3)kJmol^(-1)`C. `-0.263xx10^(3)k J mol^(-1)`D. `+0.263xx10^(3)k J mol^(-1)` |
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Answer» Correct Answer - a `Delta_(r)G^(c-)._(cell)=-n_(cell)FE^(c-)._(cell)=-12xx96500xx2.73` `=-3.16xx10^(3)k J mol^(-1)` For `n_(cell)` write the half cell reaction as `:` Cathode `: 2H_(2)O(l)+O_(2)(g)+4e^(-) rarr 4 overset(c-)(O)H(aq)` Anode `:Al(s)+4overset(c-)(O)H(aq)rarr Al(OH)_(4)^(c-)(aq)+3e^(-)` |
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| 1625. |
`Cu^(2+)+2e^(-) rarr Cu`. For this, graph between `E_(red)` versus `ln[Cu^(2+)]` is a straight line of intercept `0.34V`, then the electrode oxidation potential of the half cell `Cu|Cu^(2+)(0.1M)` will beA. `0.34+(0.0591)/(2)`B. `-0.34-(0.0591)/(2)`C. `0.34`D. `-0.34+(0.0591)/(2)` |
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Answer» Correct Answer - d `Cu^(2+)+2e^(-) rarr Cu` `E_(Cu^(2+)|Cu)=E^(c-)._(Cu^(2+)|Cu)-(0.059)/(2) log .(1)/([Cu^(2+)])` ` =E^(c-)._(Cu^(2+)|Cu)-(RT)/(2F)ln[Cu^(2+)]` Intercept`=0.34impliesE^(c-)._(Cu^(2+)|Cu)=0.34V` `implies E_(cu^(2+)|Cu)=0.34+(0.059)/(2) log 0.1=0.31V` `impliesE_(Cu|Cu^(2+))=-E_(Cu^(2+)|Cu)=-0.34+(0.059)/(2)V` |
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| 1626. |
Mobility of `H^(+)` (In aq. Medium) is high because:A. of the small size of `H^(+)`B. Of the high hydration energy of `H^(+)`C. It exhibits of Grotthus type of conductionD. Hydrogen is the lightest element |
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Answer» Correct Answer - C Ionic mobility of `H^(+)` is high because it exhibit Grotthus different. |
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| 1627. |
NERNST EQUATIONA. `(2.303)/(nF)`B. `(2.303 RT)/(F)`C. `(RT)/(nF)`D. `(2.303RT)/(nF) "log"_(10) ("reduction state")/("oxidised state")` |
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Answer» Correct Answer - B `E = E^(@) - (2.303 RT)/(F) = (2.303 xx 8.314 xx 298)/(96500) = 0.0592` |
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| 1628. |
At `pH = 2, E_(("Quinhydrone"))^(@) = 1.30 V, E_("Quinhydrone")` will be : A. `1.36 V`B. `1.30 V`C. `1.42 V`D. `1.20 V` |
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Answer» Correct Answer - C `E = E^(@) = (0.059)/(2)log[H^(+)]^(2)` `= 1.30 - (0.059)/(2)log(10^(-2))^(2)` `= 1.30 + (0.236)/(2) = 1.418 V` |
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| 1629. |
The reduction potential of a hydrogen electrode at `pH 10` at `298K` is : `(p =1` atm)A. 0.51 VB. 0.000 VC. 0.59 VD. 0.059 V |
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Answer» Correct Answer - C `E = E^(@) - (0.059)/(n) " log " _(10) [H^(+)]` Given `pH = 10 , n =1 therefore [H^(+)] = 10^(-10)` , `E_(SHE)^(@) = 0 , 10 = - "log"_(10) [ H^(+)]` `E = E^(@) - 0.059 xx log 10^(-10) = 0-0.059 xx (-10)` `= 0 + 0.59 = 0.59` V |
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| 1630. |
The combustion of butane in `O_(2)` at 1 bar and `298K` shows a decrease in free energy equal to `2.95xx10^(3)kJ mol ^(-1)` in a fuel cell. `K` and `E^(c-)` of the fuel cell areA. `9.55xx10^(482),1.096V`B. `9.55,1.096V`C. `1.023xx10^(966),2.85V`D. `5.5xx10^(484),0.55V` |
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Answer» Correct Answer - a `DeltaG=DeltaG^(c-)=-2.75xx10^(6)J mol^(-1)(` as `P=1bar` and `T=298K)` `-DeltaG^(c-)=nE^(c-)F` `E^(c-)=(2.75xx10^(6))/(26xx96500)` `[underset(4x=-10,n=26)(C_(4)H_(10)+(13)/(2)O_(2))rarrunderset(4x=16)(4CO_(2)+5H_(2)O)]` `E^(c-)=1.096V` Also, `E^(c-)=(0.059)/(n)logK_(p)implies1.096=(0.059)/(26)logK_(p)` `K_(p)=9.55xx10^(482)` |
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| 1631. |
For the given cell `Pt_(D_(2)|D^(o+))||H^(o+)|Pt_(H_(2))`, if `E^(c-)._(D_(2)|D^(o+))=0.003V,` , what will be the ratio of `D^(o+)` and `H^(o+)` at `25^(@)C` when the reaction `D_(2) + 2H^(o+) rarr 2D^(o+)+H_(2)` attains equilibriumA. `1.34`B. `1.24`C. `1.124`D. `1.45` |
