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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.
| 1201. |
The measured potential for, `Mg^(2+) + 2e^(-) hArr Mg (s)` does not depend upon:A. raising the temperatureB. increasing the concentration of `Mg^(2+)` ionsC. making the magnesium plate biggerD. purity of magnesium plate |
| Answer» Correct Answer - C | |
| 1202. |
The standard electrode potential for the reactions, `Ag^(+)(aq)+e^(-) rarr Ag(s)` `Sn^(2+)(aq)+2e^(-) rarr Sn(s)` at `25^(@)C` are `0.80 ` volt and `-0.14 ` volt, respectively. The `emf`of the cell `Sn|Sn^(2+)(1M)||Ag^(+)(1M)Ag` is `:`A. `0.66` VB. 0.80 VC. 0.94 VD. 1.08 V |
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Answer» Correct Answer - A `E_(cell)^(@) = E_(RP_(Ag))^(@) - E_(RP_(Sn))^(@)` `= 0.80 - (-0.14) = 0.94` V |
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| 1203. |
The measured reduction potential for the reaction, `Mg^(2+) + 2e^(-) hArr Mg(s)` depends upon:A. temperatureB. the concentration of `Mg^(2+)` ionsC. the purity of the magnesium plateD. the area if cross-section of magnesium plate |
| Answer» Correct Answer - A::B::C | |
| 1204. |
Calculate equilibrium constant for `I_(2)+I^(-) hArr I_(3)^(-)` at 298 K from the following information : `{:(I_(2) (aq.)+2e^(-) rarr 2I^(-),,E^(@)=0.6197" volt"),(I_(3)^(-)+2e^(-) rarr 3I^(-),,E^(@)=0.5355" volt"):}` |
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Answer» Correct Answer - 706.9 `{:(I_(2)+2e^(-) rarr 2I^(-)," "E^(@)=0.6197" volt"),(3I^(-) rarr I_(3)^(-)+2e^(-)," "E^(@)=-0.5355" volt"),(bar(I_(2)+I^(-) hArr I_(3)^(-),)" "E^(@)=0.6197-0.5355),(" "=0.0842" volt"):}` `K="antilog" [(nE^(@))/0.0591]="antilog"[(2xx0.0842)/0.0591]=706.9` |
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| 1205. |
In an electrochemical cell, anode and cathode are:A. positive electrode , negative electrodeB. negative electrode , positive electrodeC. positive and negative electrode bothD. none |
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Answer» Correct Answer - A Anode is positive electrode and cathode is -ve electrode in electrolytic cell where as anode is -ve electrode and cathode is +ve electrode in galvonic cells . |
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| 1206. |
At equilibriumA. a cell operates first reversibly and then irreversiblyB. the cell potential is negativeC. the cell potential is positiveD. a cell is exhausted |
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Answer» Correct Answer - D As the cell discharges and current flows, the product concentrations increase. At the same time, reactant concentrations increase. This increases `log Q_("cell")`,, so the cor-reaction factor becomes more negative. Thus, the overall `E_("cell")` decreases (the reaciton becomes less favourable). Eventually the cell potential approaches zero (equilibrium) and the cell "run down". The cell is completly run down `(E_("cell") = 0)`. when the cell term `(0.0592)/(n) log Q_("cell")` is equal in magnitude to `E_("cell")^(@)` . At equilibrium the reaction quotient `(Q)` is equal to the equalirium constant `(K_(eq))`. |
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| 1207. |
The reaction, `(1)/(2) H_(2) (g) + Ag Cl (s) = H^(+) (a.q) + Cl^(-) (a.q) + Ag (s)` occurs in the galvanic cell:A. `Ag|AgCl_(s) | KCl("soln") || AgNO_(3) ("soln.") | Ag`B. `Pt|H_(2)(g) | HCl ("soln.") || AgNO_(3) ("soln.")|Ag`C. `Pt | H_(2)(g) | HCl ("soln.") || AgCl (s) | Ag`D. `Pt | H_(2) (g) | KCl ("soln.") || AgCl_(s) | Ag` |
| Answer» Correct Answer - C | |
| 1208. |
In an electrochemical cell, anode and cathode are:A. positively and negatively charged ionsB. positively and negatively charged electrodesC. negatively and positively charged electrodesD. negatively and positively charged ions |
| Answer» Correct Answer - C | |
| 1209. |
