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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.
| 3051. |
A dilute aqueous solution of `Na_(2)SO_(4)` is electrolysed using platinum electrodes. The products at the anode and cathode areA. `O_(2),H_(2)`B. `SO_(2),Na`C. `O_(2),Na`D. `S_(2)O_(8)^(2-),H_(2)` |
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Answer» Correct Answer - A `H_(2)O` is more readily reduced than `Na^(+)`. It is also readily oxidized than `SO_(4)^(2-)`. Hence, the electrode reactions are: At cathode:`2H_(2)O+2e^(-)toH_(2)+2OH^(-)` At anode: `2H_(2)toO_(2)+4H^(+)+4e^(-)` |
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| 3052. |
A dilute aqueous solution of `Na_(2)SO_(4)` is electrolysed using platinum electrodes. The products at the anode and cathode areA. `O_(2),H_(2)`B. `S_(2)O_(8)^(2-),Na`C. `O_(2),Na`D. `S_(2)O_(8)^(2-),H_(2)` |
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Answer» Correct Answer - A `H_(2)O` is more readily reduced at cathode than `Na^(+).` it is also more readily oxidized at anode than `SO_(4)^(2-)`. Hence, the electrode reactions are `2H_(2)O+2e^(-)toH_(2)uarr+2OH^(-)` [at cathode], `H_(2)O to (1)/(2)O_(2)uarr+2H^(+)+2e^(-)` [at anode]. |
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| 3053. |
In an electrolyitc cell of `Ag//NO_3Ag,` when current is passed, the concentration of `AgNO_3`.A. IncreasesB. DecreasesC. Remains sameD. None of these |
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Answer» Correct Answer - C At anode `A rarr Ag^+ +e^-` At cathode `Ag^+ +e^- rarr Ag`. so concentration of `Ag+` will remanin same. |
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| 3054. |
When a solution of `AgNO_3` (1 M) is electrolysed using platinum anode and copper cathode, what are the products obtained at two electrodes? Given :`E_(cu^(2+)|cu)^(@)=+0.34 "volt", E_(O_2,H^+|H_2O)^(@)=+1.23"volt",E_(H^+|H_2)^(@)=+0.0"volt" E_(ag^+|Ag)^(@)=+0.8 "volt"`A. `CutoCu^(2+)` at anode, `Ag^+to Ag` at cathodeB. `H_2OtoO_2` at anode, `Cu^(2+)to Cu` at cathodeC. `H_2OtoO_2` at anode, `Ag^+to Ag` at cathodeD. `NO_3^-to NO_2` at anode, `Ag^+to Ag` at cathode |
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Answer» Correct Answer - C |
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| 3055. |
If aqueous solutions of `AgNO_3` is electrolysed using inert electrode the gas evolved at anode is .A. ` NO_2`B. `O_2`C. ` H_2`D. `N_2O` |
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Answer» Correct Answer - B At anode `2H_2O rarr O_2 + 4H^+ +4e^-` At cathode `Ag^+ +e^- rarr Ag`. |
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| 3056. |
A solution of sodium sulphate in qater is electrolysed using inert electrodes, The products at the cathode and anode are respectively.A. `H_2,O_2`B. `O_2,H_2`C. `O_2,Na`D. `H_2,S_2O_8^(2-)` |
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Answer» Correct Answer - D |
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| 3057. |
The two aqueous solutions, `A (AgNO_3)` and `B` (LiCI), were electrolysed using Pt electrodes. The `pH` of the resulting solutions will.A. increase in A and decrease in BB. decrease in bothC. decrease in bothD. increase in A and decrease in B |
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Answer» Correct Answer - D `AgNO_3` At cathde : `Ag^+ +e^- rarr Ag(s)` At anode: ` 2H_2 O rarrO_2 +4H^+4e^(-)` `:.` At cathode pH will increase. |
