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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.
| 751. |
Assertion:Lithium has the lowest electrode potential. Reason: Lithium ion is the strongest oxidising agent.A. If both assertion and reason are true and reason is the correct explanation of assertion .B. If both assertion and reason are true but reason is not the correct explanation of assertion .C. If assertion is true but reason is false .D. If both assertion and reason are false . |
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Answer» Correct Answer - C Lithium ion is the weakest oxidising agent while lithium metal is the most powerful reducing agent in an aqueous solution. |
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| 752. |
Which of the following is the correct order of chemical reactivity with water according to electrochemical series ?A. `K gt Mg gt Zn gt Cu`B. `Mg gt Zn gt Cu gt K`C. `K gt Zn gt Mg gt Cu`D. `Cu gt Zn gt Mg gt K` |
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Answer» Correct Answer - A Follow `E_(RP)^(@)` values in electrochemical series . |
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| 753. |
Which species is the strongest oxidising agentA. `Br^(-)`B. `Zn^(2+)`C. `Pb^(2+)`D. `Cu^(2+)` |
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Answer» Correct Answer - D `Cu^(2+) +2 e to Cu` has positive reduction potential and thus easily gets oxidised to reduce other . |
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| 754. |
Consider the reaction `M_((aq))^(n+)+n e^(-)toM_((s))`. The standard reduction potential values of the element `M_(1),M_(2) and M_(3)` are -0.34V,-3.05 and -1.66V respectively. The order of their reducing power will be:-A. `M_(1) gt M_(2) gt M_(3)`B. `M_(3) gt M_(2) gt M_(1)`C. `M_(1) gt M_(3) gt M_(2)`D. `M_(2) gt M_(3) gt M_(1)` |
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Answer» Correct Answer - D The reducing power decreases as the reduction potential increase (becomes less negative). |
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| 755. |
The potential of the cell for the reaction `M_((s))+2H^(+)(1M)toH_(2(g)),(1atm)+M^(2+)(0.1M)` is 1.500 V. the standard reduction potential for `M^(2+)//M_((s))` couple is:-A. 0.1470VB. 1.470VC. 14.70VD. None of these |
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Answer» Correct Answer - B `E_(cell)=E_(cell)^(o)=-(0.0591)/(2)log((["products"])/(["Reactants"]))` `1.5=E_(cell)^(@)-0.0296log[(0.1)/(1)]` `E_(Cell)^(o)=1.470V` `E_(Cell)^(o)=E_(H^(+)//H_(2))^(o)-E_(M//M^(2+))^(o)` `E_(M//M^(2+))^(o)=-1.470` So, `E_(M^(2+)//M)^(o)=1.470V`. |
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| 756. |
Which is the strongest reducing agentA. AgB. NaC. `Br^(-)`D. `Cu` |
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Answer» Correct Answer - B `Na to Na^(+) + e^(-) (-ve` SRP value ) has maximum oxidation potential and thus easily gets oxidised to each other . |
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| 757. |
Consider the reaction `M_((aq))^(n+) + "ne"^(-) to M`(s) The standard reduction potential values of the elements `M_(1) , M_(2) and M_(3) " are " -0.34 V , -3.05 V ` and `-1.66 V` respectively . The order of their reducing power will beA. `M_(1) gt M_(2) gt M_(3)`B. `M_(3) gt M_(2) gt M_(1)`C. `M_(1) gt M_(3) gt M_(2)`D. `M_(2) gt M_(3) gt M_(1)`. |
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Answer» Correct Answer - D Less the reduction potential , greater is the reducing power |
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| 758. |
