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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.
| 2001. |
Statement : Passes of charge through `CuSO_(4)(aq)` solution in presence of `Pt` electode increase it `pH`. Explanation : Concentration of `[OH^(-)]` in solution decreases.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 Anode: `2H^(2)O rarr 4H^(+) + O_(2) + 4e` or `2OH^(-) rarr H_(2)O + (1)/(2)O_(2) + 2e` Cathode: `Cu^(2+) + 2e rarr Cu`. |
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| 2002. |
Statement : `Pt_(H_(2))|HCl ` at `25^(@)C E_(H) = 0`. Explanation : For primary reference electrode `E_(H//H^(+))^(@) = 0`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 - B Primary reference electrode is `{:(P.t_(H_(2))),(1 atm):}|{:(HCl),(1M):}` at `25^(@)C` its `E^(@)=0` |
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| 2003. |
`{:(Column-I,Column-II),((A)"Electrolytic cell",(p)-DeltaG^(@)),((B)nFE_(cell)^(0),(q)"Concentration cell"),((C)E_(cell)=(0.059)/(n)log.(C_("cathode"))/(C_("anode")),(r)96500 "Coulombs"),((D)"Diffusion of ions",(s)"Device converting electrical energy into chemical energy"),((E)1"Farday",(u)"Salt bridge"):}` |
| Answer» Correct Answer - `A-(s),B-(p),C-(q),D-(u),E-(r)` | |
| 2004. |
Assertion `(A):` Equivalent conductance increase with dilution for an electrolyte solution. Reason `(R):` The number of ions per litre of electrolyte increases with dilution.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 incorrect.D. If `(A)` is incorrect, `(R)` is correct. |
| Answer» Correct Answer - A | |
| 2005. |
For an electrode reaction written as `M^(n+)+n e^(-)rarr M` , `E_("red")=E_("red")^(@)-(RT)/(nf)ln(1)/([M^(n+)])` `=E_("red")^(@)-(0.059)/(n)"llog"(1)/([M^(n+)])` at 298 k For the cell reaction `aA+bB rarr xX+yY` `rArr E_("cell")=E_("cell")^(@)-(RT)/(nf)"log"([X]^(x)[Y]^(y))/([A]^(a)[B]^(b))` For pure solids, liquids or gases at lampt. molarconc = 1 Std. free energy change `Delta G^(@)=-nfE^(@)` where f - for faraday = 96,500 c, n = no. of `e^(-)` `Delta G=-2.303 RT log K_(c )` where `K_(c )` is equilibrium constant `K_(c )` can be calculated by using `E^(@)` of cell. On the basis if information avaliable from the reaction `(4)/(3)Al+O_(2)rarr(2)/(3)Al_(2)O_(3) Delta G=-827 kJ//mol^(-1)` The minimum emf required to carry out an electrolysis of `Al_(2)O_(3)` isA. 2.14 VB. 4.28 VC. 6.42 VD. 8.56 V |
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Answer» Correct Answer - A `Al rarr Al^(3+)+3e^(-)` `4/3 mol Al equiv 4/3 xx 3 mol e^(-)` `equiv 4 mol e^(-)` i.e., `n=4` `DeltaG=-nFE` `-827xx1000=-4xx96500xxE` |
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| 2006. |
If conductivity is directly proportional to the cross-sectional area of the solution and the concentration of the solution in it and is inversely proportional to the length of the cessel then the constant proportionality is expressed inA. `S m^(2) mol^(-1)`B. `S^(2) m^(2) mol`C. `S m mol^(-1)`D. `S^(2) m^(2) mol^(-1)` |
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Answer» Correct Answer - C Let the constant of proportionating be `x`, then `kappa = x(A)/(l)M` Substituting the units, we have `S m^(-1) = x(m^(2))/(m)mol m^(3)` `x = S m mol^(-1)` |
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| 2007. |
The best conductor of electricity is a 1M solution ofA. `CH_(3)COOH`B. `H_(2)SO_(4)`C. `H_(3)PO_(3)`D. Boric acid |
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Answer» Correct Answer - B `H_(2)SO_(4)` is strong electrolyte . |
