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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.
| 2901. |
Three litres of 0.5 M `K_(2)Cr_(2)O_(7)` soluton have to be completely reduced in th acidic medium. The number of faradays of electricity required will be. |
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Answer» 3L of 0.5 M `K_(2)Cr_(2)O_(7)` solution will contain `0.5xx3=1.5` moles of `K_(2)Cr_(2)O_(7)`. The reduction reaction is `Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-)toCr^(3+)+7H_(2)O` thus, 1 mole of `Cr_(2)O_(7)^(2-)` requires 6 F of electricity. `therefore1.5` mole will require electricity=9F. |
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| 2902. |
Calculate the emf of the cell : `Mg|Mg^(2+)(0.1 M)||Ag^(+)(1.0xx10^(-3) M)|Ag` Given that `E_(cell)^(@)=3.15" V"`. |
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Answer» `E_(cell)^(@)=E_(cell)^(@)-(0.0591)/(n)"log"([Mg^(2+)])/([Ag^(+)]^(2))` `=3.15-(0.0591)/(2)"log"(0.1)/((1.0xx10^(-3))^(2))` `=3.15-0.02955" log "5` `=3.15-0.02955xx0.6989` `=3.15-0.0207=3.13~~3.0` |
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| 2903. |
What is the number of Faradays required to convert 1 mole of `Cr_(2)O_(7)^(2-)` into `Cr^(3+)` ions ? |
| Answer» `overset(+6" ")((Cr_(2)O_(7))^(2-))+6e^(-) 6e^(-) to overset(+3)(2Cr)` | |
| 2904. |
Predict the products of electrolysis obtained at the electrodes in each case by using platinum electrodes : (i) An aqueous solution of `AgNO_(3)` using platinum electrodes (ii) An aqueous solution of `CuSo_(4)` using attackable electrodes. |
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Answer» (i) Ag is deposited at cathode. `O_(2)` is evolved at anode. (ii) `Cu^(2+)` ions are deposited at cathode. Equivalent amount of Cu from anode goes into the solution. |
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| 2905. |
Predict the products of electrolysis of a solution of `H_(2)SO_(4)` using platinum electrodes. |
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Answer» (a) By the elctrolysis of dilute solution of `H_(2)SO_(4)` For answer, consult 18(N.C.E.R.T. Exercise). (b) By the electrolysis of strong `H_(2)SO_(4)` `H_(2)SO_(4) to underset((Cathode))(2H^(+))+underset((Anode))(SO_(4)^(2-))` `"At cathode" : " " 2H^(+)+2e^(-) to 2HorH_(2)` `"At anode" " " 2SO_(4)^(2-) to (S_(2)O_(8))^(2-)+2e^(-)` `(SO_(4)^(2-)` ions will take up `H^(+)` ions from the solution to from `H_(2)S_(2)O_(8)`which is peroxydisulphuric acid. |
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| 2906. |
The equilibrium constant for the following reaction at 298K is expressed as `x xx10^(y)` `2Fe^(3+)+2I^(-)to2Fe^(2+)+I_(2),E_(cell)^(@)=0.235V` The value of y is. |
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Answer» For the given reaction, n=2 `E_(cell)^(@)=(0.0591)/(2)logK therefore0.235=(0.0591)/(2)logK` or `logK=7.9526` or `K=8.966xx10^(7)`. |
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| 2907. |
The cell in which the following reaction occurs `2Fe^(3+)(aq)+2I^(-)(aq)to2Fe^(2+)(aq)+I_(2)(s)` has `E_(cell)^(@)=0.236V` at 298K. Calculate the standard gibbs energy of the cell reaction. (Given: `1F=96,500" C "mol^(-1)`) |
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Answer» Correct Answer - `45.548kJ" "mol^(-1)` For the given cell reaction, `n=2,DeltaG^(@)=-nFE_(cell)^(@)`. |
