3195 + Interview Questions in ELECTROCHEMISTRY IN GENERAL AWARENESS Page 10 InterviewSolution

451.	The most durable metal plating on iron to protect it againt corrosion isA. nickel platingB. copper platingC. tin platingD. zinc plating
Answer» Correct Answer - D This is because zinc has higher oxidation potential than Ni, Cu and Sn.

Discussion

452.	Arrange the following in decreasing order of their reactivity:Al,Cu, Fe,Mg and Zn.
Answer» Mg,Al,Zn Fe, Cu.

Discussion

453.	Specific conductance of a decinormal solution of KCl is 0.0112 `ohm^(-1) cm^(-1)`. The resistance of a cell containing the solution was found to be 56. What is the cell constant ?
Answer» We know that, Sp. Conductance= Cell constant `xx` Conductance or Cell constant `=("Sp. Conductance")/("Conductance")` =Sp. Conductance `xx` Resistance `=0.0112xx56` `=0.6272 cm^(-1)`

Discussion

454.	`E^(@)` for `F_(@)+2e hArr2F^(-)` is `2.7 V` Thus, `E^(@)` for `F^(-) hArr(1)/(2)Fe +e` is :A. `1.35 V`B. `-1.35 V`C. `-2.7 V`D. `2.7 V`
Answer» Correct Answer - C `E_(OP)^(@) = -E_(RP)^(@)`

Discussion

455.	Silver is electro-deposited on a metallic vessel of surface area 800 `cm^(2)` by passing a current of 0.2 ampere for 3 hours. Calculate the thickness of silver deposited given that its density is 10.47 g `cm^(-3)`. (At mass of Ag =107.92).
Answer» Correct Answer - `2.88xx10^(4) cm` The charge ,Q on n moles of electrons is given by Q=nF where Faraday constant, `" " F=96500" C" mol^(-1)` Thus, quantity of electricity required to deposit 1 mole of Ag, `Q=1" mol"xx96500"C mol"^(-1)=96500" C " =9365xx10^(4)" C " `. Quantity of electricity actually used used`="Current in amperes"xx "Time in seconds"` `=0.2Axx(3xx60xx60)s` `=0.2xx3xx60xx60" As" =2160" C " " "(because As=C)` Molar mass fo Ag`=107.92" g mol^(-1)` `9.65xx10^(4)C` of charge produce silver=107.92 g `:.` 2160 C of charge produce silver`=((107.92 g))/((9.65xx10^(4)C))xx(2160" C")=2.4156 g` Volume of silver deposited`=("Mass")/("Density")=((2.4156 g))/((10.47 g cm^(-3)))=0.2307 cm^(3)` Thickness of silver deposited`=("Volume")/("Surface area")=((0.2307 cm^(3)))/((800 cm^(2)))=2.88xx10^(-4) cm`.

Discussion

456.	Calculate the number of coulombs required to deposit 5.4 g of Al when the electrode reaction is `Al^(3+)+3e^(-) to Al` (Given , atomic mass of Al =27 g `mol^(-1), F=96500 C mol^(-1)`)
Answer» Correct Answer - `5.79xx10^(4) C` `Al^(3+)(aq)+underset(3xx96500C)(3e^(-)) to underset(27 g)(Al(s))` To deposit 27 g of Al(s), the amount of charge required `=3xx96500" C"` To deposit, 5.4 g of Al (s), the amount of charge required`=((3xx96500C))/((27 g))xx(5.4 g)=57900C=5.79xx10^(4)C`.

Discussion

457.	Calculate the number of coulombs required to deposit 40.5 g of Al when the electrode reaction is , `Al^(3+)+3e^(-)rarr Al`
Answer» Correct Answer - `4.342xx10^(5) C` `Al^(3+)(aq)+underset(3xx96500C)(3e^(-)) to underset(27 g)(Al(s))` To deposit 27 g of Al(s), the amount of charge required `=3xx96500" C"` To deposit, 40.5 g of Al (s), the amount of charge required`=((3xx96500C))/((27 g))xx(40.5 g)=4.342xx10^(5)C`.

Discussion

458.	The number of coulombs required to deposit 5.4g of Al when the electrode reaction is `Al^(3+)+3e^(-)rarrAl`A. `1.83xx10^(5)C`B. `57900C`C. `5.86xx10^(5)C`D. None of the above
Answer» Correct Answer - B 27g of Al is deposited by `=3xx96500C` `5.4g` of Al is deposited by `=(3xx96500xx5.4)/(27)=57900C`

Discussion

459.	The rate constant for the decomposition of hydrocarbons is `2.418 xx10^(-5)s^(-1)` at 546 K. If the energy of activatin is `179.9 kJ//mol.` What will be the value of per-exponential factor ?
Answer» According to Arrhenius equation, `log k = log A -(E_(a))/(2.303RT)` `k =2.418 xx10^(-5)s^(-1)` `E _(a) =179.9 KJmol ^(-1) or 179900 J mol ^(-1)` `R= 8.314 JK ^(-1)mol ^(-1)` `T=546K` `log A =log k + (E_(a))/(2.303RT)` `=log (2.418 xx10^(-5)s^(-1))+ (179900 J mol ^(-1))/(2.303 xx(8.314JK^(-1))xx546k)` `=-4.6184+17.21 =12.5916` `A="Antilog" 12.5916=3.9xx10^(12)s^(-1)` `A=3.9xx10^(12)s^(-1)`