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| 3058. |
A solution of sodium sulphate in qater is electrolysed using inert electrodes, The products at the cathode and anode are respectively.A. ` H_2, O_2`B. ` O_2 H_2`C. ` O_2, Na`D. ` O_2, SO_2` |
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Answer» Correct Answer - A At cathode : `2 H^+ 2e rarrH_2`. At anode : ` 2H^- rarr Al_2O + 1/2 O_2 +2e`. |
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| 3059. |
A solution of sodium sulphate in qater is electrolysed using inert electrodes, The products at the cathode and anode are respectively.A. ` O_2,H_2`B. ` H_2O_2`C. `O_2Na`D. `O_2,SO_2` |
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Answer» Correct Answer - B At cathode : `2H^+ +2e rarrH_2` At anode : `2OH^- rarr H_2O + 1/2 O_2 +2e`. |
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| 3060. |
Assertion `(A) :` The Daniell cell becomes dead after sometimes. Reason `(R) :` The oxidation protential of `Zn` anode decreases and that of `Cu` increases.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 | |
| 3061. |
Assertion : The Daniell cell becomes dead after sometime. Reason : Reduction potential of Zinc increases while that of copper decreases.A. If both assertion and reason are correct and reason is correct explanation for assertion.B. If both assertion and reason are correct but reason is not correct explanation for assertion.C. If assertion is correct but reason is incorrect.D. If assertion and reason both are incorrect. |
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Answer» Correct Answer - A (a) Reason is the correct explanation for assertion. |
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| 3062. |
Assertion A :Increase in the concentration of copper half cell in Daniel cell increases the emf of the cell. Reason R: According to Nernst equation. `E_(cell)=E_(cell)^(0)+(0.059)/(2)log.([Cu^(++)])/([Zn^(++)])`A. Both A and R are true and R is the correct explanation of AB. Both A and R are true but R is not a correct explanation of AC. A is true but R is falseD. Both A and R are false. |
| Answer» Correct Answer - B | |
| 3063. |
Statement-I: Equivalent conductance of all electrolytes decreases with increasing concentration. Because Statement-II: Lesser number of ions ate available per gram equivalent at higher concentration.A. Both A and R are true and R is the correct explanation of AB. Both A and R are true but R is not a correct explanation of AC. A is true but R is falseD. Both A and R are false. |
| Answer» Correct Answer - B | |
| 3064. |
The essential condition to set a standard hydrogen electrode is :(a) 298 K (b) pure and dry H2 gas at 1 atm (c) solution containing H+ at unit activity (d) all of these |
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Answer» Option : (d) all of these |
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| 3065. |
The electrode potential becomes equal to standard electrode potential when reactants and products concentaration ratio is .A. equal to `1`B. greater than `1`C. less than `1`D. none of the above |
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Answer» Correct Answer - A `E=E^@ - (0.0591)/n log. ([Product])/([Reactant])` if `(["Product "])/(["Reactant"]) =1`. Then ` E=E^@`. |
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| 3066. |