Assertion: In an electrochemical cell anode and cathode are respectively negative and positive electrodes. Reason: At anode oxidation takes place and at cathode reduction takes place.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. Anode (oxidation) for eq. `ZntoZn^(2+)+2e^(-)`, so Excess of electrons and hence negatively charged while cathode is positively charged. |
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| 1210. |
The metal oxide which decomposes on heating, is:A. `ZnO`B. `HgO`C. `Al_(2)O_(3)`D. `CuO` |
| Answer» Correct Answer - B | |
| 1211. |
Consider a cell given below. `Cu|Cu^(2+)|Cl^(-)|Cl_(2).Pt` Write the reactions that occur at anode and cathode. |
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Answer» The reaction taking place at the two electrodes are : At anode : `Cu(s) to Cu^(2+)(aq) +2e^(-)` At cathode : `Cl_(2)(g) +2e^(-) to 2Cl^(-)(aq)` Thus,Cu(s) is oxidised to `Cu^(2+)`(aq) ions at anode while `Cl_(2)`(g) is reduced to `Cl^(-)` (aq) ions at cathode. |
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| 1212. |
Find the the reducation potential of the half-cell `Pt(s)|Cu^(2+)(aq., 0.22 M), Cu^(+)(aq., 0.043 M)`A. `0.20 V`B. `-0.20 V`C. `0.30 V`D. `-0.30 V` |
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Answer» Correct Answer - A The half-cell reaction is `Cu^(2+)(aq.)+e^(-)hArr Cu^(+)(aq.)` and its `E_(cu^(2+)//Cu)^(@)` is `0.158 V` The value of `n`, the number of electrons transferred, is clearly `1`, for this half-reaction. The expression for `Q` is `C_(Cu^(+)//C_Cu^(2+))` Substituting these data into the Nernst equaltion. `E= E^(@) - (0.0592)/(n) log Q` `E_(Cu^(2+)//Cu^(+)) = E_(Cu^(2+)//Cu^(+))-(0.0592 V)/(n) "log" (C_(Cu^(+)))/(C_(Cu^(++)))` ` (0.158 v) - (0.0592 V)/(1) "log" (0.043)/(0.22)` `= 0.20 V` This answer is consistent with the chemistry of this system. The positive value of `E_("cell")` tells us that `Cu^(+)` (aq.) is stable relative to `Cu^(2+)(aq.)` and will predominte at equibirium. In this half-cell, the systeam is even further from equilibrium than in the standard state because the `[Cu^(2+)]` is sibstanitially larger than the `[Cu^(+)]`. Therefore, `E` is greater than `E^(@)`. |
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| 1213. |
Calculate the voltage `E` of the cell `Ag(s) AgIO_(3)(s)|Ag^(+) (x,M), HIO_(3) (0.300M)||Zn^(2+) (0.175M)|Zn(s)` If `K_(SP) = 3.02 xx 10^(-8)` for `AgIO_(3)(s)` and `K_(a) = 0.162` for `HIO_(3), E^(@) (Zn^(+2)//Zn) =- 0.76 V, E^(@) (Ag//Ag^(+)) =- 0.8V` |
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Answer» Correct Answer - `-1.188V` `{:(HIO_(3),hArr,H^(+)+,IO_(3)^(-)),(C,,,),(C-X,,X,X):}` `0.162 =(x^(2))/(0.3 -x)` `x = 0.153` `K_(sp) = [Ag^(+)] [IO_(3)^(-)]` `[Ag^(+)] = 1.974 xx 10^(-7)` `E_(cell) = 0.8 +(0.76) -(0.0591)/(2)log. (|Ag^(+)|^(2))/([Zn^(2+)])` `E_(cell) - 1.188` |
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| 1214. |
A first order reaction has a rate constatn `1.15 xx10^(-3) s ^(-1).` How long will 5 g of this reactant take to reduce to 3 g ? |
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Answer» Rate constant `[k] =1.15xx10^(-3) s^(-1)` initial amount `[R]_(0)=5g,` Final a amount `[R]=3g` `t=(2.303)/(R) log ""([R]_(0))/(R)` `=(2.303)/((1.15xx10^(-3) s^(-1)))log"" ((5g)/(3g))=(2.303)/([1.15xx10^(-3) s^(-1)]) [log 5- log 3]` `=(2.303)/(1.15xx10^(-3) s^(-1))(0.6990-0.4771)=2.0 xx10^(3)xx 0.2219 s=443.8s` Time `=444s.` |
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| 1215. |
At infinite dilution the molar conductance of `AI^(+3)` and `SO_(4)^(-2)` ion are 189 and `160 Omega^(-1) cm^(2) "mole"^(-1)` resoectively. Calculate the equilvalent and molar conductivity at infinite dilute of `AI_(2)(SO_(4))_(3)`. |
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Answer» `lambda_(eq.[AI_(2)(SO_(4))_(3)])^(prop) = (1)/(3) lambda_(AI^(+3))^(prop) +(1)/(2) lambda_(SO_(4)^(-2))^(prop)` `= (1)/(3) xx 189 +(1)/(2) xx 160` `= 143 Omega^(-1) cm^(2) eq^(-1)` Molar conductivity `= lambda_(eq) xx V.F. = 143 xx 6` `= 858 Omega^(-1) cm^(2) mol^(-1)` |
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| 1216. |