Discussion

460.	The electrode pptenticals for ` Cu^(2+) (aq) +e^(-) rarr Cu^+ (aq) ` and ` Cu^+ (aq) + e^- rarr Cu (s)` are `+ 0.15 V` and ` +0. 50 V` repectively. The value of `E_(cu^(2+)//Cu)^@` will be.A. `0.500" V "`B. `0.325" V "`C. `0.650" V "`D. `0.150" V "`
Answer» Correct Answer - B (b) `Cu^(2+)(aq)+e^(-) to Cu^(+)(aq),E_(1)^(@)=0.15" V "` `Cu^(+)(aq)+e^(-) to Cu(s),E_(2)^(@)=0.50" V "` `Cu_((aq))^(2+)+2e^(-) to Cu(s)` `DeltaG^(@)=DeltaG_(1)^(@)+DeltaG_(2)^(@)` `-nFE^(@)=-n_(1)FE_(1)^(@)-n_(2)FE_(2)^(@)` or `E^(@)=(n_(1)E_(1)^(@)+n_(2)E_(2)^(@))/(n)=(1xx0.15+1xx(0.50))/(2)` `=(0.15+0.50)/(2)=(0.65)/(2)` `=0.325" V "`

Discussion

461.	The electrode pptenticals for ` Cu^(2+) (aq) +e^(-) rarr Cu^+ (aq) ` and ` Cu^+ (aq) + e^- rarr Cu (s)` are `+ 0.15 V` and ` +0. 50 V` repectively. The value of `E_(cu^(2+)//Cu)^@` will be.A. 0.500 VB. 0.325 VC. 0.650 VD. 0.150 V
Answer» Correct Answer - B `Cu_((aq))^(2+) + e^(-) to Cu_((aq))^(+) , E_(1)^(@) = 0.15 V` `Cu_((aq))^(+) + e^(-) to Cu_((s)) , E_(2)^(@) = 0.50 V` `Cu^(2+) + 2e^(-) to Cu` Now , `Delta G^(@) = Delta G_(1)^(@) + Delta G_(2)^(@)` or , `- n FE^(@) =-n_(1) FE_(1)^(@) - n_(2) FE_(2)^(@)` or , `E^(@) = (n_(1) E_(1)^(@) + n_(2) E_(2)^(@))/(n) = (1 xx 0.15+ 1 xx 0.50)/(2) = 0.325` V

Discussion

462.	The electrode pptenticals for ` Cu^(2+) (aq) +e^(-) rarr Cu^+ (aq) ` and ` Cu^+ (aq) + e^- rarr Cu (s)` are `+ 0.15 V` and ` +0. 50 V` repectively. The value of `E_(cu^(2+)//Cu)^@` will be.A. ` 0.150 V`B. ` 0.325 V`C. ` 0. 500 V`D. ` 0. 650 V`
Answer» Correct Answer - B ` Cu_(aq)^(2+) + 2e rarr Cu_(aq)^(+) , Delta G_(1)^(@) =- 9 1 xx 0. 15 xx F]` ` Cu_(aq)^+ +e rarr Cu , Delta G_2^@ =- [1 xx 0. 50 xx F]` On adding `:. Cu_(aq)^(2+) = 2e rarr Cu, Delta G_3^@ =- [2 xx E_3^@ xx F]` `:. 2 E_3^@ = 0.56 V` or `E_3^2 = 0. 325 V`

Discussion

463.	Why alternating current is sued in place of direct current in measuring the electrolytic conductance?
Answer» Direct current results in the electrolysis of the electrolytic solution.

Discussion

464.	Which of the following electrolytic solutions has the least specific conductance?A. 2NB. 0.002NC. 0.02ND. 0.2N
Answer» Correct Answer - B Specific conductance is the conductance of ions per `cm^(3)` of the solution. No. of ions/cc will b e minimum in case of 0.002 N solution. Hence its specific conductance will be minimum (Specific conductance decrease with dilution because number of ions/`cm^(3)` decrease)

Discussion

465.	The electrode pptenticals for ` Cu^(2+) (aq) +e^(-) rarr Cu^+ (aq) ` and ` Cu^+ (aq) + e^- rarr Cu (s)` are `+ 0.15 V` and ` +0. 50 V` repectively. The value of `E_(cu^(2+)//Cu)^@` will be.A. 0.325VB. 0.650VC. 0.150 VD. 0.500 V
Answer» Correct Answer - A (a) `Cu^(2+)+e^(-) to Cu^(+), E_(1)^(@)=0.1 5 V, DeltaG_(1)^(@)=-n_(1)E_(1)^(@)F` `Cu^(+)+e^(-) to Cu, E_(2)^(@)=0.50 V, DeltaG_(2)^(@)=-n_(2)E_(2)^(@)F` `Cu^(2+)+2e^(-) to Cu, E^(@)=?, DeltaG^(@)=-nE^(@)F` ` DeltaG^(@) =DeltaG_(1)^(@)+DeltaG_(2)^(@)` `-nE^(@)F=-n_(1)E_(1)^(@)F-n_(2)E_(2)^(@)F` `or -2E^(@)F=-Fxx0.15+(-1Fxx0.50)` `or -2E^(@)F=-0.15F-0.50F` ` or -2FE^(@)=-F(0.15+0.50)` `:. E^(@)=(0.65)/(2)=0.325V`

Discussion

466.	The electrolytic conductance si a direct measure of .A. ` E_2 = 0 !=E_1`B. ` E_1gtE_2`C. ` E_1lt E_2`D. ` E_1` =E_2`
Answer» Correct Answer - D Because conductance os increase when the dissociation is more.

Discussion

467.	Standard electrode potential for `Sn^(4+)//Sn^(2+)` couple is `0.15 V` and that for the `Cr^(3+)//Cr` couple is `-0.74 V`. These two couples in their standard state are connected to make a cell. The cell potential will beA. `+1.19" V "`B. `+0.89" V "`C. `+0.18" V "`D. `+1.83" V "`
Answer» Correct Answer - B (b) Cell potential is : `E_(cell)=E_(cathode)^(@)-E_(anode)^(@)` `=E_(cathode)^(@)-E_(anode)^(@)` `=0.15-(-0.74)=0.15+0.74=0.89" V "`.