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| 2008. |
When a copper wire is placed I a solution of `AgNO_3`, the solution acquires blue colour. This due to the formation of .A. forms a soluble complex with `AgNO_(3)`B. is oxidized to `Cu^(++)`C. is reduced to `Cu^(++)`D. splits up into atomic form and dissolves |
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Answer» Correct Answer - B SRP of Cu is less than Ag and thus `Cu + 2 Ag^(+) to Cu^(2+) + 2 Ag` reaction occurs . |
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| 2009. |
When a copper wire is placed I a solution of `AgNO_3`, the solution acquires blue colour. This due to the formation of .A. Soluble complex of copper with `AgNO_3`B. `Cu^(+) ions`C. `Cu^(2+)` ionsD. `cu^(-)` ion by the reduction of Cu |
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Answer» Correct Answer - C Since, `Ag^(+)` ions are reduced to Ag and `E_(Ag^(+)//Ag)^@ gt E_(cu^(++)//Cu)^@ Cu` is oxidized `Cu^(++)`. |
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| 2010. |
The value of equilibrium constant for a feasible cell reaction is :A. `lt 1`B. `= 1`C. `gt 1`D. zero |
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Answer» Correct Answer - C K = antilog `((nE^(@))/0.0591)` For feasible cell, `E^(@)` is positive, hence from the above equation, `K gt 1` for feasible cell reaction. |
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| 2011. |
`E^(@)` for the electrochemical cell `Zn(s)|Zn^(2+) 1 M (Aq.)||Cu^(2+) 1 M (aq.)|Cu(s)` is 1.10 V at `25^(@)C`. The equilibrium constant for the cell reaction, `Zn(s) +Cu^(2+) (aq.) hArr Zn^(2+) (aq.)+Cu(s)` Will be :A. `10^(-37)`B. `10^(37)`C. `10^(-39)`D. `10^(39)` |
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Answer» Correct Answer - B K = antilog `[(nE^(@))/0.0591]` = antilog `[(2xx1.10)/0.0591]=1.67xx10^(37)` |
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| 2012. |
Distinguish between electrolytic and galvanic cells. |
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Answer» Electrolytic cell: 1. This device is used to bring about a non-spontaneous chemical reaction by passing an electric current. 2. It is used to bring about a chemical reaction generally for the dissociation (electrolysis) of compounds. 3. In this cell, electrical energy is converted into chemical energy. 4. In this cell, the cathode is negative and the anode is positive. 5. Electrolytic cells are irreversible. 6. Oxidation takes place at the positive electrode and reduction at the negative electrode. 7. The electrons are supplied by the external source and enter through cathode and come out through anode. 8. It is used for electroplating, electrorefining, etc. Electrochemical cell (Galvanic cell or Voltaic cell): 1. This device is used to produce electrical energy by a spontaneous chemical reaction. 2. It is used to generate electricity. 3. In this cell, chemical energy is converted into electrical energy. 4. In this cell, the cathode is positive and the anode is negative. 5. Electrochernical cells are reversible. 6. Oxidation takes place at the negative electrode and reduction at the positive electrode. 7. The electrons move from anode to cathode in the external circuit. 8. It is used as a source of electric current. |
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| 2013. |
For the cell: `Mg(s)|Mg^(2+)(aq)||Ag^(+)(aq)|Ag(s)`, calcualte the equilibrium constant of the cell reaction at `25^(@)C` and maximum work that can be obtained by operating the cell, `E_(Mg^(2+)//Mg)^(@)=-2.37V and E_(Ag^(+)//Ag)^(@)=+0.80V(R=8.31" J "mol^(-1)K^(-1))`. |
