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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.
| 1701. |
STATEMENT-1: 1 coulomb charge deposits 1 g-equivalent of a substance. STATEMENT-2: 1 faraday is charge is charge on 1 mole of electrons.A. If both (A) and (R) are correct, and (R) is the correct explanation of (A).B. If both (A) and (R) are correct, but (R) is not the correct explanation of (A).C. If (A) is correct, but (R) is incorrectD. If (A) is incorrect, but (R) is correct. |
| Answer» Correct Answer - C | |
| 1702. |
How many electrons will have a total charge of 1 Coulomb? |
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Answer» Given : Charge = 1 Coulomb Number of electrons = ? 1 Faraday = 96500 C per mol electrons ∵ 96500 C electric charge is present on 1 mol electrons ∴ 1C charge is present on \(\frac{1}{96500}\) mol electrons ∴ Number of electrons = \(\frac{1}{96500}\) x 6.022 x 1023 = 6.24 × 1018 electrons ∴ 1 Coulomb charge is present on 6.24 × 1018 electrons. |
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| 1703. |
STATEMENT-1: 1 coulomb charge deposits 1 g-equivalent of a substance. STATEMENT-2: 1 faraday is charge is charge on 1 mole of electrons.A. If both the statements are TRUE and STATEMENTS-2 is the correct explantion of STATEMENTS-1B. If both the statements are TRUE but STATEMENTS-2 is NOT the correct explanation of STATEMENTS-1C. If STATEMENTS-1 is TRUE and STATEMENTS-2 is FALSED. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE |
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Answer» Correct Answer - D |
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| 1704. |
Which of the following represents the reduction potential of silver wire dipped into `0.1M AgNO` solution at `25^(@)C`:-A. `E_("red")^(@)`B. `(E_("red")^(@) + 0.059)`C. `(E_("oxid")^(@) - 0.059)`D. `(E_("red")^(@) - 0.059)` |
| Answer» Correct Answer - D | |
| 1705. |
Which of the following represents the reduction potential of silver wire dipped into `0.1M AgNO` solution at `25^(@)C`:-A. `E_(red)^(@)`B. `(E_(red)^(@) +0.059)`C. `(E_(OX)^(@)-0.059)`D. `(E_(red)^(@)-0.059)` |
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Answer» Correct Answer - D `Ag^(+)+e^(-)rarr Ag` `E_("cell")=E^(@)-(0.059)/(1)"log"[(1)/(0.1)]` `= E_("red")^(@)-0.059` |
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| 1706. |
During discharge of a lead storage cell the density of sulphuric acid in the cell:A. increasingB. decreasingC. remains unchargedD. initially increases but decreases subsequently |
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Answer» Correct Answer - B Sulphuric acid works as an electrolyte in ideal cell due to consumption its density decreases. |
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| 1707. |
Match the items of column I to those of column I: `{:(Column-I, Column-II),((A)"Cell constant",(p) E_("cathode")^(0)+E_("anode")^(0)),((B)Anode,(q)l//A),((C)"Conductance",(r)"Mass of product by 1 coulomb of electricity"),((D)"Electrochemical equivalent",(s)("Resistance")^(-1)),((E)E_(cell)^(0),(u)"Involve oxidation"):} ` |
| Answer» Correct Answer - `A-(q),(B)-(u),C-(s),D-(r),E-(p)` | |
| 1708. |
`{:(Column-I,Column-II),((A)"Cathode",(p)"Primary cell"),((B)1"Coulomb",(q)"Secondary cell"),((C)"Dry cell",(r)6.24 xx 10^(18) "electrons"),((D)"Lead stron cell",(s)"Concentration cell"),((E)Zn|Zn^(2+)(0.10M)||Zn^(2+)(0.1M)|Zn,(u)"Positive terminal of electrochemical cell"):}` |
| Answer» Correct Answer - `A-(u),B-(r),C-(p),D-(q),E-(s)` | |
| 1709. |
Electrolysis of aqueous `CuSO_(4)` with inert electrodes gives ______A. Cu at cathode, anode gets dissolvedB. Cu at cathode, `O_(2)` at anodeC. `O_(2)` at anode, `H_(2)` at cathodeD. `O_(2)` at anode, cathode gets dissolved |
| Answer» Correct Answer - B | |
| 1710. |