Discussion

468.	Standard electrode potential for `Sn^(4+)//Sn^(2+)` couple is `0.15 V` and that for the `Cr^(3+)//Cr` couple is `-0.74 V`. These two couples in their standard state are connected to make a cell. The cell potential will beA. `+0.89 V`B. `+0.18 V`C. `+1.83 V`D. `+1.199 V`
Answer» Correct Answer - A (a) `E_(Sn^(4+)//Sn^(2+))^(@)=+0.15V` `E_(Cr^(3+)//Cr)^(@)=-0.74V` `E_(cell)^(@)=E_("cathode (RP)")^(@)-E_("anode (RP)")^(@)` `=0.15-(-0.74)` `=+0.89 V`

Discussion

469.	On the bassis of the information available from the reaction ` 4/3 Al + O_2 rarr 2/3 Al_2 O_3 . Delta G =- 82 7 k J "mol"^(-1)` of `O_2` the minimum emf required to carry out an electorlysis of `Al_2 O_3` is `(F= 96500 C "mol"^(-1))`A. ` 8.56 V`B. ` 2.14 V`C. ` 4.` 28 V`D. ` 6. 42 V`
Answer» Correct Answer - B `V= (82 7 xx 10^3)/( 4 xx 96500) = 2.14 V`

Discussion

470.	Celll equation : `A = 2B^(+) rarr A^(2+) + 2 B` `A^(2+) + 2e rarr A` `E^@ = + 0.34 V` and `log_(10) K= 15. 6 ` at `300 K` for cell reactions Find `E^@` for `B^(+) +e rarr B` Given `[( 2.303 RT)/(nF) =0.059]` at `300 K`.A. ` 0. 80 `B. ` 1.26 `C. ` -0. 64`D. ` +0.94`
Answer» Correct Answer - A `E_(cell)^@ =(0.059)/9 log k` `E_(B^+//B)^@ -E_A^@+ 2//A= (0.59)/2 log` `E_(B^+//M)^@ - 34. = (0.059)/2 xx 15 .6` ` E_(B^+//B)^@ = 0. 80`.

Discussion

471.	How many `kJ` of energy is spent when a currebnt of (4) amp passes for `200` second under a potential of `115 V` ?A. ` 52 k J`B. `72 k J`C. ` 82 k J`D. `92 k J`
Answer» Correct Answer - D ` Delta G=- nFE=-QE` `=- 4 xx200 xx 115 xx 10^(-3) h J` `Delta G=- 92 kJ`

Discussion

472.	On the bassis of standard electrode potential of redox couples given below, find out which of the following is the strongest oxdising afent . `(E^@ value : Fe^(3+)) = + 0. 77 C, I_(2(g)) \| I = + 0/ 54 V`, `Cu^(2+) \| Cu = + 34 V, Ag^(+) \|Ag = + 0. 80 V)`.A. ` Fe^(3+)`B. ` I_2(g)`C. ` Cu^(2+) `D. `Ag^+`
Answer» Correct Answer - D Reduction potential values `prop` strength of oxidising agent Strength of oxidinsing agent `Ag^+ gt I_2 gt Fe^(3+) gt Cu^(2+)`.

Discussion

473.	In which of the following electrochemical cell, overall cell reaction is : ` HgO (s) + H_2 (g) rarr Hg E + H_2O(e )`.A. `Pt \| H_2 (g) \| H^(+) (aq) \|HgO(s) \|Hg\|Hg€ \| Pt`B. `Pt\|H_2(g) \|NaOH (aq) \|HgO (s) \|HgE \|Pt`C. `Pt \| H_2(g) \|H^(+) \|\|NaOH(aq)\| HgO(s) \|Hg € \|Pt`D. ` Pt \|H-2 (g) \|H^(+) \|\|HgO(s) \|Hg€ \|Pt`
Answer» Correct Answer - B ` Pt \| H_2 (g) \| NaOH (aq) \| HgO (s) \| Hg (l) Pt1` ` LHE` reaction : ` H_2 (g) + 2 OH^(-) H_2O (l) _2e` ` rarr Hg (l) 2OH^(-)` Net cell reaction : ` HgO (s) + H_2 (g) rarr Hg (l) +H_2 O`.

Discussion

474.	In a cell that untillizes the reaction , `Zn_((s)) + 2 H_((aq))^+ rarr Zn_((aq.))^(2+) + H_(2 (g) )`, adedition of ` H_2 SO_4` to cathode compartment will :A. Increase the E and shift equilibrium to the rightB. Lower the E and shift equilibrium to the leftC. Lower the E and shift equilibrium to the leftD. Increase the E and shift equilibrium to the right
Answer» Correct Answer - A `Zn_((s))+2H_((aq))^(+)hArrZn_((aq))^(2+)+H_(2(g))` `E_(Cell) = E_(Cell)^@ - (0.95)/2 log. ([Zn^(2+)])/([H^+]^2)` When `H_2 SO_4` is added then `[H^+]` wil increase therefore ` E_(Cell)` will also increase and equilibrium will shift towards right .

Discussion

475.	The mass of copper deposited from a solution of `CuSO_4` by passage of `45 A` current for `965` second is (Mol . Wt. pf Copper `= 63.5` ).A. ` 15. 875 g`B. ` 1. 585 g`C. ` 4825 g`D. ` 96 500 g`
Answer» Correct Answer - B Current `I = 5 A` and time ` T = 965 sec` We know that equivalent weight of copper ` = ("Molecular weight")/("Valancy") = (63.5)/2` and quantitu of electricity passed in coulomb = current xx time ` =5 xxx 965 = 4825 C`. Since `96500` coulombs will deposit `(63.5)/2 g` of copper , therefore ` 4825` coulombs will deposit ` = (63. 5 xx 4825)/(96 500 xx 2) = 1. 5875 g`.