| Answer» Correct Answer - Max. work `DeltaG^(@)=611.81" kJ Eqm. Const, K"=1.891xx10^(107)` | |
| 2014. |
The value of the reaction quotient, Q, for the cell: `Ni(s)|Ni^(2+)(0.9M)||Cl^(-)(0.40M)|Cl_(2)(g0.10"atm"),Pt(s)` isA. `1.3xx10^(-1)`B. `8.0xx10^(-2)`C. `3.0xx10^(-1)`D. `3.0xx10^(-2)` |
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Answer» Correct Answer - c The half cell reaction and overall recation are `Ni rarrNi^(2+)+2e^(-)` `Cl_(2)+2e^(-)rarr2Cl^(-)` `NirarrNi^(2+)+2Cl^(-)` `therefore Q([Ni^(2+)][Cl^(-)](2))/([pCl_(2)])` `=(0.19xx(0.4)^(4))/(0.1)=0.304` `=3.04xx10^(-1)` |
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| 2015. |
Reduction electrode potentials of half cells (1) `Pt(H_(2))|H^(+)(C_(f)),(2)Pt(cl_(2))|(Cl^(-))(C_(2))` and (3) `Ag^(+)|Ag^(+)(C_(3))` on inc reasing `C_(1)C_(2)`, and `C_(3)` (all gases are at 1 atm pressure)A. will increase in all the three casesB. will decrease in all the three casesC. will increase in case (1) and (3) but decrease in case (2)D. will decrease in case (1) and (3) but increase in case (2). |
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Answer» Correct Answer - c Reduction potential in three cases are as follows: (A) `H^(+)+e^(-)rarr(1)/(2)H_(2)` On increasing `[H^(+)]E_(2)H+H_(2)` will increase `(1)/(2)Cl_(2)+2e^(-)rarrCl^(-)` On increasing `[Cl_(4)^(-)],E_(Cl_(2)//Ag)` will increase. |
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| 2016. |
One litre 1 M `CuSO_(4)` solution is electrolysed. After passing 2 F of electricity, the molarity of solution will beA. `M//3`B. `M//2`C. `M//4`D. 0 |
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Answer» Correct Answer - d `Cu^(2+)+2e^(-) rarrCu` 2F will decompose 1 mol of `Cu^(2+)` and will deposit 1 mol of copper. 1L of 1M `CuSO_(4)` solution contains 1 mol of `CuSO_(4)` therefore, molarity after electrolysis =0 |
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| 2017. |
When `CuSO_(4)` is electrolysed, using Pt electrodesA. Copper is liberated at cathode and sulphur at anodeB. Copper is liberated at cathode and oxygen at anodeC. Sulphur is liberated at cathode and oxygen at anodeD. Oxygen is liberated at cathode and copper at anode |
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Answer» Correct Answer - B (b) `CuSO_(4)overset((aq)) to Cu^(2+)(aq)+SO_(4)^(2-)(aq)` `H_(2)O overset((aq))hArr H^(+)(aq)+OH^(-)(aq)` At cathode : `Cu^(2+)(aq)+2e^(-) to Cu(s)` At anode : `4OH^(-)(aq) to 2H_(2)O(aq)+O_(2)(g)+4e^(-)` |
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| 2018. |
The standard `EMF` of a galvanic cell involving cell reaction with `n=2` is found to be `0.295 V` at `25^(@)C` . The equilibrium constant of the reaction would beA. `1.0xx10^(10)`B. `2.0xx10^(11)`C. `4.0xx10^(12)`D. `1.0xx10^(2)` |
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Answer» Correct Answer - A `E^(@)=(0.0591)/(n)"log" k_(c)` `0.295=(0.0591)/(2)"log"k_(c)` `"log" k_(c)=(0.295xx2)/(0.0591)` `"log" k_(c)=9.9830` `k_(c)="antilog" 9.9830=10^(10)` |
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| 2019. |
For the reduction of silver ions with copper metal, the standard cell potential was found to be +0.46V at `25^(@)C`. The value of standard Gibbs energy, `DeltaG^(@)` will be `(F=96500" C "mol^(-1))`A. `-98.0kJ`B. `-89.0kJ`C. `-89.0J`D. `-44.5kJ` |
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Answer» Correct Answer - B `Cu+2Ag^(+)toCu^(2+)+2Ag,E_(cell)^(@)=0.46V` `DeltaG^(@)=-nFE_(cell)^(@)=-2xx96500xx0.46J=-88780J=-89kJ` |