The amount of copper deposited by the passage of 1 mole of electrons during the electrolysis of aqueous `CuSO_(4)` solution isA. `63.5 g`B. `31.75 g`C. `159.5 g`D. `79.75 g` |
| Answer» Correct Answer - B | |
| 1711. |
When a quantity of electricity is passed through `CuSO_(4)` solution, 0.16 g of copper gets deposited. If the same quantity of electricity is passed through acidulated water, then the volume of `H_(2)` liberated at STP will be : (given atomic weight of Cu=64)A. `4 cm^(3)`B. `56 cm^(3)`C. `604 cm^(3)`D. `8 cm^(3)` |
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Answer» Correct Answer - B Number of equivalents of copper deposited `=0.16/32=0.005` Volume of `H_(2)` gas at STP `=11.2xx0.005` `=0.056` litre `=56 cm^(3)` |
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| 1712. |
When a quantity of electricity is passed through `CuSO_(4)` solution 0.16 g of copper gets deposited if the same quantity of electricity is passed through acidullated water, then the volume of `H_(2)` gas liberated at S.T.P. will be (Given at wt. of Cu=64)A. `4.0cm^(3)`B. `56cm^(3)`C. `604cm^(3)`D. `8.0cm^(3)` |
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Answer» Correct Answer - B `("Mass of Cu deposited")/("Volume of" H_("liberated"))=("Eq. wt of Cu")/("Eq. wt. of H")` `(0.18)/("Mass of" H_(2) "liberated")=(64//2)/(1)` or mass of `H_(2)` liberated `=(0.16)/(32)g=5xx10^(-3)g` `therefore` volume of `H_(2)` liberated at STP `=(22400)/(2)xx5xx10^(-3)cm^( 3)=56cm ^(3)` |
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| 1713. |
Select the correct statements `(s)` about `NHE`.A. `E^(c-)` of `SHE` is arbitrarily assumed to be zero.B. `E^(c-)` of `SHE` is equal to zero.C. `SHE` refers as `underset(1 ba r)(Pt,H_(2)(g))|underset(a=1)(H^(o+)(aq))`at `25^(@)C`.D. `SHE` is very susceptible to dissolved `O_(2),H_(20S` and all other reducing agents. |
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Answer» Correct Answer - a,c,d Factual statement |
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| 1714. |
In which of the following salt bridge is not needed ?A. `Pb|PbSO_(4)(s)|H_(2)SO_(4)|PbO_(2)(s)|Pb`B. `Cd|CdO(s)|KOH(aq)|NiO_(2)(s)|Ni`C. `Fe(s)|FeO(s)|KOH(aq),Ni_(2)O_(3)(s)|Ni`D. `Zn|ZnSO_(4)|CuSO_(4)|Cu` |
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Answer» Correct Answer - a,b,c Salt bridge is used to eliminate liquid junction potential arised due to different speed of ions present in cathodic and anodic compartments. |
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| 1715. |
Statement : `((deltaE)/(deltaT))_(P)` is called temperature coefficient of emf. Explanation : `((deltaE)/(deltaT))_(P)` may be `+ve, -ve` and depends upon heat of reaction.A. `S` is correct but `E` is wrongB. `S` is wrong but `E` is correct.C. Both `S` and `E` are correct and `E` is correct explanation of `S`.D. Both `S` and `E` are correct but `E` is not correct explanation of `S`. |
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Answer» Correct Answer - D `DeltaH = nF[T((deltaE)/(deltaT))_(P)-E],` where `((deltaE)/(deltaT))_(P)` is temperature coefficient. |
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| 1716. |
Among the following, the intersive property is (properties are) :A. molar conductivityB. electromotive forceC. resistanceD. heat capacity |
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Answer» Correct Answer - A::B `(c)` and `(d)` depends upon the size of system. |
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| 1717. |
In which of the following salt bridge is not needed ?A. `Pb|PbSO_(4(s))|H_(2)SO_(4)|PbO_(2(s))|Pb`B. `Zn|ZnSO_(4)|CuSO_(4)|Cu`C. `Cd|CdO_(s)|KOH_(aq.)|NiO_(2(s))|Ni`D. `Fe_(s)|FeO_(s)|KOH_(aq.)|Ni_(2)O_(3(S))|Ni` |
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Answer» Correct Answer - A::C::D Salt bridge is used to eleiminate liquid juncation potential arised due to different speed is ions present in cathodic and anodic compartment. |