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| 2908. |
The cell in which the following cell reaction occurs, `2Fe _((aq))^(3+)+2I_((aq))^(-)to 2Fe _((aq))^(2+)+I_(2(s))` has `E_(cell)^(0)=0.236V` at 298 K. Calculate the standard Gibbs energy and the equilibrium costant of the cell reaction. |
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Answer» Two half reactions for the given redox reaction may be written as : `2Fe_((aq))^(3+)+2e^(-)to2Fe_((aq))^(2+),2I^(-)to I_(2)+2e^(-)` 2 moles of electrons are involved in the reactin, so `n=2` `DeltaG^(0)=-nFE_(cell)^(0)=(-2mol)xx(96500mol ^(-1))(0.236V)` `=- 45548CV=-454548J` `Delta_(r)G^(0) =-45. 55kj` `log K_(c)=(DeltaG^(0))/(2.303RT)` `=((-45.55kj))/(2.303xx(8.314xx10^(-3)kJK^(-1))xx(298K))=7.983` `K_(c)="Antilig"(7.983)=9.616xx10^(7)` `K_(c)=9.616xx10^(7)` |
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| 2909. |
The cell in which the following reaction occurs: `2Fe^(3+)(aq)+2I^(-)(aq)to2Fe^(2+)(aq)+I_(2)(s)` has `E_(cell)^(@)=0.236V` at 298K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction. |
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Answer» `2Fe^(3+)+2e^(-)to2Fe^(2+)` or `2I^(-)toI_(2)+2e^(-)` `Delta_(r)G^(@)=nFE_(cell)^(@)=-2xx96500xx0.236J=-45.55kJ" "mol^(-1)` `Delta_(r)G^(@)=-2.303RT" log "K_(c)` or `"log "K_(c)=(Delta_(r)G^(@))/(2.303RT)=(-45.55" kJ "mol^(-1))/(2.303xx8.314xx10^(-3)" kJ "K^(-1)xx298K)=7.983` `thereforeK_(c)=`Antilog `(7.983)=9.616xx10^7` |
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| 2910. |
The cell in which the following reaction occurs `2Fe^(3+)(aq)+2I^(-)(aq)to2Fe^(2+)(aq)+I_(2)(aq)+I_(2)(s)` has `E_(cell)^(0)=0.236V` at 298 K. Calculate the standard gibbs energy and the equilibrium constant of the cell reaction. |
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Answer» `2Fe^(3+)(aq)+2I^(-)(aq)to2Fe^(2+)(aq)+I_(2)(s)` For the given cell `n=2` `Delta_(r)G^(@)=-nFE_(cell)^(@)` `=-2xx96500xx0.236` `=-45.55kJ" "mol^(-1)` also, `Delta_(r)G^(@)=-2.303RT" "log" "K_(C)` `implieslogK_(C)=(-Delta_(r)G^(@))/(2.303RT)` `=(-45.55)/(2.303xx8.314xx10^(-3)xx298)` `=7.983` `impliesK_(C)="antilog"(7.983)` `=9.616xx10^(7)` |
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| 2911. |
The equilibrium constant of the following redox rection at 298 K is `1 xx 10^(8)` `2Fe^(3+) (aq.) +2I^(-) (aq.) hArr 2Fe^(2+)(aq.) +I_(2) (s)` If the standard reducing potential of iodine becoming iodide is +0.54 V. what is the standard reduction potential of `Fe^(3+)//Fe^(2+)` ?A. `+ 1.006 V`B. `-1.006 V`C. `+ 0.77 V`D. `-0.77 V` |
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Answer» Correct Answer - C `E^(@)=0.059/n log_(10) K` `=0.059/2 log_(10) 10^(8)=0.236` `E_(cell)^(@)=E_("Reduced species")^(@)-E_("Oxidised species")^(@)` `0.236=E_(Fe^(3+)//Fe^(2+))^(@)-0.54` `E_(Fe^(3+)//Fe^(2+))^(@)=0.77 V` |
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| 2912. |
Standard electrode potentials are `Fe^(2+)//Fe, E^(@) = -0.44 V` `Fe^(3+)//Fe^(2+), E^(@) = +0.77 V` If `Fe^(3+), Fe^(2+)`, and Fe block are kept together, thenA. `Fe^(3+)` increasesB. `Fe^(3+)` decreasesC. `Fe^(2+)//Fe^(3+)` remains unchangedD. `Fe^(2+)` decreases. |
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Answer» Correct Answer - B As `E(Fe^(3+)//Fe^(2+))` is more than ` E(Fe^(2+)//Fe),therefore Fe^(3+)` will get reduced and Fe will get oxidsied As such `Fe^(3+)` will decrease. |