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| 2020. |
The standard emf of a galvanic cell involving 3 moles of electrons in a redox reaction is 0.59V. The equilibrium constant for the reaction of the cell isA. `10^(25)`B. `10^(20)`C. `10^(15)`D. `10^(30)` |
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Answer» Correct Answer - D `E_(cell)^(@)=(0.0591)/(n)logK" "thereforelogK=(0.59xx3)/(0.0591)=30thereforeK=10^(30)` |
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| 2021. |
The amount of copper deposited by one Faraday current will be maximum in an acidic solution of one litre ofA. 1M `Cu_(2)Cl_(2)`B. 2M Cu`(NO_(3))_(2)`C. 5M `CuSO_(4)`D. `5MCu_(3)(PO_(4))_(2)` |
| Answer» Correct Answer - A | |
| 2022. |
In a galvanic cell .A. Chemical energy is converted into electricityB. Chemical energy is converted into heatC. Electrical energy is converted into heatD. Electrical energy is converted into chemcial energy. |
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Answer» Correct Answer - C In Galvanic cell chemical energy is converted into electricity |
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| 2023. |
The standard `E_("Red")^(@)` values of A,B,C are 0.68V, - 2.54V,- 0.50V respectively. The order of their reducing power isA. `AgtBgtC`B. `AgtCgtB`C. `CgtBgtA`D. `BgtCgtA` |
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Answer» Correct Answer - D Lower is reduction potential, greater is the tendency to get oxidised and greater is reducing power. |
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| 2024. |
The activation energy for the reactio `2HI_((g))to H_(2(g))+I_(2(g))at 581` K is `209. 5kJ//mol.` Calculte the fraction of molecules having energy equal to or grater than activation energy. `[R=8.31 JK^(-1)mol ^(-1)]` |
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Answer» Fraction of molecules [x] having energy equal to or more than activation energy may be calculated as follows : `x=n//N=e^(-E_(a)//RT)` `In x= (-E_(a))/(RT)or log x=(-E_(a))/(2.303RT)` or `log x=-(209.5xx10^(3)J mol ^(-1))/(2.303 xx[8.314JK ^(-1) ]xx581 K)=-18.8323` x = Antilog `[-18.8324]=` Antilog `bar(19).1677=1.471xx10^(-19)` Fraction of molecules `=1.471xx10^(-19)` |
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| 2025. |
What will happen to the value of emf of the following cell if the cencentration of the electrolyte in the anode compartment is increased ? `Zn | Zn^(2+)(0.1 M)||Cu^(2+)(0.1 M) Cu`. |
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Answer» According to Nernst equation `E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"([Zn^(2+)])/([Cu^(2+)]` If the concentration of the `Zn^(2+)` is increased, emf of the cell will decrease. |
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| 2026. |
The rate constant of a reaction is doubled when the temperature is raised from 298 K to 308 K. Calculate the activation energy. |
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Answer» `E_(a)=?k_(2)=2k_(1),T_(1)=298K,T_(2) =308K` `log ""(k_(2))/(k_(1))=(E_(a))/(2.303R)((T_(2)-T_(1))/(T_(1)T_(2)))` `log ""(k_(2))/(k_(1))=(E_(a))/(2.303xx8.314)((10)/(298xx308))` `0.3010 =(E_(a))/(2.303xx8.314)xx(10)/(298xx308)` `E_(a)=(0.3010xx2.303xx8.314xx298xx308)/(10)=52897.7 m ol ^(-1).` `E_(a)=52.8977kJ mol ^(-1).` |
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| 2027. |
Calculate the maximum work that can be obtained from the daniell call given below, `Zn(s)|Zn^(2+)(aq)||Cu^(2+)(aq)|Cu(s)`. Given that `E_(Zn^(2+)//Zn)^(@)=-0.76V` and `E_(Cu^(2+)//Cu)^(@)=+0.34V`. |