Discussion

476.	The standard emf of a cell having one electron change is found to be `0.591 V` at `25^@ C`, The equilibrium constant of the reaction is :A. ` 1^(-10)`B. `29.5 xx 10^(-2)`C. ` 10`D. `1xx10^(1)`
Answer» Correct Answer - D `DeltaG =- nFE^@` ` DeltaG =- 2. 303 RT log K, nFE^@ = 2.303 RT log K` `log K = (nFE^@)/( 2.303RT) = (2 xx96 500 xx 0. 295)/( 2.303xx 8341 xx 198)` `log K= 9.97 = K= 1 xx 10^(10)`.

Discussion

477.	The standard emf of a cell having one electron change is found to be `0.591 V` at `25^@ C`, The equilibrium constant of the reaction is :A. `1.0 xx 10^(30)`B. `1.0xx10^5`C. `1.0 xx 10^(10)`D. `1.0 xx 10^(1)`
Answer» Correct Answer - C ` E_("cell")^@ = (0059)/1 log K_C` ` 0.591= (0.059)/1 log K_C` ` :. K_c = 1 xx10^(01)`.

Discussion

478.	The standard emf of a cell having one electron change is found to be `0.591 V` at `25^@ C`, The equilibrium constant of the reaction is :A. `1.0 xx 10^(30)`B. `1.0 xx 10^(5)`C. `1.0 xx 10^(10)`D. `1.0 xx 10^(1)`
Answer» Correct Answer - C `E_(cell)^(@) = (0.059)/(1)log K_(c)` `0.591 = (0.059)/(1)log K_(c)` `:. K_(c) = 1 xx 10^(10)`

Discussion

479.	The standard emf of a cell having one electron change is found to be `0.591 V` at `25^@ C`, The equilibrium constant of the reaction is :A. ` 1.0 xx 10^(10)`B. ` 1.0 xx 10^5`C. ` 1.0 xx 10^(1)`D. ` 1.0 xx 10^(30)`
Answer» Correct Answer - A ` E_(Cell) = E_(Cell)^@ =0 rArr 0=0.591 - (0.0591)/n log K_C` ` log K_c = (0.591xx1)/( 0.0591) = 10 , K_c = `Anti `log 10 =1 xx 10^(10)`.

Discussion

480.	The standard emf of a cell involving one electron change is found to be 0.591 V at `25^(@)C`. The equilibeium constant of the reaction is (`1F=96500 C mol^(-1)`)A. `1.0 xx10^(1)`B. `1.0xx10^(5)`C. `1.0xx10^(10)`D. `1.0xx10^(30)`
Answer» Correct Answer - C Relation between `K_(eq)` and `E_("cell")^(@)` is `E_("cell")^(@)=0.0591/n"log" K_(eq)` (at 298 K) `0591=0.0591/1log K_(eq)` `:. K_(eq)=1xx10^(10)`

Discussion

481.	Assertion`(A): ` The rusting on the surface of iron involves following reaction `:` `a. Fe(s) rarr Fe^(2+)(aq)+2e^(-)` `(` at anodic site `)` `b. O_(2)(g) +4H^(o+)(aq)+4e^(-) rarr 2H_(2)O(l)` `(` at cathodic site `)` `c. 4Fe^(2+)(aq)+O_(2)(g) +4H_(2)O(l) rarr 2Fe_(2)O_(3)(s)+8H^(o+)` `d.Fe_(2)O_(3)(aq)+xH_(2)O(l) rarr Fe_(2)O_(3).xH_(2)O` Reason `(R) : ` Rusting is accelerated in the presence of `NaCl` and `CO_(2)`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 - b

Discussion

482.	Rusting on the surface of iron involves :A. `Fe(s) rarr Fe^(2+) (aq.)+2e^(-)" (at anodic site)"`B. `O_(2)(g)+4H^(+) (aq.)+4e^(-) rarr 2H_(2)O(l)" (at cathodic site)"`C. `4Fe^(2+) (aq.)+O_(2)(g)+4H_(2)O (l) rarr 2Fe_(2)O_(3) (s)+8H^(+)`D. `Fe_(2)O_(3)(s)+ x H_(2)O (l) rarr Fe_(2)O_(3). xH_(2)O`
Answer» Correct Answer - A::B::C::D

Discussion

483.	For the cell reaction, `2Cr^(4+)+Coto2Ce^(3+)+Co^(2+)" "E_((cell))^(@)` is 1.89V and `E_(Co//Co^(2+))^(@)=-0.028`. If `E_(Ce^(4+)//Ce^(3+))^(@)`A. `-1.64V`B. `+1.64V`C. `-2.08V`D. `+2.17V`
Answer» Correct Answer - B In this cell Co is oxidised and it acts as anode and Ce acts as cathode. `E_(Cell)^(@)=E_("cathode")^(@)-E_("anode")^(@)=1.89=E_(Cell)^(@)-(-0.28)` `E_(Cell)^(@)=-1.89-0.28=1.61`volt.