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| 2913. |
What is primary cell? Give an example. |
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Answer» The cell which becomes dead after a period of time and can not be recharged back into the reactants is called primary cell. Example :Drycell,mercury cell. |
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| 2914. |
During electrolysis of aqueous `CuBr_(2)` using `Pt` electrode,A. `Br_(2)(g)` is evolved at anodeB. `Cu(s)` is deposited at cathdoeC. `Br_(2)(g)` is evolved at anode and `H_(2)(g)` at cathodeD. `H_(2)(g)` is evolved at anode |
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Answer» Correct Answer - a,b Aqueous `CuBr_(2):` At cathode `:(sqrt())Cu^(2+)+2e^(-)rarr Cu,` `E^(c-)._(Cu^(2+)|Cu)=+0.34V` `(X)2H_(2)O+2e^(-) rarr H_(2)+2overset(c-)(O)H,` ltbr. `E^(c-)._(H_(2)O|H_(2))=-0.83V` At anode `:(sqrt())2Br^(c-)rarr Br_(2)+2e^(-),` `E^(c-)._(Br^(c-)|Br_(2))=-1.09V` `(X)2H_(2)Orarr O_(2)+4H^(o+)+2e^(-), ` `E^(c-)._(H_(2)O|H_(2))=-1.23V` |
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| 2915. |
Calculate the molar ionic conductance of `Al^(3+)` ions at inifinite dilution, given that the molar conductance of `Al_(2)(SO_(4))_(3)` and molar ionic conductance of `SO_(4)^(2-)` ions at infinite dilution are `858" S "cm^(2)mol^(1) and 160" S "cm^(2)mol^(-1)` respectively. |
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Answer» Correct Answer - 189 S `cm^(2)mol^(-1)` `wedge_(m)^(@)[Al_(2)(SO_(4))_(3)]=2lamda_(m)^(@)(Al^(3+))+3lamda_(m)^(@)(SO_(4)^(2-))` |
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| 2916. |
`{:(Column-I,Column-II),((A)"Conductance",(p)Cm^(-1)),((B)"Specific condutance",(q)Ohm^(-1)cm^(2)mol^(-1)),((C)"Cell constant",(r)Ohm^(1-)),((D)"Equivalent conductance",(s)Ohm^(-1)cm^(-1)),((E)"Molar conductance",(u)Ohm^(-1)cm^(2) "equivalent"^(-1)):}` |
| Answer» Correct Answer - `A-(r),B-(s),C-(p),D-(u),E-(q) | |
| 2917. |
Specific conductivity of a 0.12 normal solution of an electrolyte is 0.023 `ohm^(-1)cm^(-1)`. Determine its equivalent conductivity. |
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Answer» Correct Answer - 200 `ohm^(-1)cm^(2)eq^(-1)` `wedge_(eq)=kappaxx(1000)/("Normality")=0.024Omega^(-1)cm^(-1)xx(1000cm^(3)L^(-1))/(0.12g" "eqL^(-1))=200Omega^(-1)cm^(2)eq^(-1)`. |
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| 2918. |
Conductivity of aqueous solution of an electrolyte depends onA. molecular mass of the electrolyteB. boiling point of solventC. degree of ionisationD. volume of the solvent |
| Answer» Correct Answer - C | |
| 2919. |
If the molar conductance values of `Ca^(2+)` and `Cl^(-)` at infinite dilution are respectively `118.88xx10^(-4) m^(2)` mho `mol^(-1)` and `77.33xx10^(-4) m^(2)` mho `mol^(-1)` then that of `CaCl_(2)` is : (in `m^(2)` mho `mol^(-1)`)A. `118.88xx10^(-4)`B. `154.66xx10^(-4)`C. `273.54xx10^(-4)`D. `1 96.21xx10^(-4)` |
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Answer» Correct Answer - C `^^_(m)^(@)(CaCl_(2))=lambda_(m)^(@)(Ca^(2+))+2lambda_(m)^(@)(Cl^(-))` `=118.88xx10^(-4)+2(77+33xx10^(-4))` `=273.54xx10^(-4)m^(2)"mol"^(-1)` |
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| 2920. |