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| 1718. |
Statement : Liquid juncation potential can be eliminated by putting a salt bridge of `KCl`. Explanation : The funcation of salt bridge is to remove liquid junction potential because the salt used has same speed of cathions and anions.A. `S` is correct but `E` is wrongB. `S` is wrong but `E` is correct.C. Both `S` and `E` are correct and `E` is correct explanation of `S`.D. Both `S` and `E` are correct but `E` is not correct explanation of `S`. |
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Answer» Correct Answer - C Explanation is correct reason for statement. |
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| 1719. |
In which of the following redox pairs, the reduction potential very, with `pH` ?A. `MnO_(4)^(-)//Mn^(2+)`B. `AmO_(2)^(2+)//AmO_(2)^(+)`C. `AmO_(2)^(2+)//Am^(4+)`D. `Zn^(2+)//Zn` |
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Answer» Correct Answer - A::C The half reaction involve `H^(+)` only in `4H^(+) + AmO_(2)^(2+) 0+ 2e rarr Am^(4+) + H_(2)O` `MnO_(4)^(-) + 8H^(+) + 5e rarr Mn^(2+) + 4H_(2)O` |
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| 1720. |
Select the correct statement(s) about the liquid juncation potential:A. The sign and magnitude of liquid junction potential depdens upon transport number of anions and cations (`t_(a)` and `t_(c))`.B. If `t_(a) = t_(c), E_(IJP) = 0`C. If `t_(a) gt t_(c) , E_(IJP)` increases the `E.M.F.` of cell and if `t_(a) lt t_(c) , E_(IJP)` decreases the `E.M.F.` of cell.D. Only two vertical lines between two electrodes indicates the existance of liquid juncation potential. |
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Answer» Correct Answer - A::B::C Only one vertical line `(I)` indicatesliquid juncation potential. Two vertical line `(II)` indicates salt bridge, i.e., no liquid junction potential. |
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| 1721. |
The concentration of potassium ions inside a biological cell is at least 20 times higher than outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simplel model for a concentration cell involving a metal `M` is `M(s)|M^(o+)(aq,0.05 ` molar`)||M^(o+)(aq,1` molar`)|M(s)` For the abov electrolytic cell, the magnitude of the cell potential is `|E_(cell)|=70mV.` If the `0.05` moolar solution of `M^(o+)` is replaced by a `0.0025` molar `M^(o+)` solution, then the magnitude of the cell potential would beA. `35 mV`B. `70 mV`C. `140 mV`D. `700 mV` |
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Answer» Correct Answer - C According to passage, `0.059 log 20 = 70 mV` therefore, when `0.05` molar solution of `M_(L)^(+)` is replaced by `0.0025` molar `M_(L)^(+)` solution. Then, `E_(cell) = -0.059 log.([M_(R)^(+)])/([M_(L)^(+)])` `= 0.059 log.(1)/(0.0025)` `= 0.059 log 400 = 0.059 log 20^(2)` `= 2 xx 0.059 log 20` `= 2 xx 70 = 140 mV` |
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| 1722. |
Which of the following are correct about liquid junction potential?A. The liquid juncation potential is due to different ionic mobility of ions.B. The liquid junction potential can be measured directlty.C. The magnitude of liquid juncation potential depends upon relative speed of ions.D. The liquid junction potential always increases the `E.M.F.` of cell. |
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Answer» Correct Answer - A::C The liquid junction potential decreases the `E.M.F.` of cell and can bo be measured directly. It is calculated. |
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| 1723. |
Select the correct statements:A. Presence of `CO_(2)` increase rate of rustingB. Silver plating is done with the help of `K[Ag(CN)_(2)]`C. In saline water, rate of rusting is retardedD. Pure metals undergo corrosion faster than impure metlas. |
