3195 + Interview Questions in ELECTROCHEMISTRY IN GENERAL AWARENESS Page 18 InterviewSolution

851.	Which of the metals does not corrode inspite of favourable standard electrode potential?A. `Ti`B. `Al`C. both (1) and (2)D. `Fe`
Answer» Correct Answer - C Aluminium , a very active metal, reacts rapidly with `O_(2)` form the air to form a surface layer of aluminium oxide, `Al_(2)O_(3)`, that is so thin that it is trasparent. This very tough, hard substance is inert to `O_(2), H_(2)O`, and most other corrosive agents in the environment. In this way objects made of `Al` form their own protective layers and need not be treted further to inhibit corrosion. However, acid rain endangers structural `Al` by dissolving this `Al_(2)O_(3)` (amphoteric oxide) coating. Similarly `Ti` (which is often used to make sports bikes) does not corrode becuase of a hard, impenetrable layer of `TiO_(2)` that adheres to the surgace and protects themetal form further oxidation.

Discussion

852.	Iron can be protected by coating with zinc or tin. If coating is broken.A. iron will corrode faster if coated with zincB. iron will corrode faster if coated with tinC. iron will corrode faster in both casesD. iron will not undergo any corrosion in both cases.
Answer» Correct Answer - B If coating is broken , iron will corrode faster if coated with tin . This is because their standard oxidation potentials are : ` E_(Zn\|Z^(2+))^(@) = + 0.76 V , E_(Fe\|Fe^(2+))^(@) = +0.44` V `E_(Sn\| Sn^(2+))^(@) = +0.14 V ` .

Discussion

853.	A metal is left exposed to the atmosphere for some time . It gets coated with green carbonate . The metal must beA. silverB. copperC. ironD. zinc .
Answer» Correct Answer - B `CuCO_(3)` is green .

Discussion

854.	For the electrochemical cell, `Mg(s)\|Mg^(2+)(aq,1 M)\|\|Cu^(2+)(aq.1 M) Cu(s)`, the standard emf of the cell is 2.70 V at 300 K. When the concentration of `Mg^(2+)` is changed to x M, the cell potential changes to 2.67 V at 300 K. What is the value of x ? (Given, F/R =11500 K `V^(-1)`,where F is the Faraday constant and R is the gas constant, the value of `"In"_((10))(10)=2.30)`.
Answer» The cell reaction is : `Mg(s)+Cu^(2+)(aq)rarr Mg^(2+)(aq)+Cu(s)` According to Nernst equation `E=E^(@)-(RT)/(2F)In([Mg^(2+)])/([Cu^(2+)])` `E_(cell)^(@)=2.70 V,E_(cell)=2.67 V,Mg^(2+)=x M, Cu^(2+)=1 M` `2.67=2.70-(Rxx300)/(2F)"In "x` `-0.03=-(Rxx300)/(2F)"In "x or "In "x=((0.03)xx2)/(300)xx(F)/(R)` In `Inx=(0.03xx2xx(11500))/(300)=2.3 or x=10`

Discussion

855.	For the electrochemical cell, `Mg(s)\|Mg^(2+) (aq. 1M)\|\|Cu^(2+) (aq. 1M)\|Cu(s)` the standard emf of the cell is 2.70 V at 300 K. When the concentration of `Mg^(2+)` is chaged to x M, the cell potential changes to 2.67 V at 300 K. The value of x is ________ . (Given `F/R=11500 kV^(-1)`. where F is the Faraday constant and R is the gas constant, ln (10) = 2.30)
Answer» Correct Answer - 10 Cell reaction : `Mg(s)+Cu^(2+) (aq) hArr Mg^(2+) (aq)+Cu(s)" "(n=2)` when the concentration of ionic species is 1 M then `E=E^(@)` `:. E^(@)=2.70 V` Using Nernst equation : `E=E^(@)-(RT)/(nF)"ln" ([Mg^(2+)])/([Cu^(2+)])` `2.67=2.70-(RT)/(2F) log_(e) [x/1]` `ln x=(0.03xx2)/300xxF/R=(0.03xx2xx11500)/(300)` `=2.30=ln 10` `x=10`

