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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.
| 3101. |
`E^(@) (SRP)` of different half cell given `{:(E_(Cu^(2+)//Cu)^(@) =0.34"volt",,,E_(Zn^(2+)//Zn)^(@) =- 0.76"volt"),(E_(Ag^(+)//Ag)^(@) = 0.8"volt",,,E_(Mg^(2+)//Mg)^(@) =- 2.37 "volt"):}` In which cell `Delta^(@)` is most negative:-A. `Zn|Zn^(2+)(1 M)||Mg^(2+)(1 M)|Mg`B. `Zn|Zn^(2+)(1 M)||Ag^(+)(1 M)|Ag`C. `Cu|Cu^(2+)(1 M)||Ag^(+)(1 M)|Ag`D. `Ag|Ag^(+)(1 M)||Mg^(2+)(1 M)|Mg` |
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Answer» Correct Answer - B (b) (a)`Zn|Zn^(2+)(1 M)||Mg^(2+)(1M)|Mg` `E_(cell)^(@)=E_(cathode)^(@)-E_(anode)^(@)=E_(Mg^(2+)//Mg)^(@)-E_(Zn^(2+)//Zn)^(@)` `=-2.37-(-0.76)=-1.61" V"` `DeltaG^(@)=-nFE_(cell)^(@)=-2xx96500(C )xx(-1.61" V")` `=+310,730" CV"=+310.730" kJ"` (b) `Zn|Zn^(2+)(1 M)||Ag^(+)(1 M)|Ag` `E_(cell)^(@)=E_(Ag^(+)//Ag)^(@)-E_(Zn^(2+)//Zn)^(@)` `=0.80-(-0.76)=+1.56" V"` `DeltaG^(@)=-nFE_(cell)^(@)=-2xx96500" C"xx1.56" V"` `=-301,080" CV"` `=-301.08" CV"` (c )`Cu|Cu^(2+)(1 M)||Ag^(+)(1M)|Ag` `E_(cell)^(@)=E_(Ag^(+)//Ag)^(@)-E_(Cu^(2+)//Cu)^(@)` `=0.80-0.34=+0.46" V"` `DeltaG^(@)=-nFE_(cell)^(@)=-2xx96500" C"xx-0.46" V"` `=-88,780" CV"=-88.780" kJ"` (d)`Ag|Ag^(+)(1 M)||Mg^(+)(1M)|Mg` `E_(cell)^(@)=E_(Mg^(2+)//Mg)^(@)-E_(Ag^(+)//Ag)^(@)` `=-2.37-0.80=-3.17" V"` `DeltaG^(@)=-nFE_(cell)^(@)=-2xx96500" C"xx(-3.17" V")` `=+611,810" CV"=+611.0810" kJ"` (b) has maximum negative `DeltaG^(@)` value. |
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| 3102. |
`E^(@) (SRP)` of different half cell given `{:(E_(Cu^(2+)//Cu)^(@) =0.34"volt",,,E_(Zn^(2+)//Zn)^(@) =- 0.76"volt"),(E_(Ag^(+)//Ag)^(@) = 0.8"volt",,,E_(Mg^(2+)//Mg)^(@) =- 2.37 "volt"):}` In which cell `Delta^(@)` is most negative:-A. `Zn(s) |Zn^(2+)(1M)||Mg^(2+) (1M)|Mg(s)`B. `Zn(s) |Zn^(2+)(1M)||Ag^(+)(1M)|Ag(s)`C. `Cu(s)|Cu^(2+)(1M)||Ag^(+)(1M)|Ag(s)`D. `Ag(s)|Ag^(+)(1M)||Mg^(2+)(1M)|Mg(s)` |
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Answer» Correct Answer - B For (A) and (D) `E_("cell")^(@)=-veimplies Delta G^(@)= +ve` `{:("for"(C),,E_("cell")^(@)= +0.46),("for"(B),,E_("cell")^(@)= +1.56):}` `Delta G^(@)` is most negative for (B). |
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| 3103. |
Electrolysis of a solution of `MnSO_(4)` in aqueous sulphuric acid is a method for the preparation of `MnO_(2)` as per reaction, `Mn_(aq.)^(2+)+2H_(2)O rarr MnO_(2(s))+2H^(+)(aq.)+H_(2(g))` Passing a current of `27` ampere for `24` hour gives one `kg` of `MnO_(2)`. What is the value of current efficiency ? Write the reaction taking place at the cathode and at the anode.A. ` 100%`B. ` 95.185%`C. ` 80%`D. ` 82. 951%` |
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Answer» Correct Answer - B ` (1000 xx2)/((55+ 32)) = ( 27 xx 24xx 3600 xx ita)/(96500) or eta=0.951 = 95. 1%`. |
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| 3104. |
The highest electrical conductivity of the following aqueous solutions is ofA. ` 0.1 M` acetice acidB. `0.1` M chloroacetic acidC. ` 0.1 M` fluoroacetic acidD. ` 1.1 M` difluoroacetic acid |
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Answer» Correct Answer - A ` E_(I^- Ag//Ag)^@ = E_(Ag^+//Ag)^@ + ( 0.50)/1 log KSP_(Agl)` `-0. 152 = 0.8 + ( 0.05)/(1) log. K_(SP_Agl)` ` log K_((SP_(Agl))` ` log K_(Sp_(Agl)) =- 16 . 13 `.. |
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| 3105. |
The highest electrical conductivity of the following aqueous solutions is ofA. ` 0.1 M` acetic acidB. ` 0. 1 M` chloroacetic acidC. `0.1 M` fluoroacetic acidD. `0.1 M` difluoracetic acid |
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Answer» Correct Answer - D Difluoroacetic acid is strong acid due to -IE of fluorine atoms. |
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| 3106. |