When water is added to an aqueous solution of an electrolyte, what is the change in specific conductivity of the electrolyte?A. Conductivity decreasesB. Conductivity increasesC. Conductivity remains sameD. Conductivity does not depend on number of ions. |
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Answer» Correct Answer - A Conductivity decreases because number of ions per unit volume decreases |
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| 2921. |
The increase in the value of molar conducitivity of acetic acid with dilution is due toA. Decrease in interionic forces and increases in `alpha`B. Increase in the degree of ionization and interionic forces.C. Increase in self ionization of water.D. None of these |
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Answer» Correct Answer - a Factual statement. |
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| 2922. |
Molar conductivity of `NH_4OH` can be calculated by the equation.A. `Lambda_(NH_4OH)^@=Lambda_(Ba(OH)_2)^@+ Lamda_(NH_4Cl)^@-Lambda_(BaCl_2)^@`B. `Lambda_(NH_4OH)^@=Lambda_(BaCl_2)^@+ Lamda_(NH_4Cl)^@-Lambda_(Ba(OH)_2)^@`C. `Lambda_(NH_4OH)^@=(Lambda_(Ba(OH)_2)^@+ 2Lamda_(NH_4Cl)^@-Lambda_(BaCl_2)^@)/2`D. `Lambda_(NH^4OH)^@=(Lambda_(NH_4Cl)^@+Lambda_(Ba(OH)_2)^@)/2` |
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Answer» Correct Answer - C `Lambda_(Ba(OH)_2)^@=Lambda_(Ba^(2+))^@+ 2Lambda_(OH^-)^@` `Lambda_(BaCl_2)^@=Lambda_(Ba^(2+))^@+2Lambda_(Cl^-)^@` `Lambda_(NH_4Cl)^@=Lambda_(NH_4)^+ +Lambda_(Cl^-)^@` After substituting the above in `Lambda_(NH_4OH)^@=(Lambda_(Ba(OH)_2)^@+2Lambda_(NH_4Cl)^@ - Lambda_(BaCl_2)^@)/2` we get, `Lambda_(NH_4OH)^@ =Lambda_(NH_4^+)^@+ Lambda_(OH^-)^@` |
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| 2923. |
Which of the following curve gives the variation of `lambda_(m)` with `sqrtC` for `CH_(3)COOH` ?A. B. C. D. |
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Answer» Correct Answer - c `Lambda_(m)` decreases with increase in concentration of solution. |
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| 2924. |
Calculate the molar conductivity of acetic acid at infinite dilution. Given that molar conductivity of `HCl. CH_(3)COONa` and NaCl is 426.1,91.0 and 126.5 `ohm^(-1)cm^(2)"mol"^(-1)` respectively.A. `390-6ohm^(-1)cm^(2)"mol"^(-1)`B. `195.3ohm^(-1)cm^(2)"mol"^(-1)`C. `585.9ohm^(-1)cm^(2)"mol"^(-1)`D. `292.95ohm^(-1)cm^(2)"mol"^(-1)` |
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Answer» Correct Answer - A The degree of dissociation `(a)=(^^_(m)(CH_(3)COOH))/(^^_(m)(CH_(3)COOH))=(11.7)/(40.9+394.1)=0.03`, |
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| 2925. |
The value of molar conductivity of `HCl` is greater than that of `NaCl` at a particular temperature becauseA. molecular mass of HCl is less than that of NaClB. velocity of `H^(+)` ions is more than that of `Na^(+)` ionsC. HCl is strongly acidic.D. ionisation of HCl is larger than that of NaCl. |
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Answer» Correct Answer - B Factual question. |
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| 2926. |
Which out of 0.1 M HCl and 0.1 M NaCl, do you expect have greater ˄∞ m and why? |
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Answer» 0.1 M HCl will have greater ˄∞ m because H+ (aq) being smaller in size than Na+ (aq) and have greater mobility. |
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| 2927. |