Discussion

856.	1.05 g of a lead ore containing impurity of Ag was dissolved in quantity of `HNO_(3)` and the volume was made 350 mL . A Ag electrode was dipped in the solution and `E_(cell)` of `Pt(H_(2))\|H^(+) (1 M)\|\|Ag^(+)\|Ag` was 0.503 V at 298 K. calculate % of lead in the ore. `E_(Ag^(+)//Ag)^(@)=0.80 V`
Answer» Correct Answer - `0.0339 %` `{:("Anode "1/2H_(2)(g) rarr H^(+) (1 M)+e^(-)),("Cathode "Ag^(+)(x)+e^(-) rarrAg(s)),(ulbar(1/2H_(2)(g)+Ag^(+) (x) hArr Ag(s)+H^(+) (1 M))" "(n=1)):}` `E^(@)=0.80-0=0.80` volt `Q=([H^(+)])/([Ag^(+)])=1/x` `E=E^(@)-0.0591/n log_(10) Q` `0.503=0.80-0.0591/1 log_(10) (1/x)` `x=9.43xx10^(-6) M` Number of moles of `Ag^(+)` in 350 mL `=(MV)/1000=(9.43xx10^(-6)xx350)/(1000)` `=3.3xx10^(-6)` Mass of `Ag=3.3xx10^(-6)xx108=3.56xx10^(-4)g` % Ag in the ore`=(3.56xx10^(-4))/1.05xx100=0.0339 %`

Discussion

857.	calculate `E^(@)` of the following half -cell reaction at 298 K: `Ag(NH_(3))_(2)^(+)+e^(-) rarr Ag+2NH_(3)` `{:(Ag^(+)+e^(-) rarr Ag,,E_(Ag^(+)//Ag)^(@)=0.80 V),(Ag(NH_(3))_(2)^(+) hArr Ag^(+)+2NH_(3),,K=6xx10^(-8)):}`
Answer» Correct Answer - 0.373 volt `{:("Anode "Ag(s) rarr Ag^(+)+e^(-)," "E^(@)=0.80" volt"),("Cathode "[Ag(NH_(3))_(2)]^(+)+e^(-) rarr Ag(s)+2NH_(3)," "E^(@)=V" volt"),(ulbar(Ag(s)+[Ag(NH_(3))_(2)]^(+) hArr Ag(s)+Ag^(+)+2NH_(3))):}` `Q=([Ag^(+)][NH_(3)]^(2))/([Ag(NH_(3))_(2)]^(+))=6xx10^(-8) (n=1) E_(cell)^(@)=(V-0.80)` At equilibrium, `E=0` `:. E=E^(@)-0.0591/n log Q` `0=(V-0.80)-0.0591/1 log (6xx10^(-8))` `V=0.373` volt

Discussion

858.	The oxidation potential of hydrogen electrode `H_(2)//H_(2)O^(+)` (aq) will be greater than zero if,A. conc. Of `[H_(3)O^(+)]` ions is 2MB. conc. Of ` [H_(2)O^(+)]` ions is 1MC. Partial pressure of `H_(2)` is 2atm.D. `E_("oxi")` can never be +ve.
Answer» Correct Answer - C Apply Nernst equation to gt answer.

Discussion

859.	Calculate the emf for the following cell at 298 K. `Cd//Cd^(2+)(0.1 M) \|\| Ag^(+)(0.1 M)//Ag` `"Given"E_(Cd^(2+)//Cd)^(@)=-0.40" V ",E_(Ag^(+)//Ag)^(@)=0.80" V "`
Answer» Correct Answer - 1.17 V Cell reaction : `Cd(s)+2Ag^(+)(0.1" M") to Cd^(2+)(0.1" M")+2Ag(s)` `E_(cell)=E_(cell)^(@)-(0.0591)/(n)"log"([Cd^(2+)])/([Ag^(+)]^(2))=[(0.8)-(-0.4)]-(0.0591)/(2)"log"(0.1)/((0.1)^(2))` `=1.20-0.02955=1.17" V "`

Discussion

860.	When the cell reaction attains a state of equilibrium, the EMF of the cell isA. zeroB. positiveC. negativeD. not definite.
Answer» Correct Answer - A When a cell reaction attains equilibrium EMF of the cell becomes zero.