For a redox reaction to proceed spontaneously in a given direction , the emf shouldA. be zeroB. have + ve signC. have - ve signD. have either +ve or -ve sign |
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Answer» Correct Answer - B A cell operates only when `E_("Cell")^(@) gt 0` |
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| 759. |
When a rod of metal A is dipped in an aqueous solution of metal B (concentration of ` B^(2+) ` ion being 1 M) at `25^@ C`, the standard electrode potentials are `A^(2+) //A =- 0.76 ` volts, `B^(2+) //B= +0.34 ` volts .A. A will gradually dissolveB. B will deposit on AC. No reaction will occurD. Water will decompose into `H_(2)` and `O_(2)`. |
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Answer» Correct Answer - A ` A + B^(2+) to A^(2+) + B` has a positive EMF. |
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| 760. |
Write expression of `E_(cell)` in each of the following cell (i) `zn(s)|ZnSO_(4)(C_(1)M)||CuSO_(4)(C_(2)M)|Cu(s),E_(cell)` (ii) `Pt|H_(2)(P_(1)atm)||HCI(C_(1)M)|AgCI(s)|Ag,E_(cell)` (iii) `Pt|Fe^(2+)(C_(1)M),Fe^(+3)(C_(2)M)||Ag^(+)(CM)|Ag,E_(cell)` |
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Answer» (i) `E_(cell)=E_(cell)^(@)-(RT)/(2F)In(C_(1))/(C_(2))` (ii) `E_(cell)=E_(cell)^(@)-(RT)/(2F)in([H^(+)]^(2)[CI^(-)]^(2))/(PH_(2))=E_(cell)^(@)-(RT)/(2F)In(c_(1))/(P_(1))` (iii) `E_(cell)=_(cell)^(@)-(RT)/(F)In(C_(2))/(CxxC_(1))` |
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| 761. |
`E^(@)` of a cell `aA=bBtocC+dD` isA. `-(RT)/(nF)"log"([C]^(C)[D]^(d))/([A]^(a)[B]^(b))`B. `-Rt"log"([a]^(A)[b]^(B))/([a]^(C)[d]^(D))`C. `-(RT)/(nF)"log"([C]^(c)[d]^(D))/([A]^(a)[B]^(b))`D. `-(RT)/(nF)"log"([C]^(c)[d]^(D))/([a]^(A)[B]^(b))` |
| Answer» Correct Answer - A | |
| 762. |
Calculate the e.m.f. of the cell in which the following reaction takes place : `Ni(s) +2Ag^(+)(0.002 M)to Ni^(2+)(0.160 M)+2Ag(s)` Given `E_(cell)^(@)`=1.05 v |
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Answer» `E_(cell)=E_(cell)^(@)-(0.0591)/(n)log[(Ni^(+2))]/([Ag^(+)]^(2)` `=1.05 -(0.0561)/(2)log(0.160)/(0.002^(2)` `=1.05-(0.0591)/(2)xx4.6` =1.05-0.14=0.91 V |
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| 763. |
`E^(@)` for the cell `Zn(s)|Zn^(2+)(aq)|Cu^(2+)(aq)|Cu(s)` is 1.1V at `25^(@)C` the equilibrium constant for the cell reaction is aboutA. `10^(-37)`B. `10^(37)`C. `10^(73)`D. `10^(73)` |
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Answer» Correct Answer - B At 298k, `E^(@)=(0.0591)/(n)"log"k` `1.1=(0.0591)/(2)"log"k` (Here n=2) log `k=(1.1xx2)/(0.0591)=37.22` `k="antilog" 37.22=1.65xx10^(37)` |
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| 764. |
When a rod of metal A is dipped in an aqueous solution of metal B (concentration of ` B^(2+) ` ion being 1 M) at `25^@ C`, the standard electrode potentials are `A^(2+) //A =- 0.76 ` volts, `B^(2+) //B= +0.34 ` volts .A. A will graduall dissolveB. Water will decompose into ` H_2` and ` O_2`C. No reaction will occurD. B will deposite on A |
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Answer» Correct Answer - D Since `E_(A^(+))^@ gt E_(B^(2+)//B)^@`. A has greater tendency to be oxidized. `A+B^(2+) rarr +B`. |
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| 765. |
Which of the following metal does not react with the solution of copper sulphateA. MgB. FeC. ZnD. Ag |
| Answer» Correct Answer - D | |
| 766. |
When an acid cell is charged, thenA. Voltage of cell increasesB. Electrolyte of cell dilutesC. Resistance of cell increasesD. None of these |
| Answer» Correct Answer - A | |
| 767. |
