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2801.

The following electrochemical cell has been set-up, `Pt(1)|Fe^(3+), Fe^(2+) (a=1)||Ce^(4+), Ce^(3+) (a=1) Pt (2)` `E^(@) (Fe^(3+) //Fe^(2+))=0.77 V, E^(@) (Ce^(4+)//Ce^(3+))=1.61 V` If an ammeter is connected between the two platinum electrodes, predict the direction of flow of current. Will the current increase or decrease with time ?

Answer» Current will flow from higher reduction potential electrode to lower reduction potential electrode, i.e., from `Pt(2)` electrode to `Pt(1)` electrode.
`E_(cell)^(@)=1.61-0.77=0.84` volt
Discussion
2802.

Zinc granules are added in excess to `500mL` OF `1.0m` nickel nitrate solution at `25^(@)C` until the equilibrium is reached. If the standard reduction potential of `Zn^(2+)|Zn` and `Ni^(2+)|Ni` are `-0.75V` and `-0.24V`, respectively, find out the concentration of `Ni^(2+)` in solution at equilibrium.

Answer» The redox change is
`{:(,Zn+,Ni^(2+),hArr,Zn^(2+)+,Ni),("mM before equilibrium",,500,,0,),("mM at equilibrium",,a,,(500-a),):}`
`:. E_(cell) = E_(OP_(Zn//Zn^(2+)))+E_(RP_(Ni^(2+)//Ni))`
`E_(cell) = E_(Zn//Zn^(2+))^(@)+E_(RP_(Ni^(2+)//Ni))^(@)+(0.059)/(2)"log"_(10)([Ni^(2+)])/([Zn^(2+)])`
At equilibrium `E_(cell) = 0`
`:. E_(OP_(Zn//Zn^(2+)))^(@)+E_(RP_(Ni^(2+)//Ni))^(@) =-(0.059)/(1)log_(10).([Ni^(2+)])/([Zn^(2+)])`
or `0.75+(-0.24)=-(0.059)/(2)log_(10).([Ni^(2+)])/([Zn^(2+)]) `
`([Ni^(2+)])/([Zn^(2+)]) = "antilog" (-(0.51 xx 2)/(0.059))=5.15 xx 10^(-18)`
`:. (a)/(500-a) = 5.15 xx 10^(-18)`
`:. a = 500 xx 5.15 xx 10^(-18)`
`:. [Ni^(2+)] = (mM)/(V) = (500 xx 5.15 xx 10^(-18))/(500)`
`= 5.15 xx 10^(-18)M`
Discussion
2803.

The EMF of a cell corresponding to the reaction, `Zn(s)+2H^(+)(aq)toZn^(2+)(0.1M)+H_(2)(g,1atm)` is 0.28 volt at `25^(@)C`. Write the half-cell reactions and calculate the pH of the solution at the hydrogen electrode. `E_(Zn^(2+)//Zn)^(@)=-0.76"volt",E_(H^(+)//H_(2))^(@)=0`

Answer» The half cell reaction are: `ZntoZn^(2+)+e^(-)` (Oxidation half reaction)
`2H^(+)+2e^(-)toH_(2)` (Reduction half reaction)
The cell may be represented as: `Zn|Zn^(2+)||H^(+)|H_(2)`
For the given cell reaction, `E_(cell)^(@)=E_(H^(+)//(1)/(2)H_(2))^(@)-E_(Zn^(2+)//Zn)^(@)=0-(-0.76V)=0.76V`
Applying Nernst eqn. to the given cell reaction, `E_(cell)=E_(cell)^(@)=(0.0591)/(2)"log"([Zn^(2+)])/([H^(+)]^(2))`
`therefore0.28=0.76-(0.0591)/(2)"log"((0.1))/([H^(+)]^(2))=0.76-(0.0591)/(2)[log0.1-2log(H^(+))]`
`=0.76-0.02955(-1+2pH)` `[becausepH=-log[H^(+)]]`
`=0.76+0.2095-0.0591pH`
or `pH=(0.5095)?(0.591)=8.62`
Discussion
2804.

An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to:A. increase in number of ionsB. increase in ionic mobility of ionsC. `100%` ionisation of electrolyte at normal dilutionD. increase in both number of ions and ionic mobility of ions

Answer» Correct Answer - B
By definition, a strong electrolyte ionizes or dissociates almost completely. At high concentration, the average distance between ions is very small. Consequently,they may re combine to form ion-pairs, which have no contribution to conductance. With dilution, the average distance between ions increases. This leads to decrease in tendency to form ion pairs. In other words, the ionic mobility increases as more and more ions arc able to participate.
Discussion
2805.