Discussion

484.	Calculate the reduction potential of a half cell consisting of a platinum electrode immersed in `2.0M Fe^(2+)` and `0.02M Fe^(3+)` solution. Given `E_(Fe^(3+)//Fe^(2+))^(@) = 0.771 V`.
Answer» The half cell reaction is : `Fe^(3+)+e rarr Fe^(2+)` (or take `Fe^(2+) rarr Fe^(3+)+e`) Thus `E_(Fe^(3+)//Fe^(2+)) = E_(Fe^(3+)//Fe^(2+))^(@) + (0.059)/(1)"log"([Fe^(3+)])/ ([Fe^(2+)])` `:. E_(Fe^(3+)//Fe^(2+)) = 0.771 + (0.059)/(1)"log"(0.02)/(2.0)` `= 0.771 +(0.059)/(1)"log"10^(-2) = 0.653 V`

Discussion

485.	One litre each of a Buffer solution is taken in two beakers connected through a salt bridge. Platinum electrode is immersed in each beaker and electrysis is carried out for `10,000` sec. using a current of `9.65` Ampere. The Buffer contain `(1)/(3)` mole each of `Na_(2)HPO_(4)` and `NaH_(2)PO_(4)` Initially. using the given data answer the following. Given: ltbr. `pK_(a) (H_(3)PO_(4)) = 4, pK_(a) (H_(2)PO_(4)^(-)) = 7 , pK_(a) (HPO_(4)^(2-)) = 10` `log 2 = 0.3, log3 = 0.48` `2.303 (RT)/(F) = 0.06 at 298K` After electrolysis if `H_(2)` gas is bubbled at 1 bar pressure on each platinum electrode, what is the `EMF` of resulting Galvenic cell.A. `1.09V`B. `0.29V`C. `0.59V`D. `0.39V`
Answer» Correct Answer - C `EMF = E_("cathode") - E_("Anode")` `E_(H^(+)//H_(2)) = E_(H^(+)//H_(2))^(@) -(0.0591)/(2)log.(1)/([H^(+)]^(2))` `E_(H^(+)//H_(2)) =- 0.059 pH` `EMF =- 0.0591 pH_("cathode") -(-0.0591 pH_("anode"))` `= (pH_("anode") -pH_("cathode")) 0.06` `= (11.91 -2.09) 0.06` `= 0.59`

Discussion

486.	20mL of solution of 0.1M acetic acid is divided into two equal parts and kept in two beakers separately. To one beaker 0mL of 0.05M NaOH. Two hydrogen the other 10 mL of 0.025 M NaOH. Two hydrogen electrodes are placed in the two solutions which are linked through a salt-bridge. what would be the measured emf?A. 59mVB. 28mVC. 14mVD. 33mV
Answer» Correct Answer - b

Discussion

487.	Two electrochemical cells are assembled in which the following reactions occur `V^(2+)+VO^(2+)+2H^(+) to 2V^(3+)+H_(2),E_(cell)^(@)=0.616V` `V^(3+)+Ag^(+)+H_(2)toVO^(2+)+2H^(+)+Ag(s),E_(cell)^(@)=0.439V` Calculate the `E^(@)` value for the half-cell reaction `V^(3+)+e^(-)toV^(2+).` (Given: `E_(Ag^(+)//Ag)^(@)=0.799`volt) lt
Answer» Two half cells for the 1st cell reaction will be (i) `V^(2+)toV^(3+)+e^(-),E_(1)^(@)(O x)` (ii) `VO^(2+)+2H^(+)+e^(-)toV^(3+)+H_(2)O,E_(2)^(@)(red.)` Two half-cells for the 2nd cell reaction will be (iii) `V^(3+)+H_(2)OtoVO^(2+)+2H^(+)+e^(-),E_(3)^(@)(O x),` (iv) `Ag^(+)+e^(-)toAg,E_(4)^(@)(Red).0.799V` (given) `E_(3)^(@)+E_(4)^(@)=0.439V` `thereforeE_(3)^(@)(O x.)=0.439-E_(4)^(@)=0.439-0.799=-0.360V` As reaction (ii) is reverse of reaction (iii) `E_(2)^(@)(Red.)=-E_(3)^(@)(O x.)=+0.360V` Further, `E_(1)^(@)(O x)+E_(2)^(@)(Red.)=0.616` or `E_(1)^(@)(O x.)=0.616-E_(2)^(@)(Red.)=0.616-0.360=0.256V` As required reaction is reverse of reaction (i) `E^(@)`(Reqd. reaction)=-0.256V

Discussion

488.	When a lead storage battery is charged it acts as:A. a fuel cellB. an electrolytic cellC. a galvanic cellD. a concentration cell.
Answer» Correct Answer - B During charging, it acts as electrolytic cell and during discharging, it ac ts as Galvanic cell.

Discussion

489.	When lead storage battery is chargedA. Lead dioxide dissolvesB. Sulphuric acid is regeneratedC. The lead electrode becomes coated with lead sulphateD. The amount of sulphuric acid decreases
Answer» Correct Answer - B During charging of a lead storage battery, the reaction at the andoe and cathode are anode: `PbSO_(4)+2e^(-)toPb+SO_(4)^(2-)` Cathode: `PbSO_(4)+2H_(2)OtoPbO_(2)+4H^(+)+SO_(4)^(2-)+2e^(-)` In both the reactions `H_(2)SO_(4)` is regenerated.

Discussion

490.	STATEMENT 1 `DeltaG^(@)=-nFE^(@)` and STATE/MENT 2: `E^(@)` should be positive fo ra spontaneous reactionA. Statement 1 is ture , Statement 2 is true Statement 2 is correct explanation for Statement 7B. Statement 1 is true Statement 2 is ture Statement 2 is NOT a correct explantion for Statement 7C. Statement 1 is true statement 2 is tureD. Statement 1 is false Statement 2 is true
Answer» Correct Answer - B

Discussion

491.	The driving force `DeltaG` diminishes to zero on the way to equilibrium, just as in any other spontaneous process. Both `DeltaG` and the corresponding cell potential `(E=-(DeltaG)/(nF))` are zero when the redox reaction comes to equilibrium. The Nernst equation for the redox process of the cell may be given as : `E=E^(@)-0.059/n log Q` The key to the relationship is the standard cell potential `E^(@)`, derived from the standard free energy changes as : `E^(@)=-(DeltaG^(@))/(nF)` At equilibrium, the Nernst equation is given as : `E^(@)=0.059/n log K` The equilibrium constant `K_(c)` for the reaction : `Cu(s)+2Ag^(+) (aq.)+2Ag(s)" "(E_(cell)^(@)=0.46 V)` will be :A. antilog 15.6B. antilog 2.5C. antilog 1.5D. antilog 12.2
Answer» Correct Answer - A