When a rod of metal A is dipped in an aqueous solution of metal B (concentration of `B^(2+)` ion being 1M) at `25^(@)C`, the standard electrode potentials are `A^(2+)//A=-0.76`volts, `B^(2+)//B=+0.34` voltsA. A will gradually dissolveB. B will deposit on AC. No reaction will occurD. Water wil decompose into `H_(2) and O_(2)` |
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Answer» Correct Answer - B Since `E_(A^(2+)//A)^(o) lt E_(B^(2+)//B)^(o)`. A has greater tendency to be oxidized. `A+B^(2+)toA^(2+)+B`. |
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| 768. |
Aluminium displaces hydrogen from acids but copper does not. A galvanic cell prepared by combining `Cu//Cu^(2+) and Al//Al^(3+)` has an e.m.f. of 2.0 V at 298 K. if the potential of copper electrode is +0.34V, that of aluminium isA. `+1.66V`B. `-1.66V`C. `+2.34V`D. `-2.3V` |
| Answer» Correct Answer - B | |
| 769. |
Aluminium displaces hydrogen from dilute HCl whereas silver does not. The e.m.f. of a cell prepared by combining `Al//Al^(3+)` and `Ag//Ag^(+)` is 2.46V. The reduction potential of silver electrode is +0.80V. The reduction potential of aluminium electrode isA. `+1.66V`B. `-3.26V`C. `3.26V`D. `-1.66V` |
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Answer» Correct Answer - D Al is more reactive than Ag, i.e., cell reaction is `Al+3Ag^(+) to Al^(3+)+3Ag` `E_(cell)=E_("cathode")^(@)-E_("anode")^(@)` `=E_(Ag^(+)//Ag)^(@)-E_(Al^(3+)//Al)^(@)` `2.46=0.80-E_(Al^(3+)//Al)^(@)` or `E_(Al^(3+)//Al)^(@)=-1.66V` |
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| 770. |
Aluminium displaces hydrogen from dilute HCl whereas silver does not. The e.m.f. of a cell prepared by combining `Al//Al^(3+)` and `Ag// Ag^(+)` is 2.46V. The reduction potential of silver electrode is +0.80V. The reduction potential of aluminium electrode isA. `+1.66V`B. `-3.26V`C. `3.26V`D. `-1.66V` |
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Answer» Correct Answer - D `Al|Al^(3+)||Ag^(+)|Ag` `E=E_(Ag^(+)._(//Ag))-E_(Al^(3+)._(//Al))` `E_(Al^(3+_(//Al))=0.80-2.46=-1.66V` |
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| 771. |
Aluminium displaces hydrogen from dilute HCl whereas silver does not. The e.m.f. of a cell prepared by combining `Al//Al^(3+) and Ag//Ag^(+)` is 2.46V. The reduction potential of silver electrode is +0.80V. The reduction potential of aluminium electrode.A. `+1.66V`B. `-3.26V`C. `3.26V`D. `-1.66V` |
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Answer» Correct Answer - D Al displaces H from HCl but silver cannot it means Al is situated above the Ag in ECS, hence Al will acts as anode and Ag will act as cathode. `E_(cell)^(o)=E_("cathode")^(o)-E_("anode")^(o)=E_(Ag^(+)//Ag)^(o)-E_(Al^(3+)//Al)^(o)` `2.46=0.8-E_(Al^(3+)//Al)^(o),E_(Al)^(o)=0.8-2.46=-1.66V`. |
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| 772. |
How much silver will be obtained by that quantity of current which displaces 5.6 litre of `H_(2)` ?A. 54gB. 13.5 gC. 20 gD. 108 g |
| Answer» Correct Answer - A | |
| 773. |
Cosoder the reactopm: `(T = 298 K)` `Cl_2 (g) + 2 BR^(-) (aq) rarr 2 Cl^(-) (aq) + Br_2 (aq.)` The emf of he cell, when `[Cl^(-) = (Br_2] = [Br^(-) ] = 0.01M and Cl_2 ` gas is at `1` atm pressure, will be : (`E^@` for the above reaction is `= 29` volt ).A. 0.54 voltB. 0.35 voltC. 0.24 voltD. `-0.29 "volt"` |
| Answer» Correct Answer - B | |
| 774. |
Cosoder the reactopm: `(T = 298 K)` `Cl_2 (g) + 2 BR^(-) (aq) rarr 2 Cl^(-) (aq) + Br_2 (aq.)` The emf of he cell, when `[Cl^(-) = (Br_2] = [Br^(-) ] = 0.01M and Cl_2 ` gas is at `1` atm pressure, will be : (`E^@` for the above reaction is `= 29` volt ).A. `0.54` voltB. `0.35` voltC. `0.24` voltD. `-0.29` volt |