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Answer» Correct Answer - c `E_(cell)=E_(OPD_(2))=E_(RPH_(2))` `(OP=` oxidation potential, `RP=` reduction potential `)` `=E^(c-)._(OP_(D_(2))|D^(o+))-(0.059)/(2)log[D^(o+)]^(2)+E_(RPH^(o+)|H_(2))+0.059log[H^(o+)]^(2)` `0=0.003-(0.059)/(2)log.([D^(o+)]^(2))/([H^(o+)]^(2))` `([D^(o+)])/([H^(o+)])=1.124` |
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| 1632. |
What is `E^(c)._(red)` for the reaction`: Cu^(2+)+2e^(-) rarr Cu` in the half cell `Pt_(S^(2)|CuS|Cu)`. if `E^(c-)._(Cu^(2+)|Cu)` is `0.34V` and `K_(sp)` of `CuS=10^(-35)`?A. `0.34V`B. `-0.6925V`C. `+0.6925V`D. `-0.66V` |
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Answer» Correct Answer - b `E_(S^(2-)|CuS|Cu)=E^(c-)._(Cu^(2+)|Cu)+(0.059)/(2)log K_(sp)CuS` `=0.34+(0.059)/(2)log10^(-35)=-0.925V` |
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| 1633. |
Fuel cells `:` Fuel cells are galvanic cells in which the chemical energy of fuel cell is directly converted into electrical energy. A type of fuel cell is a hydrogen `-` oxygen fuel cell. It consists of two electrodes made up of two porous graphite impregnated with a catalyst `(` platinum, silver, or metal oxide `).` The electrodes are placed in aqueous solution of `NaOH` . Oxygen and hydrogen are continuously fed into the cell. Hydrogen gets oxidized to `H^(o+)` which is neutralized by `overset(c-)(O)H, i.e.,` anodic reaction. At cathode, `O_(2)` gets reduced to `overset(c-)(O)H` Hence, the net reaction is The overall reaction has `DeltaH=-285.6 kJ mol^(-1)` and `DeltaG=-237.4 kJ mol^(-1)` at `25^(@)C` What is the value of `DeltaS^(c-)` for the fuel cell at `25^(@)C`?A. `-1600 J K^(-1)`B. `-160 J K^(-1)`C. `160 J K^(-1)`D. `1600 J K^(-1)` |
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Answer» Correct Answer - b `DeltaG^(c-)=DeltaH^(c-)-T DeltaS^(c-)` `implies-237.4=-285.6-298DeltaS^(c-)` `:. DeltaS^(c-)=-160J K^(-1)` |
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| 1634. |
Fuel cells `:` Fuel cells are galvanic cells in which the chemical energy of fuel cell is directly converted into electrical energy. A type of fuel cell is a hydrogen `-` oxygen fuel cell. It consists of two electrodes made up of two porous graphite impregnated with a catalyst `(` platinum, silver, or metal oxide `).` The electrodes are placed in aqueous solution of `NaOH` . Oxygen and hydrogen are continuously fed into the cell. Hydrogen gets oxidized to `H^(o+)` which is neutralized by `overset(c-)(O)H, i.e.,` anodic reaction. At cathode, `O_(2)` gets reduced to `overset(c-)(O)H` Hence, the net reaction is The overall reaction has `DeltaH=-285.6 kJ mol^(-1)` and `DeltaG=-237.4 kJ mol^(-1)` at `25^(@)C` Suppose the concentration of hydroxide ioin in the cell is doubled, then the cell voltage will beA. Reduced by halfB. Increased by a factor of 2C. Increased by a factor of 4D. Unchanged |
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Answer» Correct Answer - d `E_(cell)` is independent of `[overset(c-)(O)H]` in this case. |
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| 1635. |
How much faraday is required to produce Al from Al2O3? |
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Answer» Al3+ + 3e-→ Al (27 g) 27 g Al is deposited by 3f 40 g Al will be deposited by = 3 x 40/27 = 3.44 F |
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| 1636. |
Oxidation of succinate ion produces ethylene and carbon dioxide gases. On passing `0.2` Faraday electricty through an aqueous solution of potassium succimate, the total volume of gases (all both cathode and anode) liberated at STP (1 atm and 273 K) isA. `8.96L`B. `2.24L`C. `4.48L`D. `6.72L` |
| Answer» Correct Answer - A | |
| 1637. |
An electrolytic cell contains a solution of `Ag_(2)SO_(4)` and have platinum electrodes. A current is passed until 1.6gm of `O_(2)` has been liberated at anode. The amount of silver deposited at cathode would beA. 108 gB. 1.6 gC. 0.8 gD. 21.60 g |
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Answer» Correct Answer - D Number of equivalents of oxygen `= (1.6)/(8) = 0.2` `:.` Number of equivalents of Ag deposited `= 0.2` Mass of Ag deposited `= 0.2 xx 108 = 21.6 g` |