When 0.1 mol `MnO_(4)^(2-)` is oxidized, the quantity of electricity required to cmpletely oxidize `MnO_(4)^(2-)` to `MnO_(4)^(-)` isA. 96500 CB. `2xx96500C`C. `9650C`D. `96.50C` |
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Answer» Correct Answer - C The oxidation reaction is `MnO_(4)^(2-)toMnO_(4)^(-)+e^(-)` 1 mol of `MnO_(4)^(2-)` for oxidation requires electricity=1F=96500C `therefore0.1` mol of `MnO_(4)^(2-)` will require electricity =9650C |
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| 3107. |
During the perparation of `H_(2)S_(2)O_(s)` (per disulphuric acid) `O_(2)` gas is also released at anode as byproduct. When `9.72 L` of `H_(2)` releaseds at cathode and `2.35 L O_(2)` at anode the weight `H_(2)S_(2)O_(8)` depostited at the cathode is .A. ` 87. 12`B. ` 48. 65`C. ` 83. 42`D. ` 51. 74` |
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Answer» Correct Answer - B ` 2HSO_4^(-) rarr H_2S_2O_8 +2e` ` 20^(2-) rarr O_2 +4e` `(2.35)/(22.5) =0. 105 0.42 mole` `H^+ + 2e rarr H_2` Mole of e for ` H_2 S_2O_8 =0. 87 - 0. 42 =0. 45 ` mole ` :.` Mole of ` H_2 S_2 OO_8 =0. 225 "mole" = 43. 65 g`. |
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| 3108. |
The highest electrical conductivity of the following aqueous solutions is ofA. 0.1 M fluoroacetic acidB. 0.1 M difluoroacetic acidC. 0.1M acetic acid.D. 0.1M chl oroacetic acid. |
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Answer» Correct Answer - B Difluoroacetic acid is stronger than fluoroacetic acid of chloroacetic acid or acetic acid. Therefore 0.1M difluoroacetic acid will have highest electrical conductivity. |
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| 3109. |
In order to completely oxidize `0.1 mol` of `MnO_(4)^(2-)` to permanganate ion. The quantity of electricity required isA. `96500C`B. `2xx96500C`C. `9650C`D. `96.50C` |
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Answer» Correct Answer - c `MnO_(4)^(2-)rarr MnO_(4)^(c-)+e^(-)` `1F=1mol `of `MnO_(4)^(c-)` ` 0.1F=0.1 mol` of `MnO_(4)^(c-)` `0.1F=0.1xx96500=9650C` |
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| 3110. |
The charge required for the reduction of `1 mol` `Cr_(2)O_(7)^(2-)` ions to `Cr^(3+)` isA. 1 FB. 2 FC. 6 FD. 4 F |
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Answer» `Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-) rarr 2Cr^(3+) +7H_(2)O` 1 mol `Cr_(2)O_(7)^(2-)` requires 6 mol of electrons, i.e., 6 Faraday charge |
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| 3111. |
The charge required for the reduction of `1 mol` `Cr_(2)O_(7)^(2-)` ions to `Cr^(3+)` isA. 96500CB. `2xx96500C`C. `3xx96500C`D. `6xx96500C`. |
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Answer» Correct Answer - D `Cr_(2)O_(7)^(2-)+6e^(-)rarr2Cr^(3+)` Reduction of 1 mol of `Cr_(2)O_(7)^(2-),Cr^(3+)` required 6 m oles of electrons. |
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| 3112. |
The charge required for the reduction of 1 mole of `Cr_(2)O_(7)^(2-)` ions to `Cr^(3+)` isA. 96500 CB. `2 xx 96500 C `C. `3 xx 96500 C`D. `6 xx 96500 C`. |
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Answer» Correct Answer - D `Cr_(2)O_(7)^(2-) to 2 Cr^(3+)` i.e. `2 Cr^(6+) + 6e^(-) to 2 Cr^(3+)` . Thus 1 mole of `Cr_(2)O_(7)^(2-)` ions require 6 F |
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| 3113. |
The charge required for the reduction of 1 mole of `Cr_(2)O_(7)^(2-)` ions to `Cr^(3+)` isA. 96500 CB. `2xx96500 C`C. `3 xx96500 C`D. `6xx96500 C` |
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Answer» Correct Answer - D `overset(+6)(Cr_(2))overset(-2)(O_(7)) rarr overset(+3)(Cr)` Hence, electrons involved in this reduction will be `=2xx[(+6)-(+3)]=2xx3=6` `:.` Charge required for one mole `=6xx96500 C` |
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| 3114. |