`E^(@)Cu=0.34V,E_(Zn)^(@)=`-0.76V. A Daniell cell contains 0.1M `ZnSO_(4)` solution and 0.01 M `CuSO_(4)` solution at its electrodes. E.M.F. of the cell isA. 1.10VB. 1.04VC. 1.16VD. 1.07V |
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Answer» Correct Answer - D `Zn|Zn^(2+)||Cu^(2+)|Cu` `Zn+Cu^(2+)+rarrZn^(2+)+Cu` `E_("cell")^(@)=E_(Cu^(2+)._(//Cu))-E_(Zn^(2+)._(//Zn))^(@)` `=0.34-(-0.76)=1.10V` `E_("cell")=E_("cell")^(@)(0.0591)/(2)"log"([Zn^(2+)])/([Cu^(2+)])` `=1.10V-(0.0591V)/(2)"log"(0.1)/(0.01)` `=10.7045V` |
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| 3067. |
Assertion `(A) :` The Daniell cell becomes dead after sometimes. Reason `(R) :` The oxidation protential of `Zn` anode decreases and that of `Cu` increases.A. Both A and R are true and R is the correct explanation of AB. Both A and R are true but R is not a correct explanation of AC. A is true but R is falseD. Both A and R are false. |
| Answer» Correct Answer - A | |
| 3068. |
Assertion: E.m.f. and potential difference are same for cell. Reason: Both gives the difference in electrode potential under any condition.A. if both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is falseD. If the assertion and reason both are false. |
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Answer» Correct Answer - D Both assertion and reason are false. Potential difference is the difference between the electrode potential of the two electrodes of the cell wen cell is under operation while e.m.f.is the potential difference generated by a cell when there is zero electron flow. |
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| 3069. |
During the electrolysis of fused `NaCl` , which reaction occurs at anode ?A. Chloride ions are oxidisedB. Chloride ions are reducedC. Sodium ions are oxidisedD. Sodium ions are reduced . |
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Answer» Correct Answer - A Electrolysis of fused NaCl will give `Cl^(-)` ions at the anode i.e. `2 Cl to Cl_(2) + 2e^(-)` . |
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| 3070. |
During the electrolysis of fused `NaCl` , which reaction occurs at anode ?A. Cholride ions are oxdizedB. Choloride ions are reducedC. Sodim ions are oxdisedD. Sodim ions reduced |
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Answer» Correct Answer - A In fursed NaCl chloride ions are oxdized at anode and it is called oxidation. |
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| 3071. |
During the electrolysis of fused NaCl, which reaction occurs at anode ? |
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Answer» Correct Answer - A In fused NaCl chloride ions are oxidized at anode and it is called oxidation. |
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| 3072. |
`Delta G^@` of the cell reaction `AgCl (s) = 1/2 H_2 (g) =Ag(s) H^+ + Cl^(-)` is `-21 . 52 kJ`. ` Delta G^2` of `2 AgCl (s) + H_2 (g) = 2 Ag(s) + 2 H^(+) 2 Cl^(-)` is .A. ` -21. 52 k J`B. ` - 10 . 76 k J`C. ` - 43. 04 k J`D. ` 43 . 04 k J` |
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Answer» Correct Answer - C ` K_(C_2) = K_(C_1)^@` `:. Delta G=- nFE_0 =- RT In K_(c)` `DeltaG_12 = 2 xx (1 23. 52) =- 42. 04 kJ`. |
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| 3073. |
During the electrolysis of fused `NaCl` , which reaction occurs at anode ?A. Chloride ions are oxidizedB. Chloride ions ar reducedC. Sodium ions ar oxidizedD. Sodium ions are reduced |