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Answer» Correct Answer - B `E_("cell")= 0.29 - (0.0591)/(2) "log" ([Cl^(-)]^(2)[Br_(2)])/([Br^(-)]^(2)[Cl_(2)])` `= 0.29 - (0.0591)/(2) "log" ([10^(-2)]^(2)[10^(-2)])/([10^(-2)]^(2)[1])` `0.29 + (0.0591)/(2)xx2=0.349~~ 0.35 V` |
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| 775. |
A chemist wants to produce `CI_(2)(g)` from molten `NaCI`. How many grams could be produced if he uses a steady current of 2 ampere for 2.5 minutes:-A. `3.55 g`B. `1.775 g`C. `0.110g`D. `0.1775 g` |
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Answer» Correct Answer - C `2Cl^(-)rarr Cl_(2)+2e^(-)` `W = Z I t = (71)/(2xx96500)xx2xx2.5xx60 = 0.110 g` |
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| 776. |
A chemist wants to produce `CI_(2)(g)` from molten `NaCI`. How many grams could be produced if he uses a steady current of 2 ampere for 2.5 minutes:-A. 3.55 gB. 1.775 gC. 0.110 gD. 0.1775 g |
| Answer» Correct Answer - C | |
| 777. |
The chlorate ion can disproportinate in basic solution according to reaction, `2ClO_3^(-)iffClO_2^(-)+ClO_4^-)` what is the equilibrium concentration of perchlorate ions from a solution initially at 0.1 M in chlorate ions at 298 K? Given: `E_(Cl_4^(-)|ClO_3^(-))^(@)=0.36V "and" E_(Cl_3^(-)|ClO_2^(-))^(@)=0.33V "at"298 K`A. 0.019 MB. 0.024MC. 0.1MD. 0.19M |
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Answer» Correct Answer - A |
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| 778. |
Arrange the following in the order of their decreasing electrode potential `Mg,K,Ba,Ca`A. K,Ba,Ca,MgB. Ca,Mg,K,BaC. Ba,Ca,K,MgD. Mg,Ca,Ba,K |
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Answer» Correct Answer - D The correct decreasing electrode potential order is: K,Ba, Ca, Mg. |
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| 779. |
K, Ca and Li metals may be arranged in the decreasing order of their standard electrode potential asA. K, Ca, LiB. Ca, K, LiC. Li, Ca, KD. Ca, Li, K |
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Answer» Correct Answer - B According to electrochemical series. |
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| 780. |
Which one of the following metals cannot evolve `H_(2)` from acids or `H_(2)O` or from its compoundA. HgB. AlC. PbD. Fe |
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Answer» Correct Answer - A Hg has greater reduction potential than that of `H^(+)` and hence cannot displace hydrogen from acid. |
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| 781. |
Write the cell reactions for the following cells : (a) `Fe|Fe^(2+)||H_(2)SO_(4)|H_(2)(Pt)` (b) `(Pt)H_(2)|HCl||Cl_(2)(Pt)` |
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Answer» (a) `Fe(s)+2H^(+)(aq) to Fe^(2+)(aq)+H_(2)(Pt)` (b) `H_(2)(g)+2Cl^(-)(aq) to 2H^(+)(aq)+Cl_(2)(Pt)`. |
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| 782. |
Draw a neat and labelled diagram for `H_(2)-O_(2)` fuel cell. Write the reaction which occurs at cathode of the cell. |
| Answer» For Answer, Consult Section 2. | |
| 783. |
STATEMENT 1 : For the concentration cell `Zn(s)|Zn_(C_(2))^(+2)(aq)|Zn` for spontaneous cell reaction `C_(1)ltC_(2)` and STATEMENT 2 For concentration cell `E_(cell)=(RT)/(nF)log_(e)(C_(2))/(C_(1))` for spontaneous rection `E_(cell)=+verarrc_(2)gtc_(1)`A. Statement 1 is ture , Statement 2 is true Statement 2 is correct explanation for Statement 3B. Statement 1 is true Statement 2 is ture Statement 2 is NOT a correct explantion for Statement 3C. Statement 1 is true statement 2 is tureD. Statement 1 is false Statement 2 is true |
| Answer» Correct Answer - A | |
| 784. |