The electric charge required for electrode deposition of one gram-equivalent of a substance is :A. 1 ampere per secondB. 96,500 coulomb per secondC. 1 ampere for 1 hourD. charge on 1 mole of electron |
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Answer» Correct Answer - D Charge on 1 mole electron `= N xx e = Q` = Faraday . |
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| 3115. |
One of the methods of preparation of per disulphuric acid, `H_(2)S_(2)O_(8)`, involve electrolytic oxidation of `H_(2)SO_(4)` at anode `(2H_(2)SO_(4)toH_(2)S_(2)O_(8)+2H^(+)+2e^(-))` with oxygen ad hydrogen as by-products in such an electrolysis, 9.722 L of `H_(2)` and `2.35L` of `O_(2)` were generated at STP. What is the weighht of `H_(2)S_(2)O_(8)` formed? |
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Answer» Correct Answer - `43.45g` In beginning : At anode: `2H_(2)SO_(4) rarr H_(2)S_(2)O_(8) +2H^(4) + 2e^(-)` At anode: `2H_(2)O +2e^(-) rarr H_(2) +2OH^(-)` After some time: At anode: `2H_(2)O rarr 4H^(+) +O_(2) +4e^(-)` At anode: `2H_(2) +2e^(-) rarr H_(2) +2OH^(-)` moles of `O_(2)` evolved `=(2.35)/(22.4)` during this time moles of `H_(2)` evolved `=(2xx 2.35)/(22.4)` volume of `H_(2)` evolved with `O_(2) =(2 xx 2.35)/(22.4) xx 22.4 = 4.7 L` volume of `H_(2)` evolved with `H_(2)S_(2)O_(8) = 9.722 - 4.7 = 5.022 L` moles of `H_(2)` evolved `=(5.022)/(22.4)` moles of `H_(2)S_(2)O_(8)` formed `=(5.022)/(22.4)` wt. of `H_(2)S_(2)O_(8)` formed `=(5.022)/(22.4) xx 194 = 43.45 g` |
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| 3116. |
The charge required for the reduction of `1 mol` `Cr_(2)O_(7)^(2-)` ions to `Cr^(3+)` isA. `96500C`B. `2xx96500C`C. `3xx96500C`D. `6xx96500C` |
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Answer» Correct Answer - d `6e^(c-)+Cr_(2)O_(7)^(2-) rarr 2Cr^(3+)` So charge `=6xxF=6xx96500C` |
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| 3117. |
Aluminium oxide may be electrolysed at `1000^(@)`C to furnish aluminium metal (Atomic mass = 27 amu, 1 Faraday = 96500 Coulomb). The cathode reaction is `Al^(3+) + 3e^(-) rightarrow Al`. To prepare 5.12 kg of aluminium metal by this method would require:A. `5.49xx10^(4)C` of electricityB. `5.49xx10^(1)C` of electricityC. `5.49xx10^(7)C` of electricityD. `1.83xx10^(7)C` of electricity. |
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Answer» Correct Answer - C `Al^(3+)+3e^(-)rarrAl` `3xx96500 27g` To produce 27g of Al, electricity required `=3xx96500C` `therefore` To produce 5.12kg of Al, electricity required `=(3xx96500)/(27)xx5.12xx10^(3)C` `=5489.78xx10^(3)C=5.49xx10^(7) C` |
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| 3118. |
The electrice charge for eeletrode deposition of `1g` equivalent of a substance isA. 1 ampere per secondB. 96500 Coulombs per secondC. 1 ampere for one hourD. The charge on one mole of electrons |
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Answer» Correct Answer - d The quantity of electricity required to deposit or liberate `1 g` equivalent of any substance by passage of electric current. It is equal to the charge present on `1 mol` of electrons. One faraday is equivalent to `96500C`. |
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| 3119. |
Which one of the following is correct ?A. Equivalent conductance decreases with dilutionB. Specific conductance increases with dilutionC. Specific conductance decreases with dilutionD. Equivalent conductance increases with increase in concentration |
| Answer» Correct Answer - C | |
| 3120. |
A solution containing one mole per litre of each `Cu(NO_(3))_(2),AgNO_(3),Hg_(2)(NO_(3))_(2) and Mg(NO_(3))_(2)` is being electrolysed by using inert electrodes. The values of standard electrode potentials in volts (reduction potentials) are `Ag//Ag^(+)=+0.80,2Hg//Hg_(2)^(2+)=+0.79,Cu//Cu^(2+)=0.34,Mg//Mg^(2+)=-2.37` with increasing voltage, the sequence of deposition of metals on the cathode will beA. Ag,Hg,Cu,MgB. Mg,Cu,Hg,AgC. Ag,Hg,CuD. Cu,Hg,Ag |