An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to:A. increase in number of ionsB. increase in the ionic mobility of ionsC. 100% ionisation of the electrolyte at normal dilutionD. increase in both i.e. number of ions and ionic mobility of ions.

Answer» Correct Answer - B
Strong electrolyte is completely ionized at all concentrations . Therefore, the number of ions remains the same on dilution. However on dilution. The interionic forces decrease. Consequently the ionic mobility increases. Hence the equivalent conductance increases.
Discussion
2806.

An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to:A. increase in ionic mobility of ionsB. 100 % ionisation of electrolyte at normal dilutionC. increase in both i.e., number of ions and ionic mobility of ionsD. increase in number of ions

Answer» Correct Answer - A
Strong electrlytes are completely ionised at all concentrations . On increasing dilution the no. of ions remains the same but the ionic mobility increases and the equivalent conduction increases .
Discussion
2807.

The following electrochemical cell has been set up `:` `Pt(l)|Fe^(3+),Fe(a=1)|Ce^(4+),Ce^(3+)(a=1)|Pt(2)` `E^(c-)._(((Fe^(3+),Fe^(2+)))=0.77V` and `E^(c-)._(((Ce^(4+),Ce^(3+)))=1.61V` If an ammeter is connected between two platinum electrodes, predict the direction of the flow of current. Will the current increase or decreases with time ?

Answer» Correct Answer - decrease with time
Discussion
2808.

The formal potential of `Fe^(3+)//Fe^(2+)` in a sulphuric acid and phosphoric acid mixture `(E^@= + 0.61 V)` is much lower than the standard potential (`E^@` =+0.77 V). This is due to (i) formation of the species `[FeHPO_4]^+` (ii) lowering of potential upon complexation (iii) formation of the species `[FeSO_4]^+` (iv) high acidity of the medium.A. (i) and (ii) onlyB. (i), (ii) and (iv) onlyC. (iii) onlyD. all of these.

Answer» Correct Answer - A
In the aqueous solution, if `H_2SO_4` is present then iron will from hydrated iron sulphate rather than forming `[FeSO_4]^+` solution. Moreover, tendency of formation of `FeSO_4` or `Fe_2(SO_4)_3` is even more . Formation of `[FeHPO_4]^+` is reasonably fine. Due to the formation of complex, `[Fe^(3+)]` decreases and accordingly, potential decreases.
`E=E^@-0.0591/n "log"_10 ([Fe^(2+)])/([Fe^(3+)])`
Discussion
2809.

The emf of a cell corresponding to the reaction `Zn +2H^(+)(aq) rarr Zn^(2+) (0.1M) +H_(2)(g) 1` atm is `0.28` volt at `25^(@)C`. Calculate the `pH` of the solution at the hydrogen electrode. `E_(Zn^(2+)//Zn)^(@) =- 0.76` volt and `E_(H^(+)//H_(2))^(@) = 0`A. 7.05B. 8.62C. 8.75D. 9.57

Answer» Correct Answer - B
The half-cell reaction are `Zn to Zn^(2+) + 2e^(-)`
`2H^(+) + 2e^(-) to H_2 `
We know that
`E_(Zn//Zn^(2+))=E_(Zn//Zn^(2+))^@-"RT"/"nF""In"([Zn^(2+)])/([Zn])`
`therefore E_(Zn//Zn^(2+))=0.76-(2.303xx8.314xx298)/(2xx96500)"log"0.1/1`
=0.76-(-0.03)=0.79 V
Also, `E_(H^(+)//H_2)=E_(H^(+)//H_2)^@-"RT"/"nF""In"([H_2])/([H^+]^2)`
`=0-(2.303xx8.314xx298)/(2xx96500)"log"1/([H^+]^2)`
=0.0591 log `[H^+]` =-0.0591 pH
Since `E_"cell"=E_(Zn//Zn^(2+))+E_(H^+//H_2)`
or 0.28=0.79-0.0591 pH
or `pH=(0.79-0.28)/0.0591=0.51/0.0591`=8.62
Discussion
2810.

An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to:A. Increase in numbe of ionsB. Increase in ionic mobility of ionsC. 100% ionisation of electrolyte at normal dilutionD. Increase in both i.e., number of ions and ionic mobility of ions

Answer» Correct Answer - B
Discussion
2811.