Discussion

492.	A current of `2.68 A` is passed for `1.0` hour through an aqueous solution of `CuSO_(4)` using copper electrodes. Which of the following statements is `//` are correct ?
Answer» The electrode reaction are : `underset("1 mole")(Cu^(2+))+ underset(2xx96500 C)(2e^(-)) rarr Cu ("Cathode")` `Cu rarr Cu^(2+)+2e^(-) ("Anode")` Thus, cathode increases in mass as copper is deposited on it and the anode decreases in mass as copper from it dissolves. Charge passed through cell `=2.68xx60xx60` coulomb Copper deposited or dissolved `= 63.5/(2xx96500)xx2.68xx60xx60=3.174 g` Increase in mass of cathode = Decrease in mass of anode `=3.174 g`

Discussion

493.	When a lead storage battery is discharged:A. `SO_(2)` is evolvedB. lead sulphate is consumedC. lead is formedD. sulphuric acid is consumed.
Answer» Correct Answer - D See comprehensive review for details and reactions.

Discussion

494.	When a lead storage battery is dischargedA. lead sulphate is consumedB. oxygen gas is evolved.C. lead sulphate is formedD. lead sulphide is formed.
Answer» Correct Answer - C `Pb_((s))+PbO_(2(s))+2H_2SO_(4(aq)) to 2PbSO_(4(s)) + 2H_2O_((l))`

Discussion

495.	When a lead storage battery is discharged:A. `SO_(2)` is evolvedB. lead is formedC. lead sulphate is consumedD. sulphuric acid is consumed
Answer» Correct Answer - D The reaction occuring during operation of battery is `Pb + PbO_(2) + 2H_(2)SO_(4) to 2 Pb SO_(4) + 2H_(2)O`

Discussion

496.	When a lead-storage battery is discharged, then incorrect option(s) is/are (1) H2SO4 is consumed (2) Pb is formed (3) SO2 is evolved (4) PbSO4 is consumed
Answer» Correct option (2, 3, 4) Explanation: Pb is consumed and PbSO₄ is formed. SO₂ is not evolved.

496.

When a lead-storage battery is discharged, then incorrect option(s) is/are (1) H2SO4 is consumed (2) Pb is formed (3) SO2 is evolved (4) PbSO4 is consumed

Answer»

Correct option (2, 3, 4)

Explanation:

Pb is consumed and PbSO₄ is formed. SO₂ is not evolved.

Discussion

497.	Fully charges lead storage battery contains 1.5 L of `5 M H_(2)SO_(4)`. If 2.5 amp of currect is taken from the cell for 965 minutes, then what will be the molarity of remaining `H_(2)SO_(4)` ? Assume that volume of battery fluid to be constant :A. 4 MB. 3.5 MC. 2 MD. 4.25 M
Answer» Correct Answer - A

Discussion

498.	Which of the following is not correct fo mercury cell?A. It consists of `Zn-Hg` amalgam as anodeB. It consists of a paste of `HgO` and carbon as the cathodeC. The electrolyte is a paste of `KOH` and `ZnO`.D. It is suitable for high current devies.
Answer» Correct Answer - D The mercury battery is used extensively in medicine and electronic industries and is more expensive than the common dry cell. Commonlu used mercury cells are small button like cells. They are suitable for low current devices like hearing aids, watches, etc.

Discussion

499.	Which of the following is incorrect for a lead storage battery?A. The anodes of a lead storage battery are filled with the spongy lead alloyB. The cathodes are filled with lead dioxide.C. Both electrodes are in contact with a solution of `H_(2)SO_(4)` in waterD. The electrolytic solution is about `25%` sulphuric acid by mass.
Answer» Correct Answer - D Lead storage cell is the best known eample of a secondary battery. This voltaic cell consists of electrodes of lead alloy grids., one electrode is packed with a spongy lead to from the anode, and the other electrode is packed with lead dioxide to form the cathode. Both are bathed in an aqueous solution of sulphuric acid which is `38% H_(2)SO_(4)` by mass amd has density `1.30 g//cm^(3)`.

Discussion

500.	The cathode reaction during the charging of a lead-acid battery leads to theA. deposition of `Pb`B. conversion of `PbSO_(4)` to `PbO_(2)`C. formation of `PbSO_(4)`D. formation of `PbO_(2)`
Answer» Correct Answer - B The lead storage battery used in automobilies is really a number of galvanic cells conncected in series. The glavanic cells are designed so that the cell reaction that produce current can be reversed by application of an external current. The battery is rechargeable and thus long lived. The chemistry of the automonile battery is based on changes in the oxidation state of lead. In anode half-cell, the oxidation state of lead is changed from `0` to `+2`. In the cathode half-fell. the oxidation state is changed from `+4` to `+2`. The lead in the `0` oxidation state is in the form of a spongy alloy. The lead in the `+2` oxidation state in the form of `PbSO_(4)`. a white insoluble salt that can sometimes be seen as a coating on dead batteries. Lead in the `+4` oxidation state is in the form of `PbO_(2)`, lead dioxide which is also insoluble. The chemical system of a charged lead storage battery is not at equilibrium, and an electric current flows when the plates of lead and of lead dioxide are connected. The oxidation process that occurs in the anode is `Pb(s)+SO_(4)^(2-)(aq.) rarr PbSO_(4)(s)+2e^(-)` the reduction that occurs when the lead dioxide plate (the cathode) receives the electrons released by the oxidation of lead is `PbO_(2)+SO_(4)^(2-)(aq.)+4H^(+)(aq.)+2e^(-) rarr PbSO_(4)(s)+2H_(2)O(I)` Notice that both these process produce `PbSO_(4)`, in which lead is in the `+2` oxidation state. On charging the battery the reactions are reversed i.e. anode now undergoes reduction while cathode undergoes oxidation. As a result `PbSO_(4)(s)` on anode is converted into `Pb(s)` while `PbSO_(4)(s)` on cathode is converted into `PbSO_(4)(s)` on cathode llis converted into `PbO_(2)(s)`. The overall reaction for the cell is `Pb(s)+PbO_(2)(s)+2SO_(4)^(2-)(aq.)+4H^(+)(aq.) underset("Charge")overset("Discharge")(hArr) 2PbSO_(4)(s)+2H_(2)O(l)` As the cell discharges electricity and proceeds towards equilibrium, insoluble lead sulphate is deposited on both plates and sulphuric acid is consumed. The consumption of `H_(2)SO_(4)` reduces the density of the solution in the battery. The extent to which a battery has been discharged can be determine by measurement of the density of the solution with a hydrometer.