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Answer» Correct Answer - A::B::D Greater `H^(+)` ions more rusting will take place `Fe rarr Fe^(2+) + 2e` `O_(2) + 4H^(+) rarr 2H_(2)O` `K[Ag(CN)_(2)]` is soluble in water and electrolyte used is silver plating. Pure metal are more reactive. |
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| 1724. |
For the reduction of `NO_(3)^(c-)` ion in an aqueous solution, `E^(c-)` is `+0.96V`, the values of `E^(c-)` for some metal ions are given below `:` `i.V^(2+)(aq)+2e^(-)rarr V, " "E^(c-)=-1.19V` `ii. Fe^(3+)(aq)+3e^(-) rarr Fe, " "E^(c-)=-0.04V` `iii. Au^(3+)(aq)+3e^(-) rarr Au, " "E^(c-)=+140V` `iv. Hg^(2+)(aq)+2e^(-) rarr Hg, " "E^(c-)=+0.86V` The pair`(s)` of metals that is `//` are oxidized by `NO_(3)^(c-)` in aqueous solution is `//` areA. `V` and `Hg`B. `Hg` and `Fe`C. `Fe` and `Au`D. `Fe` and `V` |
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Answer» Correct Answer - A::B::D The species having less reduction potential with respect to `NO_(3)^(-)(E^(@) = 0.96 V)` will be oxisidised by `NO_(3)^(-)`. |
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| 1725. |
For the reduction of `NO_(3)^(c-)` ion in an aqueous solution, `E^(c-)` is `+0.96V`, the values of `E^(c-)` for some metal ions are given below `:` `i.V^(2+)(aq)+2e^(-)rarr V, " "E^(c-)=-1.19V` `ii. Fe^(3+)(aq)+3e^(-) rarr Fe, " "E^(c-)=-0.04V` `iii. Au^(3+)(aq)+3e^(-) rarr Au, " "E^(c-)=+140V` `iv. Hg^(2+)(aq)+2e^(-) rarr Hg, " "E^(c-)=+0.86V` The pair`(s)` of metals that is `//` are oxidized by `NO_(3)^(c-)` in aqueous solution is `//` areA. V and HgB. Hg and FeC. Fe and AuD. Fe and V |
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Answer» Correct Answer - A::B::D Since with `V, Fe` and `H_(2),E^(@)` of cell is positive so `DeltaG^(@)` will be negative. |
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| 1726. |
NaOH is manufctured by the electrolysis of brine solution. The products of reaction areA. `Cl_(2) and H_(2)`B. `Cl_(2) and Na-Hg`C. `Cl_(2) and Na`D. `Cl_(2) and O_(2)`. |
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Answer» Correct Answer - A Electrolysis of brine solution (NaCl sol.) gives `Cl_(2)` and `H_(2)`. |
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| 1727. |
The decomposition of a compound is found to follow first order rate law. If it takes 15 minutes for `20%` of original meterial to react, calculate the rate constant. |
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Answer» Given `t=15 min., a=100` `a-x =100-20= 80` `k=(2.303)/(t)log ""(a)/(a-X)` `k=(2.303)/(15 )log ""(100)/(80) =0.1535 [2.000-1.9031]=0.1535xx0.0969` `k=0.0148 min^(-1).` |
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| 1728. |
The standard electrode potentials of the electodes are given below. Arrange them in decreasin order of ease of oxidations `{:((1),"Electrode " - I"," E^(@) = - 2.89 V ),((2),"Electrode " - II"," E^(@) = - 0.16 V),((3),"Electrode " - III"," E^(@) = 0.77 V),((4),"Electrode " - IV"," E^(@) = - 2.93 V),((5),"Electrode " - V"," E^(@) = - 1.67 V):}`A. 3 2 5 1 4B. 4 1 5 2 3C. 3 5 1 4 2D. 3 2 1 4 5 |
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Answer» The electrode with greater negative value of standard reduction potential acts as anode. The electrode with greater positive value of standard reduction potential acts as cathode. Oxidation takes placed at anode and reduction takes place at cathode. X (i) Electrode `- IV E^(@) = -2.93V` (ii) Electrode `- I, E^(@) = -2.89 V` (iii) Electrode `- V E^(@) = -1.67 V` (iv) Electrode `- II, E^(@) = -0.16V` (v) Electrode `- III E^(@) = 0.77 V` |
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| 1729. |