Calculate the euilibrium constant for the reaction, `2Fe^(3+) + 3I^(-) hArr 2Fe^(2+) + I_(3)^(-)`. The standard reduction potential in acidic conditions are `0.77 V ` and `0.54 V` respectivelu for `Fe^(3+)//Fe^(2+)` and `I_(3)^(-)//I^(-)` couples.

Answer» Correct Answer - `K_(C) = 6.26 xx 10^(-7)`
Discussion
2812.

Calculate the euilibrium constant for the reaction, `2Fe^(3+) + 3I^(-) hArr 2Fe^(2+) + I_(3)^(-)`. The standard reduction potential in acidic conditions are `0.77 V ` and `0.54 V` respectivelu for `Fe^(3+)//Fe^(2+)` and `I_(3)^(-)//I^(-)` couples.A. `4.25xx10^7`B. `7.05xx10^5`C. `6.25xx10^5`D. `6.25xx10^7`

Answer» Correct Answer - D
`2Fe^(3+) + 3I^(-) hArr 2Fe^(2+) + I_3^(-)`
For the above change at equilibrium , E=0
Using the relation, `E=E^@-0.059/2"log"K_c`
or `0=E_(Fe^(3+)//Fe^(2+))^@-E_(I_3^- //I^-)^@-0.059/2 "log"K_c`
or 0=0.77-0.54-`0.059/2"log"K_c`
or 0=0.23-`0.059/2"log"K_c`
or `0.059/2"log"K_c=0.23` or log `K_c=(0.23xx2)/0.059`
or log `K_c` =7.796 or `K_c=6.25xx10^7`
Discussion
2813.

In the following electrochemical cell `:` `Zn|Zn^(2+)||H^(o+)|(H_(2))Pt` `E_(cell)=E^(c-)._(cell).` This will be whenA. `[Zn^(2+)]=[H^(o+)]=1M` and `p_(H_(2))=1atm`B. `[Zn^(2+)]=0.01M,[H^(o+)]=0.1M,` and `p_(H_(2))=1atm`C. `[Zn^(2+)]=1M,[H^(o+)]=0.1M,` and `p_(H_(2))=1atm`D. None of the above.

Answer» Correct Answer - a,b
Cell reaction `: Zn(s)+2H^(o+)(aq)rarr Zn^(2+) (aq)+H_(2)(g)`
Using`:E_(cell)=E^(c-)._(cell)-(0.0591)/(2)log(Q_(cell)-(P_(H_(2))[Zn^(2+)])/([H^(o+)]^(2)))`
`implies` For `E_(cell)=E^(c-)._(cell),Q_(cell)=1.0`
`a. Q_(cell)=(1xx1)/(1^(2))=1`
`b. Q_(cell)=(1xx0.01)/((0.1)^(2))=1`
`c. Q_(cell)=(1xx1)/((0.1)^(2))=100`
Discussion
2814.

Predict if the reaction between the followig is feasible: (i) `Fe^(3+)(aq) and I^(-)(aq)` (ii) `Ag^(+)(aq) and Cu` (iii) `Fe^(3+) and Br^(-1)(aq)` (iv) `Ag(s) and Fe^(3+)(aq)` (v) `Br_(2)(aq) and Fe^(2+)(aq)`. Given standard electrode potential: `E_(1//2I_(2),I^(-))^(@)=-0.541,E_(Cu^(2+),Cu)^(@)=0.34V` `E_(1//2Br_(2),Br^(-))^(@)=1.09V,E_(Ag^(+),Ag)^(@)=+0.80V,E_(Fe^(3+),Fe^(2+))^(@)=+0.77V`