Discussion

861.	Explain why zinc dissolves in dilute HCl to liberate `H_(2)`(g) but from concentrated `H_(2)SO_(4)`, the gas evolved is `SO_(2)`.
Answer» In dilute HCl, zinc reacts as follows : `Zn+2H^(+)(aq) to Zn^(2+)(aq)+H_(2)(g)` Therefore, `H_(2)`(g) ion. Is liberated. With concetration `H_(2)SO_(4)`, both `H^(+)` ions and `SO_(4)^(2-)` ions will be available. But `SO_(4)^(2-)` ions is a better oxidising agent than`H^(+)` ion. Therefore, the reaction will occur as follows : `Zn(s) +SO_(4)^(2-)(aq)+4H^(+)(aq) to Zn^(2+)(aq)`

Discussion

862.	The EMF of a cell is related to the equilibrium constant of the cell reaction asA. In `k_(c)=(nFE_("cell")^(@))/(RT)`B. `k_(c)=(nFE_("cell")^(@))/(RT)`C. `E_("cell")^(@)=(RT)/(nF)` in `k_(c)`D. `k_(c)=(RT)/(nF)` in `E_("cell")^(@)`
Answer» Correct Answer - A `E_("cell")^(@)=-nFE^(@)`

Discussion

863.	Twoelectrodes are fitted in conductance cell 1.5 cm apart while the area of cross-section of each electrode is `0.75 cm^(2)`. The cell constant isA. `1.125 cm^(-1)`B. `0.5 cm^(-1)`C. `2.0 cm^(-1)`D. `0.2 cm^(-1)`
Answer» Correct Answer - C Cell constant (x) `=l/a=1.5/0.75=2 cm^(-1)`

Discussion

864.	For a reaction `A(s)+2B^(o+) rarr A^(2+)+2B` `K_(c)` has been found to be `10^(12)`. The `E^(c-)._(cell)` isA. 0.354VB. 0.708VC. 0 .0098VD. 1.36V
Answer» Correct Answer - A `DeltaG=-RT` in `k_(c)`

Discussion

865.	The two `Pt` electrodes fitted in a conductance cell are `1.5 cm` apart while the cross`-` sectional area of each electrode is `0.75cm`. What is the cell constant?A. `1.125`B. `0.5cm`C. `2.0cm^(-1)`D. `0.2cm^(-1)`
Answer» Correct Answer - c `G^(**)=(l)/(a)=(1.5cm)/(0.75cm^(2))=2.0cm^(-1)`

Discussion

866.	For a reaction `A(s)+2B^(+) (aq) rarr A^(2+) (aq) rarr A^(2+)(aq)+2B, K_(C)` has been found to be `10^(12)`. The `E_("cell")^(@)` isA. 0.354 VB. 0.708 VC. 0.0098 VD. 1.36 V
Answer» Correct Answer - A `E_("cell")^(@) = (0.0592)/(n) log K_(c)` `therefore E_(cell)^(@) = (0.0592)/(2) log 10^(12) (0.0592 xx 12)/(2) = 0.354` V

Discussion

867.	Which of the following statement is false for fuel cell?A. They are more efficientB. They are free from pollutionC. They run till reactants are activeD. Fuel burned with `O_2`
Answer» Correct Answer - D

Discussion

868.	The metal that forms a self-protecting film of oxide to prevent corrosion is:A. NaB. AlC. CuD. Au
Answer» Correct Answer - B

Discussion

869.	The metal that forms a self protecting film of oxide to prevent corrosion, isA. CuB. AlC. NaD. Au
Answer» Correct Answer - B Aluminium forms a self protecting film of oxide to prevent corrosion.

Discussion

870.	One of the following is false for HgA. it can evolve hydrogen from `H_(2)S`B. It is a metalC. It has high specific heatD. It is less reactive than hydrogen
Answer» Correct Answer - A It cannot evolve `H_(2)` from `H_(2)S` `Hg+H_(2)Sto"No reaction"`.