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Answer» Correct Answer - C A cation havig highest reduction potential will be reduced first and so on. However, `Mg^(2+)` in aqueous solution will not be reduced `(E_(Mg^(2+)//Mg)^(o)ltE_(H_(2)O//(1)/(2)H_(2)+OH^(-)))` Instead water would be reduced in preference. |
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| 3121. |
Assume that impure copper contains only iron, silver and a gold as impurities. After passage of `140A`, for `482.5` sec, of the mass of the anode decreased by `22.260g` and the cathode increased in mass by `22.011g`. Estimate the % iron and % copper originaly present. |
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Answer» Correct Answer - `Cu =98.88%, Fe = 0.85%` At cathode only pure copper is deposited so % of `Cu = (22.011)/(22.26) xx 100 = 98/88%` Charge passed `=140 xx 482.5 = 67550` colombs `= (67550)/(96500) = 0.7` Faraday Faraday used fot reduction of `Cu =(22.011 xx 2)/(63.5) = 0.69326` so Farday used for iron `=0.7 - 69326 = 0.00674 F` mass of iron `= (0.00674 xx 56)/(2) = 0.1887 g` `% of Fe = (0.1887)/(22.26) xx 100 = 0.847%` |
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| 3122. |
A galvanic cell is set up from a zinc bar weighing `100g` and `1.0L` of `1.0M CuSO_(4)` solution. How long would the cell run if it is assumed to deliver a steady current of `1.0A. (` Atomic mass of `Zn=65)`.A. 1.1 hrB. 46 hrC. 53.6 hrD. `24.00` hr |
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Answer» Correct Answer - C 100 g Zn ` = (100)/(65)` mole = 1.53 mole 1 L of 1 M `CuSO_(4)` sol contains 1 mole of `CuSO_(4)` `therefore ` In `Zn + CuSO_(4) to Zn SO_(4) + Cu , CuSO_(4)` is the limiting reagent . To deposit completely 1 mole of Cu , electricity required `= 2 xx 96500 C` `t = (Q)/(I) = (2 xx 96500)/(1) "sec " = (2 xx 96500)/( 3600) ` hr = 53.6 hr . |
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| 3123. |
A galvanic cell is set up from a zinc bar weighing `100g` and `1.0L` of `1.0M CuSO_(4)` solution. How long would the cell run if it is assumed to deliver a steady current of `1.0A. (` Atomic mass of `Zn=65)`.A. `1.1` hoursB. 46 hoursC. `53.6` hourseD. 24 hours. |
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Answer» Correct Answer - c Mole of `Zn=(100)/(65.3)=1.53mol` Mole of `Cu^(2+)=1xx1=1 mol ` So `Zn` is in excess, hence `Cu^(2+)` will react completely. `Cu^(2+)+2e^(-) rarr Cu` `2F-=1 mol `of `Cu` `2xx96500C-=1.0Axxt(` in seconds `)` `t=(2xx96500)/(3600)h=53.6h` |
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| 3124. |
In electrolysis of dilute `H_2SO_4` using platinum electrodes .A. `Cl_2 ` is obtained at cathodeB. ` NH_3` is produced at anodeC. `h_2` is evolved at cathodeD. `O_2` is produced |
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Answer» Correct Answer - C When platinum electordes are dipped in diluted solution `H_4 SO_4` tan `H_2` is evolved at cathode. |
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| 3125. |
Aluminium oxide may be electorlysed at `1000^@C` to furnish aluminim metal (Atomic Mass `= 27 ` amu, `1 F = 96, 500 C)`. The cathode reaction is `Al^3 + 3d^(-) rarr Al^@` To prepare ` 5.12 kg` of aluminimu metal by this method woold require .A. ` 5. 49 xx 10^4 C` electricityB. `1. 83 xx 10^7 C` of elctricityC. ` 5. 94 xx 10^7 C` of electricityD. ` 5 . 49 xx 10^1 C` of electricity |
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Answer» Correct Answer - C ` 27 g ` of Al is obtained by passing a current of `3xx96500 C`. `:. 1 g ` og Al is obtained by passing a current of ` 3xx(96500)/(27) xx 4 . 12 xx 1000` `= 1.83 xx 10^7 Cxx 3= 5.49 xx 10^7 C`. |
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| 3126. |