Answer» A reaction feasible if EMF of the cell reaction is +ve
(i) `Fe^(3+)(aq)+I^(-)(aq)toFe^(2+)(aq)+(1)/(2)I_(2)` i.e., `Pt|I_(2)|I^(-)(aq)||Fe^(3+)(aq)|Fe^(2+)(aq)|Pt`
`thereforeE_(cell)^(@)=E_(Fe^(3+),Fe^(2+))^(@)-E_(1//2I_(2),I^(-))^(@)=0.77-0.54=0.23V` (Feasible)
(ii) `Ag^(+)(aq)+CutoAg(s)+Cu^(2+)(aq),` i.e., `Cu|Cu^(2+)(aq)||Ag^(+)(aq)|Ag`
`E_(cell)^(@)+E_(Ag^(+),Ag^(-))^(@)-E_(Cu^(2+),Cu)^(@)=0.80-0.34=0.46V` (Feasible)
(iii) `Fe^(3+)(aq)+Br^(-)(aq)toFe^(2+)(aq)+(1)/(2)Br_(2),E_(cell)^(@)=0.77-1.09=-0.32V` (Not feasible)
(iv) `Ag(s)+Fe^(3+)(aq)toAg^(+)(aq)+Fe^(2+)(aq)E_(cell)^(@)=0.77-0.80=-0.03V` (Not feasible)
(v) `(1)/(2)Br_(2)(aq)+Fe^(2+)(aq)toBr^(-)+Fe^(3+),E_(cell)^(@)=1.09-0.77=0.32V` (Feasible)
Discussion
2815.

Given that `E_(Fe^(2+)//Fe)^(.)=-0.44V,E_(Fe^(3+)//Fe^(2+))^(@)=0.77V` if `Fe^(2+),Fe^(3+)` and `Fe` solid are kept together thenA. the concentration of `Fe^(3+)` increasesB. the concentration of `Fe^(3+)` decreasesC. the mass of Fe increasesD. the concentration of `Fe^(2+)` decreases

Answer» Correct Answer - B
Discussion
2816.

The standard reduction potential for `Cu^(2+)|Cu` is `+0.34V`. Calculate the reduction potential al `pH=14` for the above couple . `K_(sp)` of `Cu(OH)_(2)` is `1.0xx10^(-19)`

Answer» Correct Answer - `-0.22V`
`pH=14,[H^(o+)]=10^(-14),[overset(c-)(O)H]=10^(0)=1`
`Cu(OH)_(2) rarr Cu^(2+)+2overset(c-)(O)H`
`K_(sp)=[Cu^(2+)][overset(c-)(O)H]^(2)`
`1.0xx10^(-19)=[Cu^(2+)][1]^(2),[Cu^(2+)+2e^(-)rarr Cu]`
`[Cu^(2+)]=10^(-19)M`
`E=E^(c-)-(0.0591)/(2)log .(1)/((10^(-19)))`
`=0.34-(0.0591)/(2)xx19`
`=-0.22V`
Discussion
2817.

Calculate the equilibrium constant for the reaction `:` `Fe^(2+)Ce^(4+)hArrFe^(3+)+Ce^(3+)` Given `:E^(c-)._((Ce^(4+)|Ce^(3+)))=1.44V` `E^(c-)._((Fe^(3+)|Fe^(2+)))=0.68V`

Answer» We Know, `E_(cell)^(@) = (0.059)/(1)log_(10)K_(c)`
`E_(cell)^(@) = E_(OP_(Fe^(2+)//Fe^(3+)))^(@)+E_(RP_(Ce^(4+)//Ce^(3+)))^(@)`
`= -0.68+1.44 = 0.76V`
`:. Log_(10)K_(c) = (0.76)/(0.059) = 12.8814`
`:. K_(c) = 7.6 xx 10^(12)`
Discussion
2818.

For the electrochemical cell, `(M|M^(o+))||(X^(c-)|X)`, `E^(c-)._((M^(o+)|M))=0.44V`, and `E^(c-)._((X|X^(c-))=-0.33V`. From this data, one can conclude thatA. `M+X rarr M^(o+)+X^(c-)` is a spontaneous reactionB. `M^(o+)+X^(c-) rarr M+x` is the spontaneous reactionC. `E^(c-)._(cell)=0.77V`D. `E^(c-)._(cell)=-0.77V`

Answer» Correct Answer - b,c
`E^(c-)._(cell)=0.44-(-0.33)=0.77V`
Spontaneous reaction `:M^(o+)+X^(c-)rarr M+X`
Discussion
2819.

For the electrochemical cell, `(M|M^(o+))||(X^(c-)|X)`, `E^(c-)._((M^(o+)|M))=0.44V`, and `E^(c-)._((X|X^(c-))=-0.33V`. From this data, one can conclude thatA. `M + X rarr M^+ +X^-`is the spontaneous reactionB. `M + X rarr M^+ +X^-` is the spontaneous reactionC. `E_(cell)^@ =0.. 77 V`D. ` E_(cell)^@ =0.. 77 V`

Answer» Correct Answer - B
For the give cell `M|M^+ ||X^_|X, ` the cell reaction is derived as follows :
RHS : reduction `X+e^- rarr X^-` …(i)
LHS : Oxdation ` M rarr M^+ +e^-` …(ii)
Add (i) and (ii) `M+X rarr M^+ +X^-`
The cell potential `=- 0. 11 V`
Since `E_(cell) =- ve`, the cell reaction derived above is not spontaneous . In fact , the reverse reaction will occur spontaneously.
Discussion
2820.