What is standard hydrogen electrode ? |
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Answer» The electrode whose potential is known as standard electrode (or) standard hydrogen electrode. To determine the potential of a single electrode experimentally it cobine with standard hydrogen electrode and the RMF of cell so constructed is measured with potentiometer. |
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| 1217. |
Can you store copper sulphate soolutions in a zinc pot ? |
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Answer» No, zinc pot cannot store cpooer sulphate solutions because the standard electrode potential `(E^(0))` vlaud of zinc is less than that of copper. So, zinc is stronger reducing agent then copper. `Zn _((aq))^(2) +2e^(-)to Zn_((s)), E^(0) =-0.76V` `CU_((aq))^(2+)+2e^(-) to Cu(s) ,E ^(0) =-0.34V` So, zinc will loss electrons to `Cu_((aq))^(2+)` ions and redox reaction will occur as follows. `Zn (s) +Cu_((aq))^(2+)to Zn_((aq))^(2+)+Cu(s)` |
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| 1218. |
In a Daniell cell, electrons flow from zinc electrode to copper electrode outside the cell. |
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Answer» Correct Answer - T In a Daniell cell, `Zn` is anode. So `e^(c-)` flow from `Zn` to `Cu` electrode. |
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| 1219. |
Compounds of active metals `(Zn, Na, Mg)` are reducible by `H_(2)` whereas those of noble metals `(Cu,Ag,Au)` are not reducible. |
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Answer» Correct Answer - F The reverse is true. |
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| 1220. |
The e.m.f. of the cell `Ti|Ti^(+)(0.001M)||Cu^(2+)(0.01M)|Cu` is 0.83V the emf of this cell could b e increased byA. increasing the concentration of `Ti^(+)` ionsB. increasing the concentration of `Cu^(2+)` ionsC. increasing the concentration of bothD. none of the above. |
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Answer» Correct Answer - B `E_("cell")=E_("cell")^(@)+(0.059)/(n)"log"([Cu^(+2)])/([Ti^(+)]^(2))` Increasing conc. Of `Cu^(+2)` ions increases the emf. |
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| 1221. |
For the cell`:` `TI|TI^(o+)(10^(-3)M)||Cu^(2+)(10^(-1)M)|Cu` `E_(cell)` can be increased by `a. ` Decreasing `[Cu^(2+)]." "b.` Decreasing `[TI^(o+)]` `c.` Increasing `[Cu^(2+)] " "d. Increasing by `[TI^(o+)]` |
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Answer» Correct Answer - b,c Anode reaction `:` `2Tl(s) rarr 2Tl^(o+)(1^(-3)M)+cancel(2e^(-))` Cathode reaction `:` `Cu^(2+)(10^(-1)M)+cancel(2e^(-)) rarr Cu(s)` `ulbar(2Tl+Cu^(2+)(10^(-1)M)rarr 2 Tl^(o+)(10^(-3)M)+Cu(s))` `E_(cell)=E^(-)._(cell)-(0.059)/(2)log.([Tl^(o+)]^(2))/([Cu^(2+)]^(2))` `=E^(c-)._(cell)-(0.059)/(2)log[Tl^(o+)]^(2)+(0.059)/(2)log[Cu^(2+)]` `E_(cell)` can be increased by decreasing `[Tl^(o+)]` or increasing `[Cu^(2+)]` . |
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| 1222. |
The e.m.f. of the cell `Ti|Ti^(+)(0.001M)||Cu^(2+)(0.01M)|Cu` is 0.83V the emf of this cell could b e increased byA. increasing the concentration of `TI^(+)` ionsB. increasing the concentration of `Cu^(2+)` ionsC. increasing the concentration of bothD. none of above . |
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Answer» Correct Answer - B Cell reaction `2 Tl + Cu^(2+) to 2 Tl^(+) + Cu`, `E_("cell") = E_("cell")^(@) - (2.303 RT)/("nF") "log" ([Tl^(+)]^(2+))/([Cu^(2+)])`. |
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| 1223. |
Which of the following statements are not correct?A. Same quantity of electricity deposits more of iron from ferric sulphate solution than from ferrous sulphate solutionB. Electrochemical equivalent of an element can be obtained by dividing its equivalent weight by 96,500C. 1 Faraday always liberates 1 mole of the substance at the electrodeD. A 60 watt bulb emits 60 Joules of energy per second. |