An oxide of metal `("at. wt. "= 112)` contain `12.5% O_(2)` by weight. The oxide was converted into chloride by treatment with `HCl` and electrolsed. Calculate the amount of metal that would be deposited at cathode if a current of `0.965` ampere was passed for a period of `5` hr. What is valency of metal ? |
| Answer» Correct Answer - `10.08 g," " 2 ;` | |
| 1730. |
Calculate the number of `kw-h` of electricity is necessary to produce `1.0` metric ton `(1000 kg)` of aluminium by the Hall process in a cell operating at `15.0 V`. |
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Answer» `Al^(3+)+3e rarr Al` Eq. of `Al, (w)/(E) = (i.t)/(96500)` `:. i.t = (1000 xx 10^(3) xx 96500)/(27//3) [E_(Al) = 27//3]` `i.t = 107.22 xx 10^(8)` ampere-sec or coulomb `E = "Coulomb" xx "volt" = 107.22 xx 10^(8) xx 15.0` `= 1.61 xx 10^(11) J-sec` or `E = 1.61 xx 10^(11) "watt-sec"` `= (1.61 xx 10^(11))/(10^(3)) xx (1)/(3600) kW-hr` `= 4.47 xx 10^(4) kW-h` |
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| 1731. |
During an electrolysis of conc `H_(2)SO_(4)` , perdisulphuric acid `(H_(2)S_(2)_(8))` and `O_(2)` are formed in equimolar amount. The moles of `H_(2)` that will be formed simultaneously will be `a. `Thrice that of `O_(2)" "b.` Twice that of `O_(2)` `c.` Equal to that of `O_(2)." "d.` Half of that of `O_(2)` |
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Answer» `a.` This is a special case of electrolysis where two products are being obtained at anode . At anode `:` `4 overset(-)(O)H rarrO_(2)+2H_(2)O+4e^(c-) ,,,,,,,...(i)` `2SO_(4)^(2-) rarr S_(2)O_(8)^(2-)+2e^(-)" "....(ii)` `1 mol O_(2)` requires `4F` electricity and `1 mol S_(2)O_(8)^(2-)(-=H_(2)S_(2)O_(8))` requires `2F` electricity. So, if `x mol `of `O_(2)` are being produced, electricity being passed at anode is `:` `4x( f o r O_(2))+2x(3 f o r S_(2)O_(8)^(2-))=6xF` At cathode `:` `2H^(o+)+2e^(-) rarr H_(2)" "......(iii)` `2F` electricity `-=1 mol H_(2)` is produced `implies 6xF` electricity `-=1 mol H_(2)` is produced `implies ` Moles of `H_(2)` produced at cathode `=3 mol `of `O_(2)` produced at anode. |
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| 1732. |
What is the quantity of charge needed for the reduction of `1` mole of `MnO_(4)^(-)` ion to `Mn^(2+)` ion ? |
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Answer» Correct Answer - 5 `underset(("Oxidation NO. of Mn" = +7))(MnO_(4)^(-)) + 5e rarr underset(("Oxidation no. of "Mn = 2))(Mn^(2+))` `:. "Charge" = 5 xx 96500C = 5F` |
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| 1733. |
An oxide of metal `("at. wt. "= 112)` contain `12.5% O_(2)` by weight. The oxide was converted into chloride by treatment with `HCl` and electrolysed. Find the valency of metal. |
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Answer» Correct Answer - 2 Eq. of `O_(2) =` Eq. of metal `(12.5)/(8) = (87.5)/(E)` `:. E_("Metal") = (87.5 xx 8)/(12.5) = 56` Now valency of metal `= ("At. wt.")/("E. wt.") = (112)/(56) = 2` |
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| 1734. |
A current of `2.0` ampere is passed for `5.0` hour through a molten tin salt to deposit `22.2 g` tin. What is the oxidation state of tin in salt? Atomic `wt.` of `Sn = 118.69 g`. |
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Answer» Eq. of tin `= (i.t)/(96500)` `(22.2)/(E) = (2 xx 5 xx 60 xx 60)/(96500)` `E_(t i n) = 59.5` `:.` Valence of tin `= (A)/(E) = (118.69)/(59.5) = 2` (an integer) Thus `Sn^(2+)+2e rarr Sn` or tin is in `+2` oxidation state. |
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| 1735. |
The current of 2 A is passed for 5 h through a molten tin salt to deposit 22.2 g tin. What is the oxidation state of tin in salt? [Atomic weight of `Sn=118.69 g`]A. `+2`B. `+5`C. `+3`D. `+4` |
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Answer» Correct Answer - A Equivalents of tin `=(i. t)/(96500)` or `22.2/(Eq. wt.)=(2xx5xx60xx60)/(96500)` `:.` Eq. wt. `=59.5` `:.` Valency of tin `=(At. wt.)/(Eq. wt.)=118.69/59.5=2` (an integer) |