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Answer» Correct Answer - a Fused `NaCl` will give reduction of `Na^(O+)` ion to `Na` at cathode and oxidation of `Cl^(c-)` to `Cl_(2)` anode. |
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| 3074. |
The emf of given cell `Pt- H_2 | H^+|H_2 - Pt` is :A. `(RT)/F log P_1/P_2`B. `(RT)/(2F)log ep_1/P_2`C. `(RT)/F- [email protected]/P_1`D. None of these |
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Answer» Correct Answer - B `H_2 rarr 2H^+ +2e` `2H^+ +2erarr H_2` `E_("cell") = E_(OP_(H))^@ + (RT)/(2F) log_e. ([P_H_2]_1)/([PH_2]_2) = (RT)/(2F) Log_e .P_1/P_2`. |
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| 3075. |
The passage of electricity in the Daniell cell when Zn and Cu electrodes are connected is from :A. Cu to Zn in the cellB. Cu to Zn outside the cellC. Zn to Cu outside the cellD. Zn to Cu in the cell |
| Answer» Correct Answer - B | |
| 3076. |
Which statements is true about a spontaneous cell reaction in galvanic cell?A. `E^(c-)._(cell)gt0,DeltaG^(c-)lt0,` Quotient `QltK_(c)`B. `E^(c-)._(cell)lt0,DeltaG^(c-)gt0,QltK_(c)`C. `E^(c-)._(cell)gt0,DeltaG^(c-)gt0,QgtK_(c)`D. `E^(c-)._(cell)gt0,DeltaG^(c-)gt0,QltK_(c)` |
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Answer» Correct Answer - a For a spontaneous cell reaction`:E^(c-)._(cell)gt0,DeltaG^(c-)lt0,QltK_(c)[` reaction is moving in the forward direction `]` |
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| 3077. |
Which one of the following statement is always true about the spontaneous cell reaction in a galvanic cell?A. `E_(cell) gt 0, DeltaG^(@) gt 0, Q gt K_(c)`B. `E_(cell)^(@) lt 0, DeltaG^(@) lt 0, Q lt K_(c)`C. `E_(cell)^(@) gt 0, DeltaG^(@) lt 0, Q lt K`D. `E_(cell)^(@) gt 0, DeltaG^(@) lt 0, Q gt K_(c)`. |
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Answer» Correct Answer - C For spontaneous cell reaction, `E_(cell)^(@)=+ve," i.e., "gt0,DeltaG^(@)=-ve, " i.e., "lt0 and Q lt K_(c)` (Also, remember that `-DeltaG^(@)=RT" ln "K.` -ve `DeltaG` implies that ln K will be +ve, i.e., Kgt1) |
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| 3078. |
Calculate the emf of the cell consisting the following half cells `Al//Al^(3+)(0.001M), Ni//Ni^(2)(0.50M).` Given that `E_(Ni^(2+)//Ni)^(0)=-0.25V` `E_(Al^(3+)//Al)=-1.66V(log 8xx10^(-6)=-5.0969).` |
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Answer» Given `E^(0)of Al^(+3)//Al=-1.66V` `E^(0)of Ni^(+2)//Ni=-0.25V` Cell pepresentation `Al //Al_(""(0,001M))^(+3)||Ni_(""(0.50M))^(+2)|Ni` Nernst equation is `E_(A l ^(+_3)//Al)+(0.059)/(n) log [Al ^(+3)]` `=-1.66 +(0.59)/(3)log 0.001=-1.66-0.59=-1.719V` `E _(Ni^(+2)//Ni)= E_(Ni^(+2)//Ni)+(0.059)/(n) log [Ni ^(+2)] ` `=- 0.25 +(0.059)/(2)log [0.5]` `=-0.25-0.0295xx0.3010=-0.25-0.0088795=-0.25888` EMF of cell `=E _(RHS ) -E _(LHS)` `=-0.25888-(-1.719)=1.46012V` |
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| 3079. |
The emf of given cell `Pt- H_2 | H^+|H_2 - Pt` is :A. `(RT)/F log P_1/P_2`B. `(RT)/(2F)log ep_1/P_2`C. `(RT)/F- [email protected]/P_1`D. None of these |
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Answer» Correct Answer - B `H_2 rarr 2H^+ +2e` `2H^+ +2erarr H_2` `E_("cell") = E_(OP_(H))^@ + (RT)/(2F) log_e. ([P_H_2]_1)/([PH_2]_2) = (RT)/(2F) Log_e .P_1/P_2`. |