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Answer» Correct Answer - A::C (a) is wrong because `Fe^(2+)+2e^(-)toFe,Fe^(3+)+3e^(-)toFe`. One faraday from `Fe^(2+)` will deposit `(1)/(2)` mol of Fe and from `Fe^(3+)`, it will deposit `(1)/(3)` mol of Fe. (c) is wrong because 1 F liberates one grm equivalent of the substance. |
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| 1224. |
Which are true for a standard hydrogen electrode ?A. The `H^(+)` ion concentration is 1 MB. Temperature is `35^(@)C`C. Pressure of hydrogen is 1 atmosphere.D. It contains metallic conductor which does not adsorb hydrogen. |
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Answer» Correct Answer - A::C::D (a,c,d) are correct. |
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| 1225. |
Which of the followingg are correct?A. Electrolysis of dilute NaOH solution given `H_(2)` at cathode and `O_(2)` at anode.B. Electrolysis of sulphuric acid (dilute or concentrated) gives `H_(2)` at cathode and `O_(2)` at anode.C. Electrolysis of aqueous KF solution gives fluorine at the anodeD. oxidation of copper anode occurs in the electrolysis of aqueous copper sulphate solution using solution copper electrodes. |
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Answer» Correct Answer - A::D (b) is wrong because electrolysis of concentrated `H_(2)SO_(4)` does not give `O_(2)` at the anode. (c) is wrong because electrolysis of aqueous KF solution gives `O_(2)` at the anode and not `F_(2)`. |
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| 1226. |
The `E^(@)` values corresponding to the following two reduction electrode processes are: (i) `Cu^(+)//Cu=+0.52V` (ii) `Cu^(2+)//Cu^(+)=+0.16V` Formulate the galvanic cell for their combination. What will be the standard cell potential for it? Calculate `Delta_(r)G^(@)` for the cell reaction `(F=96500" C "mol^(-1))` |
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Answer» For EMF to be +ve, oxidation should take place on electrode (ii) i.e., half-cell reactions will be `Cu^(+)+e^(-)toCu` `underline(" "Cu^(+)toCu^(2+)+e^(-)" ")` Overall cell reaction: `2Cu^(+)toCu+Cu^(2+)` Hence, the cell will be represented as: `Cu^(+)|Cu^(2+)||Cu^(+)|Cu` `E_(cell)^(@)=E_(Red)^(@)(RHS)-E_(Red)^(@)(LHS)=0.52-0.16=0.36V` `Delta_(r)G^(@)=-nFE_(cell)^(@)=-1xx96500" C "mol^(-1)xx0.36V=-34740" CV "mol^(-1)=-34740" J "mol^(-1)` (1 CV=1 J) |
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| 1227. |
Calcaulte the standard free energy change taking place in `H_(2)//O_(2)` fuel in which the following reactions occurs: (i) `O_(2)+4H^(+) to 4e^(-)to2H_(2)O,E^(@)=1.229V` (ii) `2H_(2)to4H^(+)+4e^(-),E^(@)=0.000V` |
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Answer» `E_(cell)^(@)=E_(O_(2)//H_(2)O)^(@)(Red)+E_(H_(2)//H^(+))^(@)(Ox)=1.229+0=1.229V` `DeltaG=-nFE_(cell)^(@)=-4xx96500xx1.229J=-474.4kJ`. |
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| 1228. |
A constant current of 30 amperes ispassed through an aqueous solution of NaCl for 1 hour. How many grams of NaOH will be formed in the reaction ? Also find out the volume of `Cl_(2)` evolved at S.T.P. |
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Answer» The redox reaction taking place on electrolysis is an follows : `2Cl^(-)(aq) to Cl_(2)(g)+2e^(-)` `(2H_(2)O(l)+2e^(-) to H_(2)(g)+2OH^(-)(aq))/(2Cl^(-)(aq)+2H_(2)O(l) to H_(2)(g)+Cl_(2)(g)+2OH^(-)(aq)` Step I. Calculation of the mass of NaOH formed Quantity of charge passed=(30 amp)xx(60xx60 s)=108000 Coulombs Now, 96500 C of charge from NaOH=40 g 108000 C of charge will form NaOH `=(40g)xx((108000C))/((96500C))=44.77 g=1.12 mol`. Step II. Calculation of volume of `Cl_(2)` evolved at S.T.P. According to the equation, No. of moles of `Cl_(2)` evolved`=1//2xx"No"`. of mole of NaOH formed `=1//2xx1.12=0.56 mol` 1 mole of `Cl_(2)` at S.T.P. will correspond to volume =22.4L 0.56 mole of `Cl_(2)` at S.T.P. will correspond to valume`=((22.4L))/((1 mol))xx(0.56 mol)=12.54 L` |