A solution containing `1 mol` per litre of each `Cu(NO_(3))_(2), AgNO_(3),` and` Hg_(2)(NO_(3))_(2)` is being electrolyzed by using inert electrodes. The values of standard electrode potentials in volts `(` reduction potential `)` are `Ag|Ag^(o+)=+0.80, 2Hg|Hg_(2)^(2+)=+0.79` `Cu|Cu^(2+)=+0.34,Mg|Mg^(2+)=-2.37`. With increasing voltage, the sequence of deposition of metals at the cathode will beA. `Ag,Hg,Cu,Mg`B. `Mg,Cu,Hg,Ag`C. `Ag,Hg,Cu,Mg`D. `Cu,Hg,Ag,Mg` |
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Answer» Correct Answer - a Higher the oxidation potential, more easily it is oxidized and faster is the deposition of metal. Decreasing order of `E^(c-)._(o x I dation)` is `:` `E^(c-)._(Ag|Ag^(o+)(0.80))gtE^(c-)._(2Hg|Hg_(2)^(2+))(0.79)gtE^(c-)._(Cu|Cu^(2+)(0.34))gtE^(c-)._(Mg|Mg^(2+)(-2.37))` Hence, the sequence of deposition of metals at the cathode will `Ag, Hg, ` and `Cu`... |
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| 3127. |
A solution containing one mole per litre of each `Cu(NO_(3))_(2), AgNO_(3), Hg_(2)(NO_(3))_(2)` is being electrolysed by using inert electrodes. The values of standard electrode potentials in volts (reduction potentials) are `Ag^(+)//Ag = +0.80 V, HG_(2)^(2+)//Hg = +0.79 V` `Cu^(+)//Cu = +0.34 V, Mg^(2+)//Mg = -2.37 V` With increasing valtage, the sequence of deposition of metals on the cathode will beA. `Ag, Hg, Cu`B. `Cu, Hg, Ag`C. `Ag, Hg, Cu, Mg`D. `Mg, Cu, Hg, Ag` |
| Answer» Correct Answer - A | |
| 3128. |
A solution containing one mole per litre of each `Cu(NO_(3))_(2), AgNO_(3), Hg_(2)(NO_(3))_(2)` is being electrolysed by using inert electrodes. The values of standard electrode potentials in volts (reduction potentials) are `Ag^(+)//Ag = +0.80 V, HG_(2)^(2+)//Hg = +0.79 V` `Cu^(+)//Cu = +0.34 V, Mg^(2+)//Mg = -2.37 V` With increasing valtage, the sequence of deposition of metals on the cathode will beA. `Cu, Hg, Ag`B. `Ag, Hg, Cu, Mg`C. `Ag, Hg, Cu`D. `Mg, Cu, Hg, Ag` |
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Answer» Correct Answer - C The higher the reduction potential, reater the tendency to get reduced. Since reduction potentials of the ions are in the order : `Ag^(+) gt Hg_(2)^(2+) gt Cu^(2+) gt Mg^(2+)` We expect the second option to be correct but `Mg^(2+)` is never reduced in aqueous solution as its reaction potential is much lower than that of water `(-0.83 V)`. Therefore, the correct sequence of deposition of the metals will be: Ag, Hg, Cu. |
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| 3129. |
Assume that impure copper contains only `Fe, Au` and `Ag` as impurities. After passage of `140` ampere for `482.5` sec, the mass of anode decreased by `22.260g` and the cathode increased in mass by `22.011g`. Calculate the percentage of iron and percentage of copper originally present. |
| Answer» Correct Answer - `%Cu = 98.88%, %Fe = 0.85% ;` | |
| 3130. |
A daniell cell is set up by dipping a zinc rod weighing 100 g in 1 litre of 1.0 M `CuSO_(4)` solution. How long would the cell run if it delivers a steady current of 1.0 ampere? (Atomic masses Cu=63.5,Zn=65)A. 82.47 hrsB. 53.61 hrsC. 41.23 hrsD. 26.80 hrs |
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Answer» Correct Answer - B 1L of 1.0 M `CuSO_(4)` solution contains `Cu^(2+)=1` mole `100" g Zn"=(100)/(65)` mole thus, in the cell reaction `Zn+Cu^(2+)toZn^(2+)+Cu,Cu^(2+)` ions are the limiting reagent. The cell will run till all the `Cu^(2+)` ions are deposited, i.e., 1 mole of Cu is deposited. quantity of electricity requried for deposition of 1 mole of Cu=2F Time, `t=(2xx96500C)/(1A)=2xx96500s` `=(2xx96500)/(3600)hr=53.61hr` |
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| 3131. |