| Answer» `E_(cell)^(@)=E_(cathode)^(@)-E_(anode)^(@)=(0.34" V")-(-0.76" V")=1.10" V"` | |
| 2028. |
Calculate the maximum work that can be obtained from the decimolar Daniell cell at `25^(@)C`. Given `E^(c-)._((Zn^(2+)|Zn))=-0.76V` and `E^(c-)._((Cu^(2+)|Cu))=0.34V`A. `193.0k J`B. `212.3kJ`C. `81.06kJ`D. `40.53kJ` |
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Answer» Correct Answer - B Decimolar Daniell cell is represented below `:` `Zn(s)|Zn^(2+)(0.1M)||Cu^(2+)(0.1M)|Cu(s)` `E^(c-)._(cell)=(E^(c-)._(c)-E^(c-)._(a))_(red)=0.34-(-0.76)=1.10V` `E_(cell)=E^(c-)._(cell)-(0.059)/(2)log.([Zn^(2+)])/([Cu^(2+)])` `=1.10V-(0.059)/(2)log .(0.1)/(0.1)=1.10V` `W_(max)=-DeltaG=nFE_(cell)=xx96500xx1.10` `=212300J=212.3kJ` |
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| 2029. |
The standard reduction potentials of three metals `A,B,` and `C` are `+0.5 V , -3.0V,` and `-1.2V`, respectively. The order of reducing power of these metals isA. `BgtCgtA`B. `AgtBgtC`C. `CgtBgtA`D. `AgtCgtB` |
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Answer» Correct Answer - a Higher the `E^(c-)._(o x i d ation)` or lower the `E^(c-)._(reduction)`, better is the reducing agent. So, `BgtCgtA.` |
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| 2030. |
Why is chromium used for coating iron? |
| Answer» Chromium is a non-corroding metal which forms a protective laye on iron. | |
| 2031. |
Define corrosion. Write chemical formula of rust. |
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Answer» Corrosion is a process of determination of metal as a result of its reaction with air and water, surrounding it. It is due to formulation of sulphides, oxides, carbonates, hydroxides, etc. Formula of rust- Fe2O.XH2O |
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| 2032. |
Which one of the following conditions will increase the voltage of the cell represented by the equation ?A. Increase in the dimension of Cu electrodeB. Increase in the dimension of Ag electrodeC. Increase in the concentration of `Cu^(2+)` ionD. Increase in the concentration of `Ag^(+)` ion |
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Answer» Correct Answer - D Cell voltage will increase, either by increasing the concentration of `Cu^(2+)` ion or by decreasing the concentration of `Ag^(+)`. `Q=([Cu^(2+)](/([Ag^(+)])` Lesser is the value of Q, greater is the cell volatge. |
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| 2033. |
Write any three differences between potential difference and e.m.f. |
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| 2034. |
Why an electrochemical cell stops working after sometime? |
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Answer» The reduction potential of an electrode depends upon the concentration of solution with which it is in contact. As the cell works, the concentration of reactants decrease. Then according to Le chatelier’s principle it will shift the equilibrium in backward direction. On the other hand if the concentration is more on the reactant side then it will shift the equilibrium in forward direction. When cell works concentration in anodic compartment in cathodic compartment decrease and hence E0 cathode will decrease. Now EMF of cell is E0 cell = E0 cathode – E0 anode A decrease in E0 cathode and a corresponding increase in E0 anode will mean that EMF of the cell will decrease and will ultimately become zero i.e., cell stops working after some time.