Discussion

871.	Give the cell reaction of nickel-cadimuim secondary battery.
Answer» The cell reaction of nickel-cadmium secondary battery is `Cd _((s))+ 2Ni(OH)_(3(s))to CdO_((s))+2Ni (OH)_(2(s))+H_(2)O_((l))`

Discussion

872.	Give one example for a secondary battery. Give the cell reaction.
Answer» Lead storage battery is an example of secondary battery. The cell reaction when the battery is in use are `Pb_((s))+SO_(4(aq))^(-2)to Pb SO_(4(s))+2e^(-)` (Anode) `PbO_(2(s))+SO_(4(aq))+2H_(2) SO_(4(aq))to 2 P bSO_(4(s))+2H_(2) O_((l))` (Cathode) Overall call reaction is `Pb_((s))+PbO_(2(s))+2H_(2)SO_(4(s))+2 Ni(OH)_(2(s))+H_(2)O_ ((l))`

Discussion

873.	For a reaction,`A +B to` Product : the rate law is given by `r =k [A]^(1//2)[B]^(2)` What is the order of the reaction ?
Answer» `A+B to ` product `r=k [A]^(1//2)[B]^(2)` Rate of the reaction `r=1/2 +2=2.5`

Discussion

874.	The convertion of molecules X to Y follows second order kinetics. If concentration of X is increased by three times, how will it affect the rate of formation of Y.
Answer» `x to Y` It is a second order reaction `r prop [X]^(2)` `If x=1" "r=1` `If x=3" "r=3^(2)=9` The rate of formation of y increases by 9 times.

Discussion

875.	A reaction is `50%` completed in 2 hours and `75%` conmpleted in 4 hours. What is the order of the reaction.
Answer» Given theat a reaction is `50%` completeed in 2 hrs. `75%` completed in 4 hrs. From the data half life is independent of initial concentration so it a first order reaction.

Discussion

876.	How much quantity of electricity has to be passed through 200ml of 0.5 M `CuSO_(4)` solution to completely deposit copper?A. 96500CB. `2xx9650C`C. `2xx96500C`D. `4xx96500C`
Answer» Correct Answer - B Moles of `CuSO_(4)+200xx0.5xx10^(-3)=0.1` mol `Cu^(2+)+2e^(-)rarrCu` Charge required to convert 0.1 mol of `Cu^(2+)` to `Cu=2xx96500xx10^(-3)` amp. =0.1amp

Discussion

877.	In Question `pH` of the final solution will beA. 12B. 2C. `11.7`D. 3
Answer» Correct Answer - a `[overset(c-)(O)H]=0.01=10^(-2)` `pOH=2,pH=14-2=12`

Discussion

878.	A current strength of `1.0 A` is passed for `96.5s` through `200mL` of a solution of `0.05 M KCl` . Find `a.` The amoudn of gases produced `b.` The concentration of final solution `w.r.t. overset(c-)(O)H` ions `c. pH` of the solution.
Answer» `a." "Cl^(c-)rarr (1)/(2)Cl_(2)+e^(-)" "(`Anode`)` `H_(2)O+e^(-)rarr2H_(2)+overset(c-)(O)H." ".(`Cathode`)` Electricity passed `=(96.5xx1)/(96500)=(10^(-3))/(2)mol `of `e^(-)` `:.10^(-3) mol ` of `e^(-)` produces `(10^(-3))/(2) mol` of `Cl_(2)` `=(10^(-3))/(2)xx71=35.5xx10^(-3)g` Also, `10^(-3)mol `of `e^(-)` produces `=(10^(-3))/(2) mol` of `H_(2)` `=(10_(3))/(2)xx2=10^(-3)g` Total weight of gases `=10^(-3)xx35.5+10^(-3)=0.0365g` `b.` `[overset(x-)(O)H]=(10^(-3)mol)/(200mL(volume))=(1mmol)/(200mL)=0.005M` `c. pOH=-log[0.005]=2.3impliespH=14-2.3=11.7`