STATEMENT 1 A Saturated solution of KCI is used to make salt bridge in concentration cells and STATEMENT 2 Mobility of `K^(-)` and `CI^(-)` are nearly sameA. Statement 1 is ture , Statement 2 is true Statement 2 is correct explanation for Statement 4B. Statement 1 is true Statement 2 is ture Statement 2 is NOT a correct explantion for Statement 4C. Statement 1 is true statement 2 is tureD. Statement 1 is false Statement 2 is true |
| Answer» Correct Answer - A | |
| 785. |
Find the emf of the cell `Zn(s)|Zn^(+2)(0.01 M)|KCI "saturated" |Zn^(+2)(1.0 M)|Zn(s)` |
| Answer» `E_("cell") = 0 - (0.0591)/(2)log.(0.01)/(1.0)` | |
| 786. |
`Pt |H_(2)(1 bar)H^(+)||KCI(1.0 M "saturated" | Hg_(2)CI_(2)|Hg` was used to measure the pH of 0.05 M acetic acid in 0.04 M `CH_(3)COONa` Calculate the cell potential `K_(CH_(3)COOH)=1.8xx10^(-5),E_(Hg_(2)CI_(2)//Hg,Cr)=0.28 V` |
| Answer» Correct Answer - 0.555 V | |
| 787. |
Calculate the standard reduction potential of `Cd^(2+)//Cd` electrode for the cell : `Zn(s)|Zn^(2+)("IM") || Cd^(2+)(IM)|Cd(s)` `("Given that " E_(cell)^(@)=0.36 V and E_(Zn^(2+)//Zn)^(@)=-0.76V)` |
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Answer» `E_(cell)^(@)=E_(cathode)^(@)-E_(anode)^(@)` `E_(cell)^(@)=(E_(Cd^(2+)//Cd)^(@))-(E_(Zn^(2+)//Zn)^(@))` `(E_(Cd^(2+)//Cd))=E_(cell)^(@)+(E_(Zn^(2+)//Zn)^(@))=0.36+(-0.76)=-0.40 V` |
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| 788. |
Calculate the EMF of the cel `Fe(s)+2H^(+)(1M)rarrFe^(+2)(0.001 M)+H_(2)(g)(1 "atm")("given" : E_(Fe^(2+)//Fe)^(@)=-0.44 V)` |
| Answer» `E = E^(@) - (0.059)/(n) log.([Fe^(+2)])/([H^(+2)])` | |
| 789. |
Calculate `DeltaG_(r)^(@)` of the following reaction `Ag^(+)(aq)+cI^(-)(aq)rarrAgCI(s)` Given `DeltaG_(r)^(@)(AgCI)rarr109 kJ Mol^(-1)` `DeltaG_(r)^(@)(CI^(-))rarr-129 kJ Mol^(-1)` `DeltaG_(r)^(@)(ag^(-))rarr77 kJ Mol^(-1)` (i)Represent the above reaction in form of a cel (ii) Calcualte `E^(@)` of the cel (iii) Find `log_(10)K_(sp)` of AgCI |
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Answer» `57 kj mol^(-)` (i) `Ag(s)|Ag^(+)||AgCI||CI^(-)|CI_(2),pt` (ii) `E_(cell)^(@)=0.59V` (iii)`log_(10)K_(sp)=-10` |
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| 790. |
(a) If the cell emp is -1.58 V what is the concentrtion of `Zn^(+2) (b) "If" NH_(3)` added to half cel A how emf of cel will change ? `E_(Ag^(+)//Ag)^(@)=0.8 V,E_(Zn^(+23)//Zn)^(@)=0.76 V` Antilog (0.6768) =1.4768 |
| Answer» (a) `[Zn^(+2)]=0.0241 M` (b) emp increase | |
| 791. |
The standard potential of the following cell is 0.23 V at `15^(@)C` and `0.21V` at `35^(@)C`.`Pt|H_(2) |HCl(aq) |Agcl (s)|g(s)` (i) write the cell reaction .(ii) Calculate `DeltaH^(@)` and `DeltaS^(@)` for the cell reaction by assuming that these quantities remain unchanged in the range `15^(@)C` to `35^(@)C`. (iii) calculate the solubility of `AgCl` in water at `25^(@)C`. Give , the standard reduction potential of the `(Ag^(+)(aq) //Ag(s)` is 0.80 V at `25^(@) C`. |
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Answer» (i) `1/2 H_(2)(g)+AgCI(s) H^(+)(aq)+Ag(s)+CI^(-)(aq)` (ii) `Delta^(@)=-96.5 JK^(-1) mol^(-1) DeltaH^(@)=-499987 J^(-)mol^(-1)` |
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| 792. |
`E_(Cu^(+2)//Cu)^(@)=0.34V, E_(Zn//Zn^(+))^(@)=0.76V` A cell formed by the conbination of Cu and Zn (a) when `CuSO_(4)` is added to `Cu^(+2)` compartment what is the effect on emf of cell (b) when `ZNH_(3)` is added to `Cu^(+2)` compartment what is the effect on emf of cell (c ) When `ZnSO_(4)` is added to `Zn^(+2)` compartment is the effect on emf of cell (d) When `Zn^(+2)` is diluted what is the effect on emf of cell ? |