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| 2035. |
The emf of the cell in which the following reactions, `Zn(s)+Ni^(2+) (0.1M) rarr Zn^(2+) (1.0 M) + Ni(s)` occurs, is found to 0.5105 V at 298 K. The standard emf of the cell is :A. `0.4810 V`B. `0.5696 V`C. `-0.5105 V`D. `0.5400 V` |
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Answer» Correct Answer - D `E_("cell") = E_("cell")^(@) - (0.059)/(2) log.(1)/(0.1)` |
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| 2036. |
In the cell `Zn//Zn^(+2) (c_(1))//cu^(+2)//Cu, E_(cell) - E_(cell)^(0) = 0.059 V` The ratio `(C_(1))/(C_(2)) at 298 K` will beA. 2B. 100C. `10^(-2)`D. `1`. |
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Answer» Correct Answer - C The reaction is Zn `+ Cu^(2+) (C_(2)) to Cu + Zn^(2+) (C_(1))` Alc to Nernst equation `E_("cell") = E_("cell")^(@) - (0.0591)/(n) "log"_(10) ([P])/([R])` `=0.0591 = -(0.0591)/(2) "log" (C_(1))/(C_(2))` `log (C_(1))/(C_(2)) = (2 xx 0.0591)/(0.0591) = - 2 therefore (C_(1))/(C_(2)) = Al (-2.000) = 10^(-2)` |
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| 2037. |
The emf of the cell in which the following reactions, `Zn(s)+Ni^(2+) (0.1M) rarr Zn^(2+) (1.0 M) + Ni(s)` occurs, is found to 0.5105 V at 298 K. The standard emf of the cell is :A. `0.540` VB. `0.4810` VC. `0.5696 ` VD. `0.5105`V |
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Answer» Correct Answer - D `E_("cell") = E_("cell")^(@) + (0.0592)/(2) log _(10) ([Ni^(2+)])/([Zn^(2+)])` `0.5105 = E_("cell")^(@) + (0.0592)/(2) log (1)/(1)` or `E_("cell")^(@) = 0.5105` V |
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| 2038. |
The reduction potential of a hydrogen electrode at `pH 10` at `298K` is : `(p =1` atm)A. `-0.0592` VB. `+0.0592` VC. `-0.592` VD. `0.592` V |
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Answer» Correct Answer - C `E_(OP) = E_(OP)^(@) - (0.0592)/(1) "log"_(10) ([H^(+)])/((P_(H_(2)))^(1//2))` `(1)/(2) H_(2) to H^(+) + e^(-)` `because [H^(+)] = 10^(-10) , P_(H_(2)) = 1 ` atm and `E_(OP (H_(2)))^(@) = 0` `E_(OP) = 0.592 V` `therefore E_(Red) = -0.592` V |
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| 2039. |
The reduction potential of a hydrogen electrode at `pH 10` at `298K` is : `(p =1` atm)A. `-0.59V`B. `-0.0591V`C. zeroD. `-0.591V` |
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Answer» Correct Answer - D `S.R.P. =- 0.591 pH` `=- 0.0591 xx -log(10^(-10)) =- 0.591V` |
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| 2040. |
The reduction potential of a hydrogen electrode at `pH 10` at `298K` is : `(p =1` atm)A. 0.51 voltB. 0 voltC. `-0.591 "volt"`D. 0.059 volt |
| Answer» Correct Answer - C | |
| 2041. |
According to Debye-Huckel-Onsager equation,`wedge_(m)^(c)=`_____. |
| Answer» `wedge_(m)^(c)=wedge_(m)^(@)-Asqrt(c)` | |
| 2042. |
How the reduction potential of an electrode can be decreased?A. increases with increase in concentration of ionsB. decreases with increase in concentration of ionsC. remains unaffected with increase in concentration of ionsD. increases with increase of temperature . |
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Answer» Correct Answer - D `Mn^(n+) + "ne"^(-) to M` `E = E^(@) - (RT)/(nF) "ln" (1)/([M^(n+)])` As `[M^(n+)]` increases E will increase. |
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| 2043. |
What is the use of Platinum foil in the hydrogen electrode? |
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Answer» It is used for inflow and outflow of electrons. |
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| 2044. |
Why in a concentrated solution, a strong electrolyte shows deviations from Debye-Huckel-Onsager equation? |
| Answer» In concentrated solution of a strong electrolyte, the interionic forces of attraction are large. | |
| 2045. |