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| 3080. |
During the electrolysis of fused NaCl, which reaction occurs at anode ?A. Chloride ions are oxidisedB. Sodium ions are oxidisedC. Chloride ions are reducedD. Sodium ions are reduced |
| Answer» Correct Answer - A | |
| 3081. |
The strong oxidising agent has :A. high value of reduction potentialB. high value of oxidation potentialC. low value of reduction potentialD. high tendency to lose electrons |
| Answer» Correct Answer - A | |
| 3082. |
Which statements is true about a spontaneous cell reaction in galvanic cell?A. `E_(cell)^@ gt 0, Delta G^@ lt0,` and `Q lt K`B. `E_(cell)^@ gt 0, Delta G^@ lt0`, and` Q lt K`C. `E_(cell)^@ gt 0, Delta G^@ lt0`, and `Q lt K`D. ` E_(cell) gt 0, Delta G lt0`, and `Q lt K` |
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Answer» Correct Answer - D For spontaneous reaction in every condition `E_("cell") gt 0, DeltaG lt0` and `Q` (reaction quotent ) `ltK` (equilibrimu constant). |
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| 3083. |
`E^(@) (Ni^(2+)//NI) = -0.25 "volt", E^(@) (Au^(3+)//Au) = 1.50 "volt"`. The emf of the voltaic cell, `Ni|Ni^(2+) (1.0M) || Au^(3+) (1.0 M)|Au` is :A. 1.25 voltB. `-1.75 "volt"`C. 1.075 voltD. 4.0 volt |
| Answer» Correct Answer - C | |
| 3084. |
`E^(@)(Ni^(2+)//Ni) =- 0.25` volt, `E^(@)(Au^(3+)//Au) = 1.50` volt. The emf of the voltaic cell `Ni|Ni^(2+) (1.0M)||Au^(3+)(1.0M)|Au` is:-A. `1.25` voltB. `-1.75` voltC. `1.75` voltD. `4.0`volt |
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Answer» Correct Answer - C `E_("cell")^(@)= E_("cathod")^(@)-E_("anode")^(@)=1.50 - (-0.25)= 1.75 V` |
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| 3085. |
Emf of the cell `Ni| Ni^(2+) ( 0.1 M) | Au^(3+) (1.0M)` Au will be `E_(Ni//Ni(2+))^@ = 0.5=25. E_(Au//Au^(3+))^@ = 1.5 V`.A. ` 1. 75 V`B. ` +1. 7795 V`C. ` = 0 . 775 V`D. `- 1.7795 V` |
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Answer» Correct Answer - B Cell reaction: `3Ni + 2 A u^(+3) rarr 3 Ni^(+2) = 2 Au` `E_("cell") = E_("cell")^@ - (0.0591)/6 log. ([Ni^(+2)]^2)/([Au^(+3)]^2)` ` =(0.25+ 15) - (0.0591)/6 log. ((0.1)^3)/((1)^2)` ` = 1.75 - (0.0591) /2 log.(.1)` `=1. 75 + 0. 295 = +1,7795 V`. |
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| 3086. |
Assertion: `Ni//Ni^(2+)(1.0M)||Au^(3+)(1.0M)|Au`, for this cell emf is 1.75V if `E_(Au^(3+)//Au)^(o)=1.50` and `E_(Ni^(2+)//Ni)^(o)=0.25V` Reason: Emf of the cell `=E_("cathode")^(o)-E_("anode")^(o)`A. if both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is falseD. If the assertion and reason both are false. |
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Answer» Correct Answer - A Both assertion and reason are true and reason is the correct explanation of assertion. `E_(Au^(3+)//Au)^(@)-E_(Ni//Ni^(2+))^(@)=1.50-(+0.25)=1.25V` |
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| 3087. |
for the electrochemical cell: `Ag|AgCl(s)|KCl(aq)||AgNO_(3)(aq)|Ag`. The overall cell reaction isA. `Ag^(+) + KCl to AgCl (s) + K^(+)`B. `Ag + AgCl to Ag + (1)/(2) Cl_(2)`C. ` AgCl(s) to Ag^(2+) + Cl^(-)`D. `Ag^(+) + Cl^(-) to AgCl(s)` |
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Answer» Correct Answer - C `Ag Cl(s) + e^(-) to Ag + Cl^(-) (aq)` `Ag to Ag^(+) (aq) + e^(-)` Overall reaction : `AgCl_((s)) to Ag^(+) + Cl^(-)` |
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| 3088. |