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| 1229. |
When a lead storage battery is dischargedA. `SO_(2)` is evolvedB. Lead sulphate is consumedC. Lead is formedD. Sulphuric acid is consumed |
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Answer» Correct Answer - D `Pb+PbO_(2)+2H_(2)SO_(4) underset("recharge")overset("Discharge")hArr2PbSO_(4)+2H_(2)O` Sulphuric acid is consumed on discharging. |
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| 1230. |
When a copper wire is placed in a solution of `AgNO_(3)`, the solution acquirs blue colour. This is due to the formation ofA. `Cu^(2+)` ionsB. `Cu^(+)` ionsC. Soluble complexof copper with `AgNO_(3)`D. `Cu^(-)` ion by the reduction of Cu |
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Answer» Correct Answer - A Since, `Ag^(+)` ions are reduced to Ag and `E_(Ag^(+)//Ag)^(o) gt E_(Cu^(++)//Cu)^(o)` Cu is oxidised to `Cu^(++)`. |
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| 1231. |
The specific conductivity of `N//50` KCl solution at 298 K is `2.768xx10^(-3)`mho per cm. The resistance of this solutions at 298 K when measured in a particular cell is 250.2 ohm. The resistance of `M//100 CuSO_(4)` solution at 298 K measured with the same cell was 8331 ohm. Calculate the molar conductivity of copper sulphate solution. |
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Answer» Correct Answer - `8.31 S cm^(2)mol^(-1)` Step I. Calculation of cell constant `R=250.2" ohm", k=2.768xx10^(-3)ohm^(-1) cm^(-1)` Cell constant `(l//a)=kxxR=(2.768xx10^(-3)ohm^(-1)cm^(-1))xx(250.2" ohm")=0.6925 cm^(-1)`. Step II. Calculation of molar conductane`(Lambda_(m))=(kxx1000)/(M)` Specific conductance (k)`=("cell constant")/(R )=((0.6925 cm^(-1)))/((8331" ohm"))=8.31xx10^(-6) ohm^(-1)cm^(-1)` Molarity of solution (M)=0.01 M `:. "Molar conductance" (Lambda_(m))=((8.31xx10^(-5)ohm^(-1) cm^(-1))xx(1000" cm"^(3)))/((0.01" mol"))` `= 831xx10^(-2)ohm^(-1) cm^(2) mol^(-1)=8.31" S " cm^(2) mol^(-1)`. |
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| 1232. |
At `25^(@)C` specific coductivity of a normal solution of KCl is 0.002765 mho. The resistance of cell is 400 ohms. The cell constant is:-A. 0.815B. 1.016C. 1.106D. 2.016 |
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Answer» Correct Answer - C Cell constant`=("Specific conductivity")/("Observed conductance")` `=(0.002765)/(1//R)=0.002765xx400=1.106` |
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| 1233. |
The reduction potential values of M, N and O are +2.46 V, -1.13 V, -3.13 V respectively. Which of the following orders is correct regarding their reducing property ?A. `O gt N gt M`B. `O gt M gt N`C. `M gt N gt O`D. `M gt O gt N` |
| Answer» Correct Answer - A | |
| 1234. |
The tendency of an electrode to lose electrons is known asA. Electrode potentialB. Reduction potentialC. Oxidation potentialD. E.M.F. |
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Answer» Correct Answer - C The magnitude of the electrode potential of a metal is a measure of its relative tendency toloose or gain electrons. i.e., it is a measure of the relative tendency to undergo oxidation (loss of electrons) or reduction (gain of electrons). `MtoM^(n+)+n e^(-)` (oxidation potential) `M^(n+)+n e^(-)toM` (reduction potential). |
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| 1235. |
One faraday of electricity will liberate one gram mole of the metal from the solution ofA. `BaCl_(2)`B. `CuSO_(4)`C. `AlCl_(3)`D. `NaCl` |
| Answer» Correct Answer - D | |
| 1236. |
On passing 0.1 faraday of electricity through fused sodium chloride, the amount of chlorine liberated is (At. Mass of `Cl=35.45`)A. 35.45gB. 70.9gC. 3.545gD. 17.77g |
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Answer» Correct Answer - C `Cl^(-)rarr(1)/(2)Cl_(2)+e^(-)` I .e., 1 Faraday liberates `Cl_(2)=(1)/(2)"mol"=35.45g`. `therefore` 0.1 Faraday will liberate `Cl_(2)=3.545` g . |
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| 1237. |