The mass of copper that will be deposited at cathode in electrolysis of `0.2M` solution of copper sulphate when a quantity of electricity equal to that required to liberate `2.24L` of hydrogen from `0.1M` aqueous `H_(2)SO_(4)` is passed `(` atomic mass of `Cu=63.5)` will beA. `1.59g`B. `3.18g`C. `6.35g`D. `12.70g` |
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Answer» Correct Answer - c `1F=1Eq of H_(2)=11.2L of H_(2)` `11.2L of H_(2)=1Eq of H_(2)` `2.24 L of H_(2)=(2.24)/(11.2)=0.2Eq` `:. 0.2 Eq of Cu` will be deposited. `1Eq of Cu=(63.5)/(2)xx0.2=6.35g` |
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| 3132. |
A solution containing `1 mol` per litre of each `Cu(NO_(3))_(2), AgNO_(3),` and` Hg_(2)(NO_(3))_(2)` is being electrolyzed by using inert electrodes. The values of standard electrode potentials in volts `(` reduction potential `)` are `Ag|Ag^(o+)=+0.80, 2Hg|Hg_(2)^(2+)=+0.79` `Cu|Cu^(2+)=+0.34,Mg|Mg^(2+)=-2.37`. With increasing voltage, the sequence of deposition of metals at the cathode will beA. ` Ag, Hg, Cu , Mg`B. ` Mg, Cu, Ag`C. ` Cu, Hg, Ag`D. ` Ag, Hg, Cu` |
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Answer» Correct Answer - D A cation having highest reduction potential will be reduced first and so on. However, ` Mg^(2+)` in aqueous solution will not be reduced `(E_(Mg^(2+)//Mg)^(@) lt E_(H_2 O.//(1)/(2)H_2+OH^-))`. Instead wated would be reduced in prefence . |
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| 3133. |
A `Zn` rod weighing `25g` was kept in `100 mL` of `1M CuSO_(4)` solution. After a certain time the molarity of `Cu^(2+)` in solution was `0.8`. What was molartiyof `SO_(4)^(2-)`? What was the weight of `Zn` rod after cleaning ? `("At. Weight of "Zn = 65.4.)` |
| Answer» Correct Answer - `23.692 g`, no charge molarity | |
| 3134. |
A certain current liberates `0.5g` of hydrogen in 2 hours. How many grams of copper can be liberated by the same current flowing for the same time in a copper sulphate solution ?A. `12.7g`B. `15.9g`C. `31.8g`D. `63.5g` |
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Answer» Correct Answer - b `("Weight of" H_(2))/("Ew of" H_(2))=("Weight of Cu")/(E_(w)" of Cu")` `(0.5)/(1)=(x)/(63.5//2)` `:.x=15.9g` |
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| 3135. |
A certain current liberates `0.5g` of hydrogen in 2 hours. How many grams of copper can be liberated by the same current flowing for the same time in a copper sulphate solution ?A. 12.7 gmB. 15,9 gmC. 31.8 gmD. 63.5 gm |
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Answer» Correct Answer - B `("Weight of Cu")/("Weight of "H_(2))=("Eq. weight of "Cu)/("Eq. weight. Of H")` `("Weight of Cu")/(0.50)=(63.6//2)/(1)` Weight of Cu=15.9gm |
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| 3136. |
A certain current liberates `0.5g` of hydrogen in 2 hours. How many grams of copper can be liberated by the same current flowing for the same time in a copper sulphate solution ?A. ` 12.7 g`B. ` 15.9 g`C. ` 31.8 g`D. ` 45 g` |
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Answer» Correct Answer - C `("Weight of" Cu)/("Weight of" H_2) = (Eq."weifht" of Cu)/(Eq."weight" . Of H)` `("Weight of" Cu)/(0.50) = ( 63.6//2)/1` Weight of ` Cu = 15.9 g`. |
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| 3137. |
A cell constituted by two electrodes A `(E^(@)A//A+0.35V)` and `B(E^(@)B//B=-0.42V)` Has value of `E_("Cell")^(@)` equal toA. 0.07VB. 0.77VC. `0.77V`D. `-0.07V`. |
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Answer» Correct Answer - B The given electrode potentials are oxidation potentials . First of all they are dhanged to reduction potential `E_("cell")^(@)=E_(c)^(@)-E_(a)^(@)` `E_(B)^(@).^(+)._(//B)-E_(A)^(@).^(+)._(//A)` `=0.42-(-0.35)=0.77V` |
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| 3138. |