Why in a concentrated solution, a strong electrolyte shows deviations from Debye-Huckle- Onsagar equation? |
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Answer» Because interionic forces of attractions are large. |
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| 2046. |
The reduction potential of hydrogen half cell will be negative if:A. `p(H_(2))=1atm and [H^(+)]=1.0M`B. `p(H_(2))=2atm and [H^(+)]=1.0M`C. `p(H_(2))=2atm and [H^(+)]=2.0M`D. `p(H_(2))=1atm and [H^(+)]=2.0M` |
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Answer» Correct Answer - A For hydrogen electrode `H^(+)(aq)+e^(-)to(1)/(2)H_(2)(g)` `E=E^(@)-(0.0591)/(1)"log"((p_(H_(2)))^(1//2))/([H^(+)])` `=-0.0591"log"((p_(H_(2)))^(1//2))/([H^(+)])" "(becauseE^(@)=0)` E will be -ve when `((p_(H_(2)))^(1//2))/([H^(+)])` is +ve, i.e., `(p_(H_(2)))^(1//2)gt[H^(+)]` which is so in case of (b). |
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| 2047. |
The decomposition of `N_(2)O_(5)` in `C Cl_(4)` at 318K has been studies by monitoring the concentration of `N_(2)O_(5)` in the solution. Initially the concentration of `N_(2)O_(5)` is `2.33 mol L^(-1)` and after 184 minutes, it is reduced to `2.08 mol L^(-1).` The reaction takes placed according to the equation `2N_(2)O_(5)(g)to 4 NO _(2)(g) +O_(2)(g)` Calculate the average rate of this reaction in terms of hours, minutres and seconds. What is the rate of proudction of `NO_(2)` during this period ? |
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Answer» Average Rate `=1/2 {-(Delta[N_(2)O_(5)])/(Deltat)}=-1/2[((0.28-2.33)molL^(-1))/(184 min)]` `=6.79 xx10^(-4) mol L^(-1) //min` `=(6.79xx10^(-4) mol L^(-1)min ^(-1))xx(60 min //1h)` `=4.07 xx10^(-2) mol L^(-1) //h` `=6.79 xx10^(-4)mol L^(-1)xx 1min//60s =1.13 xx10^(-5)mol L^(-1)s^(-1)` It may be remembered that Rate `=1/4 {(Delta[NO_(2)])/(Deltat)}` `(Delta[NO_(2)])/(Deltat)=6.79xx10^(-4)xx4 mol L^(-1) min ^(-1)` `=2.72xx10^(-3)mol L^(-1) min^(-1)` |
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| 2048. |
What is Half life of a reaction ? |
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Answer» Half life of a reaction : The time in which the concentration of a reactant is reduced to one half of its initial concentration is called half life period. It is represented by `t_(1//2).` `t_(1//2)of n^(th)` order reaction is given bgy `t_(1//2) prop [R_(0)]^(1-n)` Where `R_(0)` is initial concentration `t_(1//2) =([R_(0)])/(2k)` for a zero order reaction: `" "t_(1//2)=(0.693)/(k)` for order reaction |
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| 2049. |
What is a second order reaction ? Give one example. |
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Answer» Second order reaction : The reactions in which the rate of reaction depends on changing concentration of two subgstances is called second order reaction. Ex: Alkaline hydrolysis of ester. |
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| 2050. |
What is Arrhenius equation ? Derive an equation which describes the effect of rise of temperature (T) on the rate constant (k) of a reaction. |
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Answer» The temperature dependence of the rate of reaction can be explained by Arrhenius equation. `k=A.e ^(-Ea//RT)` A = Arrhenius factor. `E_(a)=` activation Energy R = gas constant T = Temperature (K) `K = A.e ^(-E_(a)//RT)` ln `k= ln A- E_(a)//RT` `2.303log k=2.303log A -E_(a)//RT` `T_(1), T_(2)` are Temperatures `k_(1)` is rate constant at temperature `T_(1)` `k_(2)` is rate constant at temperature `T_(2).` `therefore 2.303 (log K_(2)-log K_(1))=(-E_(a))/(R) ((1)/(T_(2))-(1)/(T_(1)))` `log ""(k_(2))/(k_(1))=(E_(a))/(2.303R)[(1)/(T_(1))-(1)/(T_(2))]` The above equation describes the effect to temperature (T) on rate constant (k) |
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