When the same electric current is passed through the solution of different electrolytes in series, the amounts of elements deposited on the electrodes are in the ratio of theirA. densitiesB. electrochemical equivalentsC. atomic massesD. atomic numbers . |
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Answer» Correct Answer - B `W = ZQ.` If Q is same , `W prop Z`. |
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| 1238. |
What is the charge on one mole of electrons ?A. `6.28 xx 10^(18)` coulombB. `1.6 xx 10^(-19)` coulombC. `9.65 xx 10^(4)` coulombD. None of the above . |
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Answer» Correct Answer - C Charge on one mole of monovalent ions = 1 Faraday = 96500 C . |
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| 1239. |
Ione faraday of electricity will liberate one gram atom of a metal from a solution ofA. `AuCl_(3)`B. `CuSO_(4)`C. `BaCl_(2)`D. `KCl` |
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Answer» Correct Answer - D `KCl hArr K^(+) + Cl^(-)` and `K^(+) + e^(-) rarr K` Thus, one mole `KCl` need one mole electrons to liberate one gram metal atom. One mole electron charge is equal to `96500 C (1 F)`. Hence, `1 F` charge will liberate one gram atom of metal from solution of `KCl`, |
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| 1240. |
In which one of the following one faraday of electricity will liberate 1/2 gram -atom of the metal?A. `AlCl_(3)`B. `FeCl_(3)`C. `CuSO_(4)`D. `NaCl`. |
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Answer» Correct Answer - C `Cu^(2+) + 2e^(-) to Cu`. So 2 F liberate 1 g atom of Cu . |
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| 1241. |
Quantity of electricity is measured inA. ampere secB. ampereC. ampere `sec^(-1)`D. `"ampere"^(-1)` sec |
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Answer» Correct Answer - A Quantity of current is charge , i.e. coulomb or ampere sec. |
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| 1242. |
The questions consist of two atatements each, printed as Assertion and Reason. While answering these questions you are required to choose any one of the following four responses : Speed of ions depends upon nature of ions only. Mobility of ions depends upon mass, charge and size of ions only.A. If both the assertion and reason are true but the reason is ont the correct explanation of assertionB. If both the assertion and reason are true but the reason is not the correct explanation of assertion.C. If the assertion is true but reason is false.D. If assertion is false but reason is rue |
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Answer» Correct Answer - d Speed of ions depends upon nature of ions and potential gradient . |
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| 1243. |
The questions consist of two atatements each, printed as Assertion and Reason. While answering these questions you are required to choose any one of the following four responses : Assertion : If standard reduction potential for the reaction `Ag^+ + e^(-) -> Ag` is `0. 80 ` volt, then for the reaction `2Ag^+ +2e^(-) ->2Ag`, it will be `1, 60` volt .Reason : If concentration of `Ag^+` ions is doubled , the standard electrode potential is also doubled.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 - D (d) Correct assertion : If standard reduction potential for the reaction, `Ag^(+)+e^(-) to Ag` is 0.80 volt, then for the reaction `2Ag^(+)+2e^(-) to 2Ag` is also +0.80 volt. Correct reason : The effect of `E^(@)` value for an electrode cannot change. |
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| 1244. |
The questions consist of two atatements each, printed as Assertion and Reason. While answering these questions you are required to choose any one of the following four responses : Gavanised iron does not rust . Zinc has a more negative electrode potential than iron .A. If both the assertion and reason are true but the reason is ont the correct explanation of assertionB. If both the assertion and reason are true but the reason is not the correct explanation of assertion.C. If the assertion is true but reason is false.D. If assertion is false but reason is rue |