Can a nickel spatula be used to stir a solution of copper sulphate ? Justify your answer.(E˚Ni2 /Ni = -0.25VE˚ Cu2 +/Cu = 0.34V) |
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Answer» No, Because Reduction potential of Ni is less than Cu. Ni will replace the Cu from CuSO4. |
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| 2928. |
Can a nicked spon be used to stir a solution of copper sulphate ? Support your answer with reaon. (Given : `E_(Ni^(2+)//Ni)^(@)=-0.25" V", E_(Cu^(2+)//Cu)^(@)=0.34" V"`) |
| Answer» No, its is not possible. Nickel is a stronger reducing agent than copper ? It will reduce `Cu^(2+)` ions of `CuSO_(4)` to copper. The nickel spoon will be gradually consumed in the reaction. | |
| 2929. |
A current of `2` ampere passing for `5` hr through a molten tin salt deposits `22.2 g` of t tin. Find the oxidation number of tin in the salt. |
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Answer» Correct Answer - 2 `Sn^(n+) + n e rarr Sn` For Sn `" "(w)/(E) = (i xx t)/(96500)` `(22.2)/(118//n) = (2 xx 5 xx 60 xx 60)/(96500) " " :. n = 2` |
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| 2930. |
A 1 M solution of `H_2SO_4` is electrolysed. Select correct statement in respect of products obtain at anode and cathode respectively: Given : `2SO_4^(2-) toS_2O_8^(2-)+2e^-,E^(@)=-1.23V` `H_2O(l)to2H^+(aq)+1//2O_2(g)+2e^-,E^(@)=-1.23V`A. concentration of `H_2SO_4` remain constant, `H_2,O_2`B. concentration of `H_2SO_4` remain constant, `O_2,H_2`C. concentration of `H_2SO_4` remain constant, `O_2,H_2`D. concentration of `H_2SO_4` remain constant, `S_2O_8^(2-),H_2` |
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Answer» Correct Answer - B |
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| 2931. |
When water is electrolysed, hydrogen and oxygen gas are produced. If 1.008 g of `H_(2)` is liberated at the cathode. What mass of `O_(2)` is formed at the anode?A. 32gB. 16gC. 8gD. 4g |
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Answer» Correct Answer - C `(W_(1))/(W_(2)) = (E_(1))/(E_(2))` `(1.008)/(W_(2)) = (1.008)/(8)` `:. W_(2) = 8 g` Where, `E_(1) and E_(2)` are equivalent masses of hydrogen and oxygen respectively |
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| 2932. |
The ion of least limiting molar conductivity among the following is(a) SO2-4(b) H+(c) Ca2+(d) CH3COO- |
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Answer» (d) CH3COO- The ion of least limiting molar conductivity among the following is CH3COO- |
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| 2933. |
What weight of copper will be deposited by passing 2 faradays of electricity through a cupric salt `(` atomic weight of `Cu=63.5)` ?A. `2.0`B. `3.175`C. `63.5`D. `127.0` |
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Answer» Correct Answer - c `Cu^(2+)+2e^(-) rarr Cu` `2F-=1mol =63.5g` |
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| 2934. |
What weight of copper will be deposited by passing 2 faradays of electricity through a cupric salt `(` atomic weight of `Cu=63.5)` ?A. 63.5gB. 31.75gC. 127gD. 2.0g. |
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Answer» Correct Answer - A `Cu^(2+)+2e^(-)rarrCu` 2mole 1 mole 2F 63.5gm |
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| 2935. |
What weight of copper will be deposited by passing 2 faradays of electricity through a cupric salt `(` atomic weight of `Cu=63.5)` ?A. 2.0 gmB. 3.175 gmC. 63.5 gmD. 127.0 gm |