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Answer» (a) increases (b) decreases (c ) decreases (d) increases |
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| 793. |
The standard oxidation potential of `Zn` referred to SHE is `0.76V` and that of `Cu` is `-0.34V` at `25^(@)C`. When excess of `Zn` is added to `CuSO_(4),Zn` diplaces `Cu^(2+)` till equilibrium is reached. What is the ratio of `Zn^(2+)` to `Cu^(2+)` ions at equilibrium? |
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Answer» Correct Answer - `[Zn^(2+)]//[Cu^(2+)] =1.941 xx 10^(37)` `E_(cell)^(@) = (0.0591)/(2)log.([Zn^(2+)])/([Cu^(2+)])` `1.1 =(0.0591)/(2)log.([Zn^(2+)])/([Cu^(2+)])` `([Zn^(2+)])/([Cu^(2+)]) =1.941 xx 10^(37)` |
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| 794. |
Calculate the emf of the following concentration cell at `25^(@)C`: `Ag(s)|AgNO_(3) (0.01 M)||AgNO_(3) (0.05 M)|Ag (s)`A. 0.828VB. 0.0413VC. `-0.0413` VD. `-0.828` V |
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Answer» Correct Answer - B `E=E^@-0.0591/n` log Q `E^@`=0 for all concentration cells =`0-0.0591/1 "log"(0.01/0.05)`=0.0413 V |
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| 795. |
During electrolysis of molten NaCl, some water is added, What will happen :A. Electrolysis will stop.B. Hydrogen will evolveC. Some amount of caustic soda will be formedD. A fire is likely. |
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Answer» Correct Answer - B::C::D (b,c, d) At cathode : `Na^(+)+e^(-) to Na` At anode : `Cl^(-) to 1//2 Cl_(2)+e^(-)` On adding water, `Na+H_(2)O to NaOH+1//2 H_(2)` The reaction is highly exothermic. As a result, `H_(2)` evolved is likely to catch fire. |
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| 796. |
The standard reduction potential `E^(@)(Bi^(3+)//Bi)` and `E^(@)(Cu^(2+)//Cu)` are `0.226V` and `0.344V` respectively. A mixture of salts of bismuth and copper at unit concentration each is electrolysed at `25^(@)C`. To what value can `[Cu^(2+)]` be brought down before bismuth starts to deposit in electrolysis. |
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Answer» Correct Answer - `[Cu^(2+)] = 10^(-4)M` `E_(Cu^(2+)//Cu) = E_(Cu^(2+)//Cu)^(@) +(0.059)/(2) lg c[Cu^(2+)]` `[Cu^(2+)] = 10^(-4)m` |
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| 797. |
The Standard reduction potential values, `E^(@)(Bi^(3+)//Bi)` and `E^(@)(Cu^(2+)//Cu)` are 0.226 V and 0.344 V respectively. A mxiture of salt of bismut and copper at unit concentration each is electrolysed at `25^(@)C` to what value can `[Cu^(2+)]` be brought down before bismuth starts to deposit, in electrolysis. |
| Answer» Correct Answer - D | |
| 798. |
Match the followingColumn-IColumn-II(A) Calomel electrode(p) Electrolyte concentration cell(B) Zn-Cd(C1) |CdCl2| Zn-Cd(C2)(q) Metal-insoluble anion half cell(C) Quinhydrone electrode(r) Electrode concentration cell(D) Pt|H2(1 atm)|H+(C1)||H+(C2)|H2(1 atm)|Pt(s) Redox half cell |
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Answer» A→(q), B→(r), C→(s), D→(p) |
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| 799. |
Find the e.m.f. of the following cell at `18^(@)C` taking the degree of dissocution of `0.2 N AgNO_(3)` and `0.05 AgNO_(3)` solution as `0.75` and `0.95` respectively. `Ag|0.2N AgNO_(3)||0.05 AgNO_(3)|Ag` |
| Answer» Correct Answer - `-0.029 V ;` | |
| 800. |
There is blue colour formation if:A. Cu electrode is placed inside `AgNO_(3)` solutionB. Cu electrode is placed inside `ZnSO_(4)` solutionC. Cu electrode is placed inside dil `HNO_(3)`D. Cu electrode is placed inside dil `H_(2)SO_(4)` |
| Answer» Correct Answer - A::C | |