Discussion

879.	A current strength of `1.0A` is passed for `96.5 s` through `100mL` of a solution of `0.05 M KCl`. The concentration of the final solution with respect to `overset(c-)(O)H` ions isA. `0.005M`B. `0.05M`C. `0.01M`D. `0.001M`
Answer» Correct Answer - c Aqueous `KCl` on electrolsysi gives `Cl_(20` at anode and `H_(2)` at cathode and `overset(c-)(O)H` ions in the solution. Equivalent of current `=(1xx96.5)/(96500)=0.001Eq-=1mEq` `:. [overset(c-)(O)H]=(1mEq)/(100mL)=0.01M`

Discussion

880.	A current of `1.0A` is passed for `96.5s` trhough a `200mL` solution of `0.05 M LiCl` solution. Find `a.` The volume of gases produced at STP `b.` The `pH` of solution at the end of electrolysis
Answer» Number of faradays passed `=(It)/(96500)=(1.0xx96.5)/(96500)=10^(-3)F` Cathode`: 2H^(o+)+2e^(-) rarr H_(2) ` `(Li^(o+)` will remain in solution `)` `2F-=1 mol `of `H_(2)` or `1xx10^(-3)F-=0.5xx10^(-3)` mole of `H_(2)` `-=0.5xx10^(-3)xx22400mL Cl_(2)` at `STP` `=11.2mL Cl_(2)` at STP In solution, `Li^(o+)` and `overset(c-)(O)H` are left. To calculate the `pH` of solution, first calculate the millimoles of `H^(o+)` ions electrolyzed. `implies` mmoles `H^(o+)` ions electrolyzed `=mEq` of `H^(o+)` ions electrolyzed `=` Number of faradays passed `=10^(-3)F` Since `H_(2)O` produces equal number of `H^(o+)` and `overset(c-)(O)H` ions,mmoles `overset(c-)(O)H` ions left in excess `=10^(-3)` `implies [overset(c-)(O)H]~~(10^(-3))/(200//1000)=5xx10^(-3)M` `[` Neglect `overset(c-)(O)H` from dissociation of `H_(2)O]` `implies pOH=-log (5xx10^(3))=3-log 5=2.3` `impliespH=14-pOH=14-2.3=11.7`

Discussion

881.	If the temperature coefficient of `EMF `if `-0.125V K^(-1), DeltaS` for the given cell at `25^(@)C` is `:` `Fe\|Fe^(2+)(aq)\|\|Cd^(2+)(aq)\|Cd`A. `-26.125 k J K^(-1)`B. `-24.125k J K^(-1)`C. `-22.125 kJ K^(-1)`D. `-20.125 k J K^(-1)`
Answer» Correct Answer - b `n_(cell)=2,((delE)/(deltT))=-0.125 VK ^(-1)` `DeltaS=nF((delE)/(delT))` `=2xx96500Cxx-0.125 V K^(-1)` `=-24125 J K^(-1)` `=-241.25 kJ K^(-1)`

Discussion

882.	Which of the following is `(` are `)` function `(s)` of salt bridge ?A. It completes the electrical circuit with electrons flowing from one electrode to other through wires and flow of ions between the two compartments through salt bridge.B. It prevents the accumulation of the ions.C. Both `(a)` and `(b)`D. None of the above.
Answer» Correct Answer - C Factual statement

Discussion

883.	`Cu^(2+)+2e^(-) rarr Cu.` On increasing `[Cu^(2+)]`, electrode potentialA. IncreasesB. DecreasesC. No changeD. First increases, then decreases
Answer» Correct Answer - a `Cu^(2+)(aq)+2e^(-)rarrCu(s)` ltbr. `E_(Cu^(2+)\|Cu)=E^(c-)._(Cu^(2+)\|Cu)-(0.059)/(2)log .(1)/([Cu^(2+)])` `=E^(c-)._(Cu^(2+)\|Cu)+(0.059)/(2)log.[Cu^(2+)]` If `[Cu^(2+)]` increases, `E_(Cu^(2+)\|Cu)` also increases.