A `Zn` rod weighing `1.0g` is taken in `100mL` of `1M CuSO_(4)` solution . After some time, `[Cu^(2+)]` in solution `=0.9M(` atomic weight of `Zn=65.5g)`. Which of the following statements is `//` are correct ? `a. 0.655g` of `Zn` was lost during the reaction. `b. 0.327 g `of `Zn` was lost during the reaction . `c.` There is no change in the molarity of `SO_(4)^(2-)` ion. `d.` There is a change in the molarity of `SO_(4)^(2-)` ion. |
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Answer» Correct Answer - a,c `a. mEq.` of `Cu^(2+)` before reaction ltbr. `=100xx1xx2 ` `(n` factor for `Cu^(2+)=2)` `=200` `mEq` of `Cu^(2+)` after reaction `=100xx0.9xx2=180` `mEq` of `Cu^(2+)` lost `=200-180=20` `mEq` of `Zn` lost `=20` Weight of `Zn` lost `=mEqxx10^(-3)xxEw` of `Zn` `=90xx10^(-3)xx(65.5)/(2)` `(n` factor for `Zn=2)` `=0.655g` `c.` There is no change in the molartiy of `SO_(4)^(2-)` ions, since `SO_(4)^(2-)` ions have not been oxidized and there is no change is the volume of solution . |
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| 3139. |
A solution of sodium sulphate was electrolyzed using some inert electrode. The product at the electrodes areA. `O_(2) ,H_(2)`B. `O_(2),Na`C. `O_(2),SO_(2)`D. `O_(2),S_(2) ,O_(8)^(2)` |
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Answer» Correct Answer - A The reactions are similar to those of water electrolysis producing `H_(2)` at cathode , `O_(2)` at anode |
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| 3140. |
A certain current liberates `0.5g` of hydrogen in 2 hours. How many grams of copper can be liberated by the same current flowing for the same time in a copper sulphate solution ?A. 12.7gmB. 15.9gmC. 31.8gmD. 63.5gm |
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Answer» Correct Answer - A `0.5g=(0.5)/(2)"mol of" H_(2)` 0.25 mol of `H_(2)` is liberated b y electricity ltb rgt `=2xx96500xx0.25C` Ca liberated by `2xx96500xx0.25C` `=(2xx96500xx0.25)/(2xx96500)= 0.2` mol Amount of `Cu=0.2xx63.5=12.7g` |
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| 3141. |
A Current of 9.65 ampere flowing for 10 minutes deposits 3.0g of the metal which is monovalent. The atomci mass of the metal isA. 10B. 50C. 30D. 96.5 |
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Answer» Correct Answer - B `Q="If"=9.65xx10xx60=5790C` ` underset("monovalent metal") (M^(+))+underset("IF" 96500C)(e)rarrunderset("1mole")(M)` With 5790C , amount of metal depsoited=3gm With 96500C, amount of metal deposited. `=(3)/(5790)xx96500` =50 g (equivalent weight) Atomic mass=`50xx` valency=`50xx1=50` amu |
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| 3142. |
If the half-cell reaction ` A = E^(-) rarr A^-` has a large negative reduction potential, it follows that .A. A is readily oxidisedB. A is readily reducedC. `A^(-)` is readily oxidisedD. `A^(-)` is readily reduced. |
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Answer» Correct Answer - C (c ) `A^(-)` ion is readily oxidised to A by losing an electron. |
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| 3143. |
If the half-cell reaction ` A = E^(-) rarr A^-` has a large negative reduction potential, it follows that .A. A is readily reducedB. A is readily oxidizedC. `A^(-)` is readily reducedD. `A^(-)` is readily oxidized. |
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Answer» Correct Answer - D Gain of electrons is reduc tion and A has a large negative reduction potential . This means the reduction potential of A is very smaller. A is difficult to reduce to `A^(-)`. Therefore `A ^(-)` is easy to oxidise to A. |
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| 3144. |
If the half-cell reaction ` A = E^(-) rarr A^-` has a large negative reduction potential, it follows that .A. A is readily reducedB. A is readily oxidisedC. `A^(-)` is readily reducedD. `A^(-)` is readily oxidised |