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Answer» Correct Answer - C `Cu^(2+)+2e^(-)toCu` 2 faradays will deposit=1 g atm of Cu=63.5g. |
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| 2936. |
The amount of electricity that can deposit 108g of silver from silver nitrate solution isA. 1 ampereB. 1 coulombC. 1 faradayD. 2 ampere. |
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Answer» Correct Answer - C `Ag^(+)+underset(1"mole1F")(e^(-))rarrunderset(1"mole 108g")(Ag)` |
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| 2937. |
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. `Fe` and `Au`B. `Hg` and `Fe`C. `V` and `Hg`D. `Fe` and `V` |
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Answer» Correct Answer - b,c,d `NO_(3)^(c-)+e^(-) rarr ?, E^(c-)._(red)=0.96V` Compare the standard reduction potential of the given metals with that of `NO_(3)^(c-)` reduction. `E^(c-)._(reduction)` for `NO_(3)^(c-)` is greater than `E^(c-)._(reduction)` of `(i),(ii),` and `(iv).` So, `NO_(3)^(c-)` will be able to oxidize `Fe, Hg` and `V.` So pairs are as in `(b),(c),` and `(d)`. |
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| 2938. |
Consider `Zn+Cu^(2+)rarrZn^(2+)+Cu` If the standard emf is `E_("cell")=2.0V & F=96500 C` Find `DeltaG^(@) (KJ "mol")`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 mol^(-1)` `=-11.4kJ//mol` |
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| 2939. |
In the electrolysis of `7.2L` aqueous solution of `CuSO_(4)`, a current of `9.65A` passed for 2 hours. `a.` Calculate the weight of `Cu` deposited at cathode. `b.` Calculate the volume of `O_(2)` produced at anode at `27^(@)C` and `1 atm` pressure. `c.` Calculate the `pH` of the solution after electrolysis. `[` Atomic mass of `Cu=63g]` |
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Answer» First method `(CuSO_(4) rarr Cu^(2+)+SO_(4)^(2-))` At cathode`:` `Cu^(2+)+2e^(-) rarr Cu(s)` At anode `:` `2H_(2)O rarr O_(2)+4H^(o+)+4e^(-)` `(` Since the oxidation potential of `H_(2)Ogt` oxidation potential of `SO_(4)^(2-)`, so oxidation of `H_(2)O` occurs at anode and `SO_(4)^(2-)` and `H^(o+)` ions remain in the solution to give an acidic solution `).` Number of faradays `=(Ixxt)/(96500C)=(9.65A xx 2 xx 3600 s )/(96500C mol^(-1))` `a.` Now from cathode `: (Cu^(2+)+2e^(-)rarr Cu)` `2e^(-) =2F=1` mol of `Cu=63 g ` of `Cu` `0.72F=(63g xx 0.72F)/(2)=22.68g` `b.` Now from anode `: (2H_(2)Orarr O_(2)+4H^(o+)+4e^(-))`. `4e^(-)=4-=1 mo l `of `o_(2)` `0.72F=(1xx0.72)/(4)=0.18 mol` Now using gas equation `: PV=nRT` `V=(nRT)/(P)=(0.18xx0.082xx300)/(1)=4.428L` of `O_(2)` `c.` Now from anode` (2H_(2)O rarr O_(2)+4H^(o+)+4e^(-))` `4e^(-)=4F=4 mol `of `H^(o+)` `[H^(o+)]=("Mol")/("Volume in L")=(0.72 mol)/(7.2L)=10^(-1)M` `:. pH=-log [10^(-1)]=1` `A.` At cathode`:` `1F=1` equivalent of `Cu=(63)/(2) `of `Cu` `0.7F=(63)/(2)xx0.72=22.68g` of `Cu` deposited `b.` At anode `:` `1F=1` equivalent of `O_(2)` `0.72F=0.72` equivalent of `O_(2)=(0.72)/(4)=0.18 mol ` of `O_(2)` `[{:(Since mol=("Equivalent")/("n factor")`or `("Equivalent")/("charge"),=,(0.72)/(4)),(2O^(-2) rarrO_(2)+4e^(-1)(n fact o r=4),=,):}]` Now for volume of `O_(2)`, process as in first method. `c." "1F=1` equivalent of `H^(o+)` ions. `0.72F=0.72` equivalent of `H^(o+)=0.72mol `of `H^(o+)` `[` Since` n` factor for `H^(o+)=1]` `:. [H^(o+)]=(0.72 mol)/(7.2L)=10^(-1)M` `:. pH=1` |