Discussion

884.	Consider the following `E^o` values : `E^o _Fe^(3+)//FE^(2+)o = + 0.77 V` `E_(Sn^(2+)//Sn) =- 0.14 V` Under standard conditions the potential for reaction `Sn(s) +2Fe^(3+) (aq) rarr 2Fe^(2+) (sq) + Sn^(2+) (aq)` is.A. ` 0.63 V`B. ` 1.40 V`C. ` 1.68 V`D. ` 0.91 V`
Answer» Correct Answer - D For ` Sn_(s) + 2Fe_(aq)^(3+) rarr 2 Fe_(aq)^(2+) +Sn_(aq)^(2+)` `E_("cell")^@ =E_(Sn//Sn^(2+))^@ + E_(Fe^(3+)//Fe^(2+))^@ =(0.14) + (0.77)` `E_("cell")^@ =0. 91 V`.

Discussion

885.	Consider the following ` E^@` values . `E_(Fe^(3+)//Fe^(2+)^@ = + 0.77 V`, `E_(Sn^(2+)//Sn)^@=- 15. V` The `E_(cell)^@` for the reaction , `Sn (s) + 2Fe_(aq)^(3+) rarr 2 Fe_(aq.)^(2+) + Sn_(aq.)^(2+) is :A. `-0.58 V`B. `-0.30 V`C. `+0.30V`D. `+0.58 V`
Answer» Correct Answer - c At anode `: Fe(s) rarr Fe^(2+)(aq)+2e^(-)` `(` hald oxidation `)` At cathode `: Sn^(2+)(aq)+2e^(-) rarr Sn(s)` `(` half reduction `)` `ulbar(Cell reaction : Fe(s)+Sn^(2+)(aq)rarr Fe^(2+)(aq)+Sn(s))` `E^(c-)._(cell)=(E^(c-)._(red))_(c)-(E^(c-)._(red))_(a)` `=(-0.14)-(0.44)=0.30V.`

Discussion

886.	`Sn^(2+)` and `Fe^(3+)` cannot exist in the same solution.
Answer» Correct Answer - T Because redox reaction will occur. `Sn^(2+)+2Fe^(3+) rarr Sn^(4+)+2Fe^(2+)`

Discussion

887.	Metals always liberate `H_(2)(g)` from acids.
Answer» Correct Answer - F Only those metals which are above hydrogen in electrochemical series liberate `H_(2)(g)`.

Discussion

888.	The passage of current liberates `H_2` at cathode and `Cl_2` at anode. The solution isA. copper chloride in waterB. NaCl in waterC. `H_2SO_4`D. Water
Answer» Correct Answer - B Since discharge potential of watier is greater than that of sodium so water is reduced at cathode instead of `Na^(+)`. Cathode: `H_(2)O+e^(-)to(1)/(2)H_(2)+OH^(-)` Anode: `Cl^(-)to(1)/(2)Cl_(2)+e^(-)`.

Discussion

889.	Which of the following compounds will not undergo decomposition on passing electricity through aqueous solutionA. SugarB. Sodium chlorideC. Sodium BromideD. Sodium Acetate
Answer» Correct Answer - A Because in it covalent bonding is present.

Discussion

890.	The same quantity of electricity is passed through `Al_(2)(SO_(4))_(3)` and `AgNO_(3)` solution with platinum electrodes. If `n` number of `Al` atoms are deposited on the cathode, `3n` number of `Ag` atoms will be deposited on the cathode.
Answer» Correct Answer - T `1F=96500C=1 Eq of Al^(3+)=1Eq of Ag^(o+)` `=(1)/(3) mol of Al^(3+)-=(1)/(1) mol of Al^(o+)` `=1 mol of Al^(3+)=3 mol of Ag^(o+)` `=n ` atoms of `Al^(3+) = 3n` atoms of `Ag^(o+)`

Discussion

891.	How many coulombs are required for the oxidation of `1 mol` of `H_(2)O` to `O_(2)`?A. `93000C`B. `1.93xx10^(5)C`C. `9.65xx10^(40C`D. `19.3xx10^(2)C`
Answer» Correct Answer - b `H_(2)Orarr 2H^(+)+2e^(-)+(1)/(2)O_(2)` `2F-=1 mol of H_(2)O=(1)/(2) mol of O_(2)` `:. 2xx96500C=1.93 xx 10^(5)C`

Discussion

892.	On electrolysis of a solution of dilute `H_(2)SO_(4)` between platinum electrodes , the gas evolved at the anoe isA. `SO_(2)`B. `SO_(3)`C. `O_(2)`D. `H_(2)`
Answer» Correct Answer - c Oxidation potential of `H_(2)Ogt` Oxidation potential of `SO_(4)^(2-)` So oxidation of `H_(2)O` occurs at anode and gives `O_(2)(g)`.