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| 1736. |
Value of standard electrode potential for the oxidation of `Cl^(-)` ions is more positive than that of water, even then in the electrolysis of aqueous sodium chloride, why is `Cl^(-)` oxidised at anode instead of water? |
| Answer» Oxidation of `H_(2)O` to `O_(2)` is kinetically so slow that it needs some overvoltage. Hence, `Cl^(-)` is oxidized instead of water. | |
| 1737. |
Value of standard electrode potential for the oxidation of `Cl^(-)` ions is more positive than that of water, even then in the electrolysis of aqueous sodium chloride, why is `Cl^(-)` oxidsied at anode instead of water? |
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Answer» Under the condition of electrolysis of aqueous sodium chloride, oxidation of water at anode requires over potential So, `Cl^(-)` is oxidized at anode instead of water Possible oxidation half cell reactions occuring at anode are `Cl^(-)(aq)rarr(1)/(2)Cl_(2)(g)+e^(-),E_("cell")^(@)=1.36V` `2H_(2)O(l)rarrO_(2)(g)+4H^(+)(aq)+4e^(-) E_("cell")^(@)=1.23V` Species hav ing lower `E_("cell")^(@)` cell undergo oxidation first than the higher value but oxidation of `H_(@)O` to `O_(2)` is kinetically so slow that it needs some ov ervoltage. |
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| 1738. |
What is electrode potential? |
| Answer» The potential difference between the electrdoe and the electrolyte in an electrochemcial cell is called electrode potential. | |
| 1739. |
Unlike dry cell, the mercury cell has a constant cell has a constant cell potential throughout its useful life, why? |
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Answer» Correct Answer - Life time of any cell depends upon ions present in cell, ions are not involved in the overall cell reaction of mercury cell. Hence, mercury cell has a constant cell potential throughout its useful life |
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| 1740. |
The electrolysis of a solution resulted in the formation of `H_2`(g) at the anode. The solution is :A. AgCl(aq)B. `H_2SO_4(aq)`C. highly concentrated NaCl(aq) solutionD. `CuCl_2(aq)` |
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Answer» Correct Answer - B |
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| 1741. |
A galvanic cell consists of a metallic zinc plate immersed in 0.1 M Zn `(NO_(3))_(2)` solution and metallic plate of lead in 0.02M Pb `(NO_(3))_(2)` solution. Calculte the emf of the cell. Write the chemical equation for the electrode reactions and represent the cell. (Given `E_(Zn^(2+)//Zn)^(@)=-0.76V,E_(Pb^(2+),Pb)^(@)=-0.13V`) |
| Answer» Correct Answer - 0.6094V, `Zn+Pb^(2+)toZn^(2+)+Pb;Zn|Zn^(2+)(0.1M)||Pb^(2+)(0.02M)|Pb` | |
| 1742. |
Consider an electrochemical cell in which the following reaction occurs and predict which changes will decrease the cell voltage : `Fe^(2+)(aq)+Ag^(+)(aq)toAg(s)+Fe^(3+)(aq)`A. IB. II and IIIC. IID. I and II |
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Answer» Correct Answer - D |
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| 1743. |
Iron and nickel are used to make an electrochemical cell by using slat bridge to join a half-cell containing 1.0 M solution of `Fe^(2+)(aq)` in which a strip of iron has been immersed to a second half-cell which contains 1.0 M `Ni^(2+)(aq)` solution in which a strip of nickel has been immersed. A voltmeter is connected between the two metal strips. (i) In which cell does reduction occurs? (ii) Write the half-cell reactions involved. (iii) Which metal is the anode? (iv) In which direction are the electrons passing through the voltmeter? (v) What would be effect ont he voltmeter reading if `Fe^(2+)` concentration were increased? (vi) What will be the voltmeter readind when the cell reaches equilbirium? Given that the standard electrode potential of `Fe^(2+)//Fe and Ni^(2+)//Ni` electrodes are -0.44 and -0.25 volt respectively. |