The solution of `CuSO_(4)` in which copper rod is immersed is diluted to times. The reduction electrode potentialA. increases by 29.5 m VB. decreases by 29.5 m VC. increases by 529 m VD. decreases by 592 m V |
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Answer» Correct Answer - D A/C to Nernst equation `E = E^(@) + (0.0592)/(n) "log " [M^(n+)]` `[M^(n+)]` decreases by 100 times `therefore E = E^(@) + (0.0592)/(2) "log" (1)/(100) = E^(@) + (0.0592)/(2) "log" 10^(-2)` `E = E^(@) - (0.0592)/(2) xx 2 = E^(@) = -0.0592 V ` `E = E^(@) - 592` m V E decreases by 529 m V |
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| 3089. |
Which of the following is highly corrosive salt ?A. ` FeCl_2`B. ` +1.1V`C. ` +0.46 V`D. ` +0.76 V` |
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Answer» Correct Answer - d HgC has corrosive action. If is highly poisonous . It sublimes on heating. It si, therfore, known as corrosive sublimate. |
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| 3090. |
`E^(@)(Ni^(2+)//Ni) =- 0.25` volt, `E^(@)(Au^(3+)//Au) = 1.50` volt. The emf of the voltaic cell `Ni|Ni^(2+) (1.0M)||Au^(3+)(1.0M)|Au` is:-A. `+ 1.25` VB. `-1.75` VC. `+ 1.75` VD. `+4.0` V |
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Answer» Correct Answer - C `E_(cell)^(@) = E_(RHS)^(@) = E_(LHS)^(@)` `= 1.50 - (-0.25) = 1.75` V `E_(cell) = E_("cell")^(@)- (0.0592)/(n) log_(10) ([P])/([R])` Cell reaction is 3 Ni + ` Au^(3+) to 3 Ni^(2)` + 2 Au , n = 6 and `[Ni^(2+)] = [Au^(3+)] = 1` `therefore E_(cell) = 1.75 - (0.0592)/(6) log _(10) ([Ni^(2+)]^(2))/([Au^(3+)]^(2)) = 1.75` V |
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| 3091. |
The standard electrode potentials of four elements A, B, C and D are `-3.05, 1.66, -0.40` and 0.80 volt. The highest chemical activity will be shown by :A. AB. BC. CD. D |
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Answer» Correct Answer - A More negative is `E_(RP)^(@)` , more is reactivity of metal . |
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| 3092. |
The standard oxidation potential `E^@` for the half cell reaction are `Zn rarr Zn^(2+) + 2e^-` `E^@=+ 0.76 V ` `Fe rarr Fe^(2+) + 2e^-` `E^@=+ 0.41 V ` EMF of the cell rection is `Zn+Fe^(2+) rarr Zn^(2+)+Fe `A. `1.17 ` VB. `0.35` VC. `-1.17 V `D. `-0.35` V |
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Answer» Correct Answer - B Here oxidation potentials are given `E_(cell)^(@) = E_("OX (LHS)")^(@) - E_("OX (RHS)")^(@)` The cell is - Anode (LHS) `to` Higher SOP Cathode (RHS) `to` Lower SOP `E_(cell)^(@) = 0.76 - 0.41 = 0.35 V ` |
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| 3093. |
The standard oxidation potential `E^@` for the half cell reaction are `Zn rarr Zn^(2+) + 2e^-` `E^@=+ 0.76 V ` `Fe rarr Fe^(2+) + 2e^-` `E^@=+ 0.41 V ` EMF of the cell rection is `Zn+Fe^(2+) rarr Zn^(2+)+Fe `A. `-0.35` VB. `+0.35` VC. `+1.17` VD. `-0.17` V |
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Answer» Correct Answer - B `E_("cell")^(@) = E_(OP_(Zn))^(@) - E_(OP_(Fe))^(@) = 0.76 - 0.41 = 0.35` V |
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| 3094. |
Which is correct about silver platingA. Anode - pure AgB. Cathode - object to be electroplatedC. Electrolyte `- Na [Ag (CN)_(2)]`D. Electrolyte `- AgNO_(3)` |
| Answer» Correct Answer - A::B::C | |
| 3095. |