Discussion

893.	On electrolysis of a solution of dilute `H_(2)SO_(4)` between platinum electrodes , the gas evolved at the anoe is
Answer» Correct Answer - C In between dilute `H_(2)SO_(4)` and platinum electrode `O_(2)` gas evolve at anode.

Discussion

894.	In electrolysis of dilute `H_(2)SO_(4)` using platinum electrodesA. `H_(2)` is evolved at cathodeB. `NH_(3)` is produced at anodeC. `Cl_(2)` is obtained at cathodeD. `O_(2)` is produced
Answer» Correct Answer - A When platinum electrodes are dipped in dilute solution `H_(2)SO_(4)` than `H_(2)` is evolved at cathode.

Discussion

895.	The standard reduction potential at 298K for the following half reaction are given against each `Zn^(2+)(aq)+2e^(-)hArrZn(s)" "-0.762` `Cr^(3+)(aq)+2e^(-)hArrCr(s)" "-0.740` `2H^(+)(aq)+2e^(-)hArrH_(2)(g)" "0.000` `Fe^(3+)(aq)+2e^(-)hArrFe^(2+)(aq)" "0.770` which is the strongest reducing agentA. Zn(s)B. Cr(s)C. `H_(2)(g)`D. `Fe^(2+)(aq)`
Answer» Correct Answer - A More negative is the reduction potential, higher will be the reducing property, i.e., the power to give up electrons.

Discussion

896.	An electrolytic cell contains a solution of `Ag_(2)SO_(4)` and have platinum electrodes. A current is passed until 1.6gm of `O_(2)` has been liberated at anode. The amount of silver deposited at cathode would beA. 107.88 gmB. 1.6 gmC. 0.8 gmD. 21.60 gm
Answer» Correct Answer - D At cathode: `Ag^(+)+e^(-)toAg` At anode: `2OH^(-)tH_(2)O+(1)/(2)O_(2)+2e^(-)` `E_(Ag)=(108)/(1)=108,E_(O_(2))=((1)/(2)xx32)/(2)=8` `(W_(Ag))/(E_(Ag))=(W_(O_(2)))/(E_(O_(2))),W_(Ag)=(1.6xx108)/(8)=21.6gm`

Discussion

897.	How many moles of electrons are required for the reduction of :(i) 3 moles of Zn2+ to Zn, (ii) 1 mol of Cr3+ to Cr?How many Faradays of electricity will be required in each case?
Answer» (i) Given : For reduction of 3 mol Zn²⁺ to Zn; Number of moles of electrons required = ? Reduction half reaction, Zn²⁺ + 2e^- → Zn ∵ 1 mole of Zn²⁺requires 2 moles of electrons ∴ 3 moles of Zn²⁺ will require, ∵ 3 × 2 = 6 moles of electrons ∴ 1 mole of electrons = 1 F 6 moles of electrons = 6 F (ii) Given : Reduction of 1 mol of Cr³⁺ to Cr : Reduction half reaction, Cr³⁺ + 3e^-→ Cr Hence 1 mole of Cr³⁺ will require 3 moles of electrons ∵ 1 mole of electrons = 1 ∴ 3 moles of electrons = 3 F ∴ (i) 6 mol electrons and 6 Faradays. (ii) 3 mol electrons and 3 Faradays.

897.

How many moles of electrons are required for the reduction of :(i) 3 moles of Zn2+ to Zn, (ii) 1 mol of Cr3+ to Cr?How many Faradays of electricity will be required in each case?

Answer»

(i) Given :

For reduction of 3 mol Zn²⁺ to Zn;

Number of moles of electrons required = ?