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Answer» Correct Answer - a Zinc metal which has a more negative electrode potential tah iron will provide electorns in prefernce of the iron, and therfore corrodes first, Only when all the zinc has been oxidised then the iron start to rust . |
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| 1245. |
A current strength of `96.5 A` is passed for `10s` through `1L` of a solution of `0.1 M` aqueous `CuSO_(4)`. Calculate the `pH` of the solution. |
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Answer» Aqueous solution of `CuSO_(4)` on electrolysis gives gives `Cu` at cathode and `O_(2)` at anode and `H^(o+)` ions in the solution. Eq of current `=(96.5Axx10s)/(96500C)=0.01Eq=10^(-2)Eq` `:. 10^(-2)Eq` of `Cu=10^(-2)`Eq of `O_(2)=10^(-2)`Eq of `H^(o+)=10^(-2)F` `:. [H^(o+)]=(10^(-2)Eq)/("Volume of solution in L")=(10^(-2)Eq)/(1L)` `=10^(-2)N `or `M` `:. pH-log (10^(-2))=2` |
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| 1246. |
The volume of hydrogen gas liberated at STP when a current of 5.36 ampere is passed through dil. `H_(2)SO_(4)` for 5 hours will beA. 5.6 litresB. 11.2 litresC. 16.8 litresD. 22.4 litres |
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Answer» Correct Answer - B Quantity of electricity passed `= 5.36 xx 5 xx 60 xx 60 s = 96480` coulombs = 1 Faraday `2H^(+) + 2e^(-) to H_(2)` i.e., 2 Faraday , liberate `H_(2)` at STP = 22.4 litres `therefore` 1 Faraday will liberate `H_(2)` at STP = 11.2 litres . |
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| 1247. |
One faraday of electricity deposits one mol of Na from the molten salt but `(1)/(3)` mol of Al from an aluminium salt. Why? |
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Answer» The reactions at cathode for the deposit of Na and Al are `Na^(+)+e^(-)toNa and Al^(3+)+3e^(-)toAl` Thus, 1 faraday deposits 1 mol of Na whereas 3 faradays are required for depositing one mol of Al. |
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| 1248. |
Consider the following four electrodes P = Cu2+(.0001 M/Cu(s) Q = Cu2+ (0.1 M)/Cu(s) R = Cu2+(0.01 M)/Cu(s) S = Cu2+(0.001 M)Cu(s) If the standard reduction potential of Cu2+ /Cu is +.34V the reduction potential in volts of the above electrode follows the order. |
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Answer» If the standard reduction potential of Cu2+/Cu is +.34V the reduction potential in volts of the above electrode follows the order. i.e., Q > R > P |
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| 1249. |
An electric current of 100 ampere is passed through a moltan liquid of sodium chloride for 5 hours. Calculate the volume of chlorine gas liberated at the electrode at NTP. |
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Answer» The reaction taking place at anode is : `underset(71.0 g)(2Cl^(-)) rarr underset((71.0 g),("1 mole"))(Cl_(2))+underset(2xx96500" coulomb")(2e^(-))` `Q=I xx t=100 xx 5xx 60xx 60` coulomb The amount of chlorine liberated by passing `100xx5xx60xx60` coulomb of electric charge `=1/(2xx96500)xx100xx5xx60xx60=9.3264` mole Volume of `Cl_(2)` librated at `NTP=9.3264xx22.4=208.91 L` |
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| 1250. |
Point out the correct statement about `Zn-CuSO_(4)` cell.A. The flow of elec trons occurs from copper to zinc.B. The value of `E^(@)` of copper electrode is less than that of zinc electrode.C. Zinc is anode white Cu is cathode electrodeD. All the statement are correct. |
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Answer» Correct Answer - C Zinc has lower reduction potential than copper electrode. |
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