Discussion

Explore topic-wise InterviewSolutions in .

The most durable metal plating on iron to protect it againt corrosion isA. nickel platingB. copper platingC. tin platingD. zinc plating

Arrange the following in decreasing order of their reactivity:Al,Cu, Fe,Mg and Zn.

Specific conductance of a decinormal solution of KCl is 0.0112 `ohm^(-1) cm^(-1)`. The resistance of a cell containing the solution was found to be 56. What is the cell constant ?

`E^(@)` for `F_(@)+2e hArr2F^(-)` is `2.7 V` Thus, `E^(@)` for `F^(-) hArr(1)/(2)Fe +e` is :A. `1.35 V`B. `-1.35 V`C. `-2.7 V`D. `2.7 V`

Silver is electro-deposited on a metallic vessel of surface area 800 `cm^(2)` by passing a current of 0.2 ampere for 3 hours. Calculate the thickness of silver deposited given that its density is 10.47 g `cm^(-3)`. (At mass of Ag =107.92).

Calculate the number of coulombs required to deposit 5.4 g of Al when the electrode reaction is `Al^(3+)+3e^(-) to Al` (Given , atomic mass of Al =27 g `mol^(-1), F=96500 C mol^(-1)`)

Calculate the number of coulombs required to deposit 40.5 g of Al when the electrode reaction is , `Al^(3+)+3e^(-)rarr Al`

The number of coulombs required to deposit 5.4g of Al when the electrode reaction is `Al^(3+)+3e^(-)rarrAl`A. `1.83xx10^(5)C`B. `57900C`C. `5.86xx10^(5)C`D. None of the above

The rate constant for the decomposition of hydrocarbons is `2.418 xx10^(-5)s^(-1)` at 546 K. If the energy of activatin is `179.9 kJ//mol.` What will be the value of per-exponential factor ?

The electrode pptenticals for ` Cu^(2+) (aq) +e^(-) rarr Cu^+ (aq) ` and ` Cu^+ (aq) + e^- rarr Cu (s)` are `+ 0.15 V` and ` +0. 50 V` repectively. The value of `E_(cu^(2+)//Cu)^@` will be.A. `0.500" V "`B. `0.325" V "`C. `0.650" V "`D. `0.150" V "`

The electrode pptenticals for ` Cu^(2+) (aq) +e^(-) rarr Cu^+ (aq) ` and ` Cu^+ (aq) + e^- rarr Cu (s)` are `+ 0.15 V` and ` +0. 50 V` repectively. The value of `E_(cu^(2+)//Cu)^@` will be.A. 0.500 VB. 0.325 VC. 0.650 VD. 0.150 V

The electrode pptenticals for ` Cu^(2+) (aq) +e^(-) rarr Cu^+ (aq) ` and ` Cu^+ (aq) + e^- rarr Cu (s)` are `+ 0.15 V` and ` +0. 50 V` repectively. The value of `E_(cu^(2+)//Cu)^@` will be.A. ` 0.150 V`B. ` 0.325 V`C. ` 0. 500 V`D. ` 0. 650 V`

Why alternating current is sued in place of direct current in measuring the electrolytic conductance?

Which of the following electrolytic solutions has the least specific conductance?A. 2NB. 0.002NC. 0.02ND. 0.2N

The electrode pptenticals for ` Cu^(2+) (aq) +e^(-) rarr Cu^+ (aq) ` and ` Cu^+ (aq) + e^- rarr Cu (s)` are `+ 0.15 V` and ` +0. 50 V` repectively. The value of `E_(cu^(2+)//Cu)^@` will be.A. 0.325VB. 0.650VC. 0.150 VD. 0.500 V

The electrolytic conductance si a direct measure of .A. ` E_2 = 0 !=E_1`B. ` E_1gtE_2`C. ` E_1lt E_2`D. ` E_1` =E_2`

Standard electrode potential for `Sn^(4+)//Sn^(2+)` couple is `0.15 V` and that for the `Cr^(3+)//Cr` couple is `-0.74 V`. These two couples in their standard state are connected to make a cell. The cell potential will beA. `+1.19" V "`B. `+0.89" V "`C. `+0.18" V "`D. `+1.83" V "`

Standard electrode potential for `Sn^(4+)//Sn^(2+)` couple is `0.15 V` and that for the `Cr^(3+)//Cr` couple is `-0.74 V`. These two couples in their standard state are connected to make a cell. The cell potential will beA. `+0.89 V`B. `+0.18 V`C. `+1.83 V`D. `+1.199 V`