Reduction half reaction,

Zn²⁺ + 2e^- → Zn

∵ 1 mole of Zn²⁺requires 2 moles of electrons

∴ 3 moles of Zn²⁺ will require,

∵ 3 × 2 = 6 moles of electrons

∴ 1 mole of electrons = 1 F 6 moles of electrons = 6 F

(ii) Given :

Reduction of 1 mol of Cr³⁺ to Cr :

Reduction half reaction,

Cr³⁺ + 3e^-→ Cr

Hence 1 mole of Cr³⁺ will require 3 moles of electrons

∵ 1 mole of electrons = 1

∴ 3 moles of electrons = 3 F

∴ (i) 6 mol electrons and 6 Faradays.

(ii) 3 mol electrons and 3 Faradays.

Discussion

898.	The valency of the metal , if its at.mass and ECE respectively are 63.56 and `3.29 xx 10^(-4)` isA. 1B. 2C. 3D. 4
Answer» Correct Answer - B Electro chemical equivalent = `("Eq.mass")/(96,500)` Eq. wt = `("At.mass")/("Valency (n)")` `3.29 xx 10^(-4) = (63.56)/(3.24 xx 10^(-4) xx 96,500) = 2.002 = 2`

Discussion

899.	A current of 6 amperes is passed through AlCl3 solution for 15 minutes using Pt electrodes, when 0.504 g Al is produced. What is the molar mass of Al ?
Answer» Given : Electric current = I = 6 A Time = t = 15 min = 15 × 60 s = 900 s Mass of Al produced = 0.504 g Molar mass of Al = ? Reduction half reaction, \(Al^{3+}_{(aq)}\) + 3e^- ⟶Al_(aq) Quantity of electricity passed = Q = I × t = 6 × 900 = 5400 C Number of moles of electrons = \(\frac{Q}{F}\) = \(\frac{5400}{96500}\) = 0.05596 mol From half reaction, ∵ 3 moles of electrons deposit 1 mole Al ∴ 0.05596 moles of electrons will deposit, \(\frac{0.05596}{3}\) = 0.01865 mol Al Now, ∵ 0.01865 mole Al weighs 0.504 g ∴ 1 mole Al will weigh, \(\frac{0.504}{0.01865}\) = 27 g Hence molar mass of Al is 27 g mol^-1 ∴ Molar mass of Al = 27 g mol^-1

899.

A current of 6 amperes is passed through AlCl3 solution for 15 minutes using Pt electrodes, when 0.504 g Al is produced. What is the molar mass of Al ?

Answer»

Given :

Electric current = I = 6 A

Time = t = 15 min = 15 × 60 s = 900 s

Mass of Al produced = 0.504 g

Molar mass of Al = ?

Reduction half reaction,

\(Al^{3+}_{(aq)}\) + 3e^- ⟶Al_(aq)

Quantity of electricity passed = Q = I × t

= 6 × 900

= 5400 C

Number of moles of electrons = \(\frac{Q}{F}\) = \(\frac{5400}{96500}\)

= 0.05596 mol

From half reaction,

∵ 3 moles of electrons deposit 1 mole Al

∴ 0.05596 moles of electrons will deposit,

\(\frac{0.05596}{3}\)

= 0.01865 mol Al

Now,

∵ 0.01865 mole Al weighs 0.504 g

∴ 1 mole Al will weigh,

\(\frac{0.504}{0.01865}\) = 27 g

Hence molar mass of Al is 27 g mol^-1

∴ Molar mass of Al = 27 g mol^-1

Discussion

900.	The same quantity of electricity that liberated 2.158 g silver was passed through a solution of a gold salt and 1.314 g of gold was deposited. The equivalent mass of silver is 107.9. Calculate the equivalent mass of gold. What is the oxidation state of gold salt ? (At. mass of gold =197)A. 16 AB. 26.5 AC. 56 AD. 100 A
Answer» Correct Answer - A i= ? W = 39.49 , t = 3600 sec . E for `Au^(3+) = (197)/(3) = 65.66` `W = (E)/(96,500) xx ixxt` `i= (W xx 96,500)/(Exxt) = (39.4 xx 96,500)/((197)/(3) xx 3,600) = 16 A`

Discussion

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