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| 1638. |
Which of the following is not a non-electrolyte ?A. UreaB. GlucoseC. EthanolD. Acetic acid |
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Answer» Correct Answer - D The substatnce whose aqueous solutions allow the passage of electric cuurrent and are chemically decomposed are termeelectrolytes. Electrolytic substance are classified as strong or weak accoding to how redily they dissociate into conducting ions. Acetic acid is a weak electrolyte . Glucose, ethanol and urea are non-electrolytes. |
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| 1639. |
The quantity electricity needed separately for the electrolysis of 1 M solution of `ZnSO_(4), AlCl_(3)` and `AgNO_(3)` completely is in the ratio of :A. `2:3:1`B. `2:1:1`C. `2:1:3`D. `2:2:1` |
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Answer» Correct Answer - A (a) `Zn^(2+)+underset(2F)2e^(-) to underset(1mol)(Zn)` `Al^(3+)+underset(3F)3e^(-) to underset(1mol)(Al)` `Ag^(+)+underset(1F)e^(-) to underset(1 mol)(Ag)` The number of Faradays needed are in the ratio : `2F:3F:1F "or" 2:3:1`. |
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| 1640. |
Assertion: Galvanised iron does not rust. Reason: Zinc has a more negative electrode potential than iron.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 - A R is the correct explanation of A (More negative electrode potential of zinc means higher oxidation potential. Hence, zinc is more easily oxidized than iron). |
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| 1641. |
Chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules (approximately `6.023xx10^(23)`) are present in a few grams of any chemical compound varying with their atomic/molecular masses. To handle such large numbers conveniently, the mole concept was introduced. this concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry and radiochemistry. The following example illustrates a typical case, involving chemical/electrochemical reaction, which requires a clear understanding of the mole concept. A 4.0 molar aqueous solution of `NaCl` is prepared and 500 mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrodes (atomic mass: Na=23, Hg=200, 1 Faraday=96500 coulombs). Q. The total number of moles of chlorine gas evolved isA. 0.5B. `1.0`C. `2.0`D. `3.0` |
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Answer» Correct Answer - B 500 ml of 4.0 molar NaCl solution contains 2 mole of NaCL. The chlorine content of this sample will evolve as chlorine gas. The number of mole of NaCl=numer mole of `Cl^(-)=2`mole `therefore`Number of mole of `Cl_(2)` gas evolved`=(2)/(2)=1` mole. |
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| 1642. |
Chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules (approximately `6.023xx10^(23)`) are present in a few grams of any chemical compound varying with their atomic/molecular masses. To handle such large numbers conveniently, the mole concept was introduced. this concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry and radiochemistry. The following example illustrates a typical case, involving chemical/electrochemical reaction, which requires a clear understanding of the mole concept. A 4.0 molar aqueous solution of `NaCl` is prepared and 500 mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrodes (atomic mass: Na=23, Hg=200, 1 Faraday=96500 coulombs). Q. if the cathode is a Hg electrode, the maximum weight (g) of amalgam formed from this solution isA. 200B. 225C. 400D. 446 |
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Answer» Correct Answer - D 500 ml of 4.0 molar NaCl has 2 mole of NaCl. By electrolysis we can get a maximum of 2 mole of sodium which can combine with exactly 2 moles of mercury to give amalgam. `therefore` The maximum weight of amalgam which can be formed from this solution =weight of 2 mole of sodium+weight of 2 mole of mercury `=2xx23+2xx200=446`g. |
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| 1643. |