On the bassis of the information available from the reaction ` 4/3 Al + O_2 rarr 2/3 Al_2 O_3 . Delta G =- 82 7 k J "mol"^(-1)` of `O_2` the minimum emf required to carry out an electorlysis of `Al_2 O_3` is `(F= 96500 C "mol"^(-1))`A. ` 8.56 V`B. ` 2.14 V`C. ` 4.` 28 V`D. ` 6. 42 V`

Celll equation : `A = 2B^(+) rarr A^(2+) + 2 B` `A^(2+) + 2e rarr A` `E^@ = + 0.34 V` and `log_(10) K= 15. 6 ` at `300 K` for cell reactions Find `E^@` for `B^(+) +e rarr B` Given `[( 2.303 RT)/(nF) =0.059]` at `300 K`.A. ` 0. 80 `B. ` 1.26 `C. ` -0. 64`D. ` +0.94`

How many `kJ` of energy is spent when a currebnt of (4) amp passes for `200` second under a potential of `115 V` ?A. ` 52 k J`B. `72 k J`C. ` 82 k J`D. `92 k J`

The mass of copper deposited from a solution of `CuSO_4` by passage of `45 A` current for `965` second is (Mol . Wt. pf Copper `= 63.5` ).A. ` 15. 875 g`B. ` 1. 585 g`C. ` 4825 g`D. ` 96 500 g`

The standard emf of a cell having one electron change is found to be `0.591 V` at `25^@ C`, The equilibrium constant of the reaction is :A. ` 1^(-10)`B. `29.5 xx 10^(-2)`C. ` 10`D. `1xx10^(1)`

The standard emf of a cell having one electron change is found to be `0.591 V` at `25^@ C`, The equilibrium constant of the reaction is :A. `1.0 xx 10^(30)`B. `1.0xx10^5`C. `1.0 xx 10^(10)`D. `1.0 xx 10^(1)`

The standard emf of a cell having one electron change is found to be `0.591 V` at `25^@ C`, The equilibrium constant of the reaction is :A. `1.0 xx 10^(30)`B. `1.0 xx 10^(5)`C. `1.0 xx 10^(10)`D. `1.0 xx 10^(1)`

The standard emf of a cell having one electron change is found to be `0.591 V` at `25^@ C`, The equilibrium constant of the reaction is :A. ` 1.0 xx 10^(10)`B. ` 1.0 xx 10^5`C. ` 1.0 xx 10^(1)`D. ` 1.0 xx 10^(30)`

The standard emf of a cell involving one electron change is found to be 0.591 V at `25^(@)C`. The equilibeium constant of the reaction is (`1F=96500 C mol^(-1)`)A. `1.0 xx10^(1)`B. `1.0xx10^(5)`C. `1.0xx10^(10)`D. `1.0xx10^(30)`

Rusting on the surface of iron involves :A. `Fe(s) rarr Fe^(2+) (aq.)+2e^(-)" (at anodic site)"`B. `O_(2)(g)+4H^(+) (aq.)+4e^(-) rarr 2H_(2)O(l)" (at cathodic site)"`C. `4Fe^(2+) (aq.)+O_(2)(g)+4H_(2)O (l) rarr 2Fe_(2)O_(3) (s)+8H^(+)`D. `Fe_(2)O_(3)(s)+ x H_(2)O (l) rarr Fe_(2)O_(3). xH_(2)O`

For the cell reaction, `2Cr^(4+)+Coto2Ce^(3+)+Co^(2+)" "E_((cell))^(@)` is 1.89V and `E_(Co//Co^(2+))^(@)=-0.028`. If `E_(Ce^(4+)//Ce^(3+))^(@)`A. `-1.64V`B. `+1.64V`C. `-2.08V`D. `+2.17V`

Calculate the reduction potential of a half cell consisting of a platinum electrode immersed in `2.0M Fe^(2+)` and `0.02M Fe^(3+)` solution. Given `E_(Fe^(3+)//Fe^(2+))^(@) = 0.771 V`.

When a lead storage battery is charged it acts as:A. a fuel cellB. an electrolytic cellC. a galvanic cellD. a concentration cell.

When lead storage battery is chargedA. Lead dioxide dissolvesB. Sulphuric acid is regeneratedC. The lead electrode becomes coated with lead sulphateD. The amount of sulphuric acid decreases

A current of `2.68 A` is passed for `1.0` hour through an aqueous solution of `CuSO_(4)` using copper electrodes. Which of the following statements is `//` are correct ?

When a lead storage battery is discharged:A. `SO_(2)` is evolvedB. lead sulphate is consumedC. lead is formedD. sulphuric acid is consumed.

When a lead storage battery is dischargedA. lead sulphate is consumedB. oxygen gas is evolved.C. lead sulphate is formedD. lead sulphide is formed.

When a lead storage battery is discharged:A. `SO_(2)` is evolvedB. lead is formedC. lead sulphate is consumedD. sulphuric acid is consumed

When a lead-storage battery is discharged, then incorrect option(s) is/are (1) H2SO4 is consumed (2) Pb is formed (3) SO2 is evolved (4) PbSO4 is consumed

Fully charges lead storage battery contains 1.5 L of `5 M H_(2)SO_(4)`. If 2.5 amp of currect is taken from the cell for 965 minutes, then what will be the molarity of remaining `H_(2)SO_(4)` ? Assume that volume of battery fluid to be constant :A. 4 MB. 3.5 MC. 2 MD. 4.25 M

Which of the following is not correct fo mercury cell?A. It consists of `Zn-Hg` amalgam as anodeB. It consists of a paste of `HgO` and carbon as the cathodeC. The electrolyte is a paste of `KOH` and `ZnO`.D. It is suitable for high current devies.

The cathode reaction during the charging of a lead-acid battery leads to theA. deposition of `Pb`B. conversion of `PbSO_(4)` to `PbO_(2)`C. formation of `PbSO_(4)`D. formation of `PbO_(2)`