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| 2940. |
If a spoon of copper metal is placed in a solution of ferrous sulphate .A. `Cu` will precipitate outB. iron will precipitateC. `Cu` and `Fe` will precipitateD. no reaction will take place |
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Answer» Correct Answer - D Reduction potential of `Cu^(2+)` is greater than r.p. of `Fe^(2+)` so it can not dusplace `Fe^(2+)` from `FeSO_(4)`. |
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| 2941. |
In the electrolysis of `CuSO_(4)` , the reaction `Cu^(2+) + 2 e^(-) to Cu` , takes place atA. anodeB. cathodeC. in solutionD. none |
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Answer» Correct Answer - B Reduction always occurs at cathode during electrolysis . |
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| 2942. |
The efficiency of a fuel cell is given by:A. `(DeltaS)/(DeltaG)`B. `(DeltaR)/(DeltaG)`C. `(DeltaG)/(DeltaS)`D. `(DeltaG)/(DeltaH)` |
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Answer» Correct Answer - D Efficiency of a fuel cell `=(DeltaG)/(DeltaH)` |
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| 2943. |
In an electrolytic cell current flowsA. from cathode to anode in outer circuitB. from anode to cathode outside the cellC. from cathode to anode inside the cellD. none |
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Answer» Correct Answer - A Follow electrolysis . |
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| 2944. |
The most reactive metal among the following is:A. AlB. NiC. PbD. Cu |
| Answer» Correct Answer - A | |
| 2945. |
If a spoon of copper metal is placed in a solution of ferrous sulphate:A. Cu will precipitate outB. iron will precipitateC. Cu and Fe will precipitateD. no reaction will take place |
| Answer» Correct Answer - D | |
| 2946. |
The efficiency of a fuel cell is given by:-A. `(DeltaH)/(DeltaG)`B. `(DeltaG)/(DeltaS)`C. `(DeltaG)/(DeltaH)`D. `(DeltaS)/(DelaG)` |
| Answer» Correct Answer - C | |
| 2947. |
Among `Na, Hg, S, Pt` and graphite, which can be used as electrodes in electrolystic cells having aqueous solutions?A. Hg and PtB. Hg, Pt and graphiteC. Na and SD. Na, Hg and S |
| Answer» Correct Answer - B | |
| 2948. |
Among `Na,Hg,S,Pt` and graphite which can be used as electodes in electrolytic cell having aqueous solutions?A. `Na` and `S`B. `Hg,pT` and `S`C. `Na,Hg,` and `S`D. `Hg,Pt,` and graphite |
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Answer» Correct Answer - d Inert electrodes are used in electrolysis. Graphite, `Hg, Pt`. |
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| 2949. |
Which of the substances Na,Hg,S,Pt and graphite can be used as electrodes in electrolytic cells having aqueous solutionsA. Na,Pt and graphiteB. Na and HgC. Pt and graphite onlyD. Na and S only |
| Answer» Correct Answer - C | |
| 2950. |
Among `Na,Hg,S,Pt` and graphite which can be used as electodes in electrolytic cell having aqueous solutions?A. `Na` and `S`B. `Na, Hg` and `Pt`C. `Na, Hg` and `S`D. `Hg, Pt` and graphite |
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Answer» Correct Answer - D `Na` canot be used in aqueous solution as it reacts violently with water. |
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