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Answer» Correct Answer - (i) Nickel half cell (ii) `Ni^(2+)+2e^(-)toNi` (iii) Fe is anode (iv) Iron to nickel (v) Voltmeter reading decreases (vi) V=0 |
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| 1744. |
What are compelex reactions ? Name one complex reaction. |
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Answer» A sequence of elementery reactions, reactants give the products, the reactions are called complex reactions. Eg: Oxidation of Ethane to `CO_(2) and H_(2)O` passes through a series of intermediate steps in which alcohol, aldehyde and acid are formed. |
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| 1745. |
In electrolysis of aqueous copper sulphate, the gas at anode and cathode is . |
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Answer» Correct Answer - A Cathode: `2H_(2)O+2e^(-)toH_(2)+2OH^(-)` Anode: `H_(2)Oto2H^(+)+(1)/(2)O_(2)+2e^(-)` |
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| 1746. |
Use of electrolysis is .A. ElectroplatingB. ElectroefiningC. Both (a) and (b)D. None of these |
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Answer» Correct Answer - C Electrolysis use for electroplating and electrorefining. |
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| 1747. |
The following data were obtained during the first order thermal decomposition of `N_(2)O_(5)` (g) at constant volume : `2N_(2) O_(5)(g)to 2N_(2)O_(4)(g) +O_(2)(g)` Calculate the rate constnat. |
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Answer» Let the pressure of `N_(2)O_(5)(g)` decrease by 2x atm. As two moles of `N_(2)O_(5)` decompose to give two moles of `N_(2)O_(4)` (g) and one mole of `O_(2)(g),` the pressure of of `N_(2)O_(4)` (g) increases by 2x atm and that of `O_(2)(g)` increases by x atm. `{:(, 2N_(2)O_(5)(g),to,2N_(2) O_(4)(g)2N_(2) O_(4)(g),+, O_(2)(g)),("Start" t=0, 0.5atm,,0 atm ,,0atm),("At time t" , (0.5-2x)atm,,2xatm,,x atm):}` `P_(t) =P_(N_(2)O_(5))+P_(N_(2)O_(4))+P_(O_(2))=(0.5 -2x) +2x+x=0.5 x` `x = -0.5+x` `x=P_(t)=0.5` `P_(N_(2)O_(5))=0.5 -2x =0.5 -2 (P_(t)-0.5)=1.5 -2P_(t)` At `t=100s,P_(t)=0.512atm` `P_(N_(2)O_(5))=1.5-2xx0.152=0.476atm` Rate constant `k=(2.303)/(t)log ""(P_(i))/(P_(A))=(2.303)/(100s)log ""(0.5atm )/(0.476 atm )` `=(2.303)/(100s) xx0.0216` `4.98xx10^(-4) s^(-1)` |
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| 1748. |
A first order reaction is found to have a rate constant, `k=5.5 xx10^(-14)s^(-1).` Find the half- life of the reaction. |
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Answer» Half-life for a first order reaction is `t_(1//2)=(0.693)/(k)` `t_(1//2)=(0.693)/(5.5xx10^(-14)S^(-1))=1.26xx10^(13)S` |
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| 1749. |
The oxide which is not reduced by hydrogen inA. `Ag_(2)O`B. `K_(2)O`C. `Fe_(2)O_(3)`D. `P_(4)O_(10)` |
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Answer» Correct Answer - B On the basis of electrochemical series `K_(2)O` is not reduced by hydrogen. |
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| 1750. |
During the electrolysis of aqueous solution of `HCOOK,` the number of gases obtained at cathode, anode, and total number of gases areA. `1,2,3`B. `1,2,2`C. `2,1,3`D. `2,1,2` |
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Answer» Correct Answer - b At Cathode`:` `H_(2)O+e^(-) rarr overset(c-)(O)H+(1)/(2)H_(2)(one)` At anode `:` `2HCOO^(c-) rarrr 2CO_(2)+H_(2)(` two`)` `H_(2)` and `CO_(2)` at anode and `H_(2)` at cathode. So, number of gases as in `(b)`. |
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