Rusting or iron is catalysed by which of the followingA. FeB. `O_(2)`C. ZnD. `H^(+)` |
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Answer» Correct Answer - D Rusting of iron is catalysed by `[H^(+)]`. |
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| 1644. |
Which metal is used as a coating on steel to prevent corrosion where the cell reactions are:-A. Fe is oxidised to `Fe^(2+)` and issolved oxygen oxygen in water is reduced to `overset(Ө)(O)H`B. Fe is oxidised to `Fe^(3+) and H_(2)O` is reduced to `O_(2)^(2-)`C. Fe is oxidised to `Fe^(2+) and H_(2)O` is reduced to `O_(2)^(-)`D. Fe is oxidised to `Fe^(2+) and H_(2)O` is reduced to `O_(2)` |
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Answer» Correct Answer - A `FetoFe^(2+)+2e^(-)` (anode reaction) `O_(2)+2H_(2)O+4e^(-)to4OH^(-)` (cathode reaction) ltbgt The overall reaction is:- `2Fe+O_(2)+2H_(2)Oto2FE(OH)_(2)` `Fe(OH)_(2)` may be dehydrated to iron oxide FeO, or further oxidized to `Fe(OH)_(3)` and then dehydrate to iron rust, `Fe_(2)O_(3)`. |
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| 1645. |
After electrlysis of `NaCI` solution with inert electrodes for a certain period of time `600mL` of the solution was left. Which was found to be `1N` in `NaOH`. During the same time, `31.75 g` of `Cu` deposited in the copper voltameter in series the electrolytic cell. calculate the percentage yield of `NaOH` obtained. |
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Answer» Equivalent mass of `NAOH =40` Amouunt of `NAOH` formed `=40/1000xx600=24 g` `31.75 g` of `Cu=1 g`-equivalent of `Cu`. During the same period, 1 g-equivalent of `NaOH` should have been formed. 1 g-equivalent of `NaOH = 40 g` % yield `=24/40xx100=60` |
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| 1646. |
The number of coulombs required for the deposition of 107.870 of silver isA. 96500B. 48250C. 193000D. 10000 |
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Answer» Correct Answer - A `underset("1mole")(Ag^(+))+underset(1F)(e^(-))rarrAg` To deposit 1 mole (or 107.870g) of silver. Electricity required=1F=96500C. |
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| 1647. |
The number of coulombs required for the deposition of 107.870 of silver isA. 96500B. 48250C. 1,93,000D. 10000 |
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Answer» Correct Answer - A Equivalent weight of silver=107.870g. 1 Faraday=96500 coulomb. |
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| 1648. |
Use `E^(@)` values to calculate `Delta G^(@)` for the reaction `Fe^(2+) + Ag^(+) rarr Fe^(3+) + Ag` `E_(Ag//Ag)^(@) = 0.80` volt and `E_(Pt//Fe^(3+))^(@) = 0.77` volt |
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Answer» Correct Answer - `-2895 J` `E_(cell)^(@)=0.03" volt", DeltaG^(@)=-nFE_(cell)^(@)` |
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| 1649. |
The specific conductance of 0.02 M HCl is 8.2 × 10-3 Ω-1 cm-1. Hence its molar conductivity is :(a) 164 Ω-1 cm2 mol-1(b) 6.1 × 103Ω-1 cm mol-1(c) 239.6 S cm2 mol-1(d) 410 S cm2 mol-1 |
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Answer» Option : (d) 410 S cm2 mol-1 |
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| 1650. |
A conductance cell was calibrated by filling it with a 0.02 M solution of potassium chloride (specific conductance `= 0.2768 ohm^(-1) m^(-1)`) and measuring the resistance at 298 K which was found to be 457.3 ohm. The cell was then filled with a calcium chloride solution containing 0.555 g of `CaCl_(2)` per litre. The measured resistance was 1050 ohm. Calculate the molar conductivity of `CaCl_(2)` solution. |
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Answer» Correct Answer - `0.0241` mho `m^(2) mol^(-1)` Determine cell constant with KCl data. Its value is `126.6 m^(-1)`. Sp. Cond. Of `CaCl_(2)` soln. = Cell const. `xx 1/("Resistance")` `=126.6xx1/1050=0.1206 ohm^(-1) m^(-1)` Molar conductance `=("Sp. Cond.)/("Conc.)` Conc. `=0.555/111.0=0.005" mol dm"^(-3) =5" mol m"^(-3)` |
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