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Answer» Correct Answer - D Negative `E_(RP)^(@)` means more `E_(OP)^(@)` , i.e., more is the tendency to get itself oxidised . Thus `A^(-)` is readily oxidised . |
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| 3145. |
If the half-cell reaction ` A = E^(-) rarr A^-` has a large negative reduction potential, it follows that .A. A is readily reducedB. A is readily oxdisedC. `A^-` is readily reducedD. A is readily oxidised |
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Answer» Correct Answer - B Negative electrode potential (reduction potential ) indicates lesser tendency for the reduction. Hence A is readily oxidized. |
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| 3146. |
If the half-cell reaction `A+e^(-)toA^(-)` has a large negative reduction potentials, it follows that:A. A is readly reducedB. A is readily oxidisedC. `A^(-)` is readily reducedD. `A^(-)` is readily oxidised |
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Answer» Correct Answer - D |
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| 3147. |
`wedge^(@)._(m)` of `BaCl_(2) , H_(2)SO_(4)`, and `HCl(aq)` solutions are `x_(1),x_(2),` and `x_(3)` respectively. `wedge_(m(BaSO_(4)))` is `:`A. `x_(1)+x_(2)-x_(3)`B. `x_(1)-x_(2)-x_(3)`C. `x_(1)+x_(2)-2x_(3)`D. `x_(1)-2x_(2)+x_(3)` |
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Answer» Correct Answer - C `wedge^(@)._(m(BaCl_(2)))=lambda_(m)^(@)._((Ba^(2+)))+2lambda^(@)._(m(Cl^(c-)))` `wedge^(@)._(m(H_(2)SO_(4)))=2lambda^(@)._(m(H^(o+)))+lambda^(@)._(m(SO_(4)^(2-)))` `wedge^(@)._(m(HCl))=wedge^(@)._(m(BaCl_(2)))-2wedge^(@)._(m(HCl))` `x_(1)+x_(2)-2x_(3)` |
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| 3148. |
Calculate `DeltaG_(298)^(@)` and `DeltaS_(298)^(@)` for the reaction : `2H_(2(g))+O_(2(g)) rarr 2H_(2)O_((l)): DeltaH_(298)^(@) = -136.64kcal`. Given that `O_(2(g))+4H^(+)+4e rarr 2H_(2)O_((l)), E^(@) = 1.23V` `2H_(2(g)) rarr 4H^(+)+4e, E^(@) = 0.00V` |
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Answer» Given `{:(O_(2(g))+4H^(+)+4erarr2H_(2)O_(4), E^(@)=1.23V),(2H_(2(g))rarr4H^(+)+e^(-), E^(@)=0.00V):} /("On adding" 2H_(2(g))+O_(2(g))rarr2H_(2)O_((l)))` , `E^(@)=1.23+0 = 1.23V)` `because DeltaG_(298)^(@) = -nE^(@)F` `= 4 xx 1.23 xx 96500 = -474780J` `= -474.78J = -113.58 kcal` Also `DeltaG_(298)^(@) = DeltaH^(@)-TDeltaS^(@)` `DeltaS_(298)^(@) = [(DeltaG^(@)-DeltaH^(@))/(T)]= (DeltaH^(@)-DeltaG^(@))/(T)` `= (-136.64-(-113.58))/(298)` `= -0.77 kcal` |
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| 3149. |
To find the standard potential of `M^(3+)//M` electrode, the following cell is constituted : `Pt|M|M^(3+) (0.0018 mol^(-1) L)||Ag^(+) (0.01 mol^(-1) L(|Ag` The emf of this cell is found to be 0.42 volt. Calculate the standard potential of the half reaction `M^(3+)+3e^(-) rarr M`. `E_(Ag^(+)//Ag)^(@)=0.80` volt. |
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Answer» The cell reaction is `M+3Ag^(+) rarr 3Ag+M^(3+)` Applying Nernst equation, `E_(cell)=E_(cell)^(@)-0.0591/3"log" ([M^(3+)])/([Ag^(+)]^(3))` `0.42=E_(cell)^(@)-0.0591/3"log"((0.0018))/((0.01)^(3))=E_(cell)^(@)-0.064` `E_(cell)^(@)=(0.42+0.064)=0.484` volt `E_(cell)^(@)=E_("Cathode")^(@)-E_("Anode")^(@)` or `E_("Anode")^(@)=E_("Cathode")^(@)-E_(Cell)^(@)` `=(0.80-0.484)=0.32` volt |
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| 3150. |
For the reduction of `NO_(3)^(-)` ion in an aqueous solution `E^(@)` is `+0.96V`. Values of `E^(@)` for some metal ions are given below `V^(2+)(aq)+2e^(-)hArrV,E^(@)=-1.19V t t` `Fe^(3+)(aq)+3e^(-)rarrFe: E^(@)=-0.04V` `Au^(3+)(aq)+3e^(-)rarrAu, E^(@)=+1.40V` `Hg^(2+)(aq)+3e^(-)rarrHg, E^(@)=+0.86V` The pari(s) of metals that is/are oxidised by `NO_(3)^(-)` in aqueous solution is (are)A. V and HgB. Hg and FeC. Fe and AuD. Fe and V |
| Answer» Correct Answer - a,b,d | |