For the following cell, `Zn(s)|ZnSO_(4)(aq)||CuSO_(4)(aq)|Cu(s)` When the concentration of `Zn^(2+)` is 10 times the concentration of `Cu^(2+)`, the expression for `DeltaG` ( in `Jmol^(-1)`) is [F is faraday constant: R is gas constant, T is temperature, `E^(@)(cell)=1.1V`]A. 2.303RT+1.1FB. 1.1FC. 2.303RT-2.2FD. `-2.2F` |
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Answer» Correct Answer - C `DeltaG=DeltaG^(o)+2.303RTlog_(10)Q,Q=([Zn^(2+)])/([Cu^(2+)])` `=-2F(1.1)+2.303RTlog_(10)10` `=2.303RT-2.2F` Given `E_(Cl_(2)//Cl^(-))^(o)=1.36V,E_(Cr^(3+)//Cr)^(o)=-0.74V` `E_(Cr_(2)O_(7)^(2-)//Cr_(3+))^(o)=1.33V,E_(MnO_(4)^(-))^(o)//Mn^(2+)=1.51V` |
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| 3096. |
For the following cell `Zn(s)|ZnSO_(4)(aq)||CuSO_(4)(aq)|Cu(s)` when the concentration of `Zn^(2+)` is 10 times the concentration of `Cu^(2+)`, the expression for `DeltaG` (in 1 `mol^(-1)`) is [F is faraday constant, R is gas constant, T is temperature, `E^(@)(cell)=1.1V`]A. `2.303RT+1.1F`B. `1.1F`C. `2.303RT-2.2F`D. `-2.2F` |
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Answer» Correct Answer - C Cell reaction is `Zn+Cu^(2+)toZn^(2+)+Cu` `DeltaG=DeltaG^(@)+2.303RTlogQ` `=DeltaG^(@)+2.303RTlog([Zn^(2+)])/([Cu^(2+)])` `DeltaG^(@)=-nFE_(cell)^(@)=-2F(1.1)` `thereforeDeltaG=-2F(1.1)+2.303RTlog10` `=2.303RT-2.2F` |
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| 3097. |
For the following cell, `Zn(s)|ZnSO_(4)(aq)||CuSO_(4)(aq)||Cu(s)` When the concentration of `Zn^(2+)` is 10 times the concentration of `Cu^(2+)`, the expression for `DeltaG` (in J `"mol"^(-1)`) [F is Faraday constant, R is gas constant] T is temperaure, `E^(@)("cell")=1.1V`A. 2.303 RT +1.1 FB. 1.1 FC. 2.303 RT -2.2 FD. `-2.2 F` |
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Answer» Correct Answer - C (c ) `DeltaG=DeltaG^(@)+2.303" RT "log_(10)Q,Q=([Zn^(2+)])/([Cu^(2+)])` `=-2F(1.1)+2.303" RT "log_(10)10` `=2.303" RT"-2.2" F"` |
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| 3098. |
For the cell: `Zn(s)|ZnSO_(4)(aq)||CuSO_(4)(aq)Cu(s)`, calculate standard cell potential if standard state reduction electrode potentials for `Cu^(2+)//Cu and Zn^(2+)//Zn` are +0.34V and -0.76V respectively. |
| Answer» Correct Answer - 1.10 V | |
| 3099. |
The conductivity of a saturated solution of `Ag_(3)PO_(4)` is `9 xx 10^(-6) S m^(-1)` and its equivalent conductivity is `1.50 xx 10^(-4) Sm^(2) "equivalent"^(-1)`. Th `K_(sp)` of `Ag_(3)PO_(4)` is:-A. `4.32 xx 10^(-18)`B. `1.8 xx 10^(-9)`C. `8.64 xx 10^(-13)`D. None of these |
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Answer» Correct Answer - A `lambda_(eq)+K +(1000)/(N)` `1.50xx10^(-4)xx10^(4)=9xx10^(-5)xx10^(-2)xx(1000)/(N)` `N= 6xx10^(-5)` `S=M=(N)/(n_(f))=(6xx10^(-5))/(3)=2xx10^(-5)mol//L` . `{:(Ag_(3)PO_(4)=,3 Ag^(+)+,PO_(4)),(,3S,S):}` `K_(sp)=(3S)^(3).S=27.S^(4)=27xx(2xx10^(-5))^(4)` `=4.32xx10^(-18)` . |
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| 3100. |
Consider the following standard electrode potentials and calculate the eqillibrium constant at `25^(@)`C for the indicated disproportional reaction : `3 Mn^(2+)(aq)toMn(s)+2Mn^(3+)(aq)` `Mn^(3+)(aq)+e^(-)toMn^(2+)(aq), E^(@)=1.51 V` `Mn^(2+)(aq)+2e^(-)toMn(s), E^(@)=-1.185 V``A. `1.2xx10^(-43)`B. `2.4xx10^(-73)`C. `6.3xx10^(-92)`D. `1.5xx10^(-62)` |
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Answer» Correct Answer - C |
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