For the electrochemical cell, `(M)|M^(+))||(X^(-)|X)`. `E^(@)(M^(+)//M)=0.44V` and `E^(@)(X//X^(-))=0.33V`. From this data one can deduce thatA. `M+XrarrM^(+)+X^(-)` is the spontaneous reactionB. `M^(+)+X^(-)rarrM+X` is the spontaneous reactionC. `E_("cell")=0.77V`D. `E_("cell")=-0.77`V.

Answer» Correct Answer - B
`M|M^(+)||X^(-)|X`
The two half cell reactions for the cell are
`MrarrM^(+)+e^(-)`
or `X+e^(-)rarrX^(-)`
`M+XrarrM^(+)+X`
`E_("cell")^(@)=E_(X//X^(-))^(@)-E_(M^(+)//M)`
`=0.33V-0.44V=-0.11V`
As `E_("Cell")^(@)` is negative the reaction involved is non spon taneus. the reverse reaction, howev er, will be spontaneous.
`M^(+)+X^(-)rarrM+X`
Discussion
2821.

Which of the followig is false?A. Saline water slows down rustingB. In daniell cel, if concentrations of the solutions are doubled, the emf of the cell is also doubled.C. EMF of a cell is an intensive whereas free energy change, `DeltaG` is extensive.D. Galvanised iro sheets remain protected from rusting even if a crack is developed.

Answer» Correct Answer - A::B::D
Saline water speeds up rusting. Hence, (a) is false.
`Zn+Cu^(2+)toZn^(2+)+Cu,E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"([Zn^(2+)])/([Cu^(2+)])`
If `[Zn^(2+)]` and `[Cu^(2+)]` are doubled, there is no effect on EMF, hence (b) is wrong.
Discussion
2822.

The standard emf of the cell, `Cd(s) |CdCI_(2) (aq) (0.1M)||AgCI(s)|Ag(s)` in which the cell reaction is `Cd(s) +2AgCI(s) rarr 2Ag(s) +Cd^(2+) (aq)` is `0.6915 V` at `0^(@)C` and `0.6753V` at `25^(@)C`. The `DeltaH` of the reaction at `25^(@)C` is,-A. `-176 kJ`B. `-234.7 kJ`C. `+123.5 kJ`D. `-167.26 kJ`

Answer» Correct Answer - D
`DeltaG =- nFE^(@) = DeltaH^(@) - T DeltaS^(@)`
`=-2 F xx 0.695 = DeltaH^(@) - 273 DeltaS^(@)` ...(i)
`=- 2F xx 0.6753 = DeltaH^(@) - 298 DeltaS` ..(ii)
Solve both equation.
Discussion
2823.

The emf of the cell involving the reaction `2Ag^(+) (aq)+H_(2)(g) rarr 2Ag(s)+2H^(+) (aq)` is `0.080 V`. The standard oxidation potential od silver electrode isA. 0.80 VB. 0.40 VC. `-0.80 V`D. 0.20 V

Answer» Correct Answer - C
From the equation, give cell represented as
`H_(2)|H^(+) (aq) ||Ag^(+) (aq)| Ag`
Hence, `E_("cell")=E_(Ag^(+)//Ag)-E_(H^(+)//1/2 H_(2))`
or `0.80=E_(Ag^(+)//Ag)-0.0`
`:. E_(Ag^(+)//Ag)=0.80 V`
Hence, `E_("oxi")=-0.80 V`
Discussion
2824.

The emf of the cell involving the following reaction, `2Ag^(+) +H_(2) rarr 2Ag +2H^(+)` is `0.80` volt. The standard oxidation potential of silver electrode is:-A. `-0.80` voltB. `0.80` voltC. `0.40` voltD. `-0.40` volt

Answer» Correct Answer - A
`E_("cell")^(@)=E_("cathode")-E_("andoe")0.80= E_(Ag^(+)//Ag)^(@)-E_(H_(2)//H^(+))^(@)`
`0.80 = E_(Ag^(+)//Ag)-0`
`E_("oxidation")^(@)= - E_("red")^(@)= -0.80 V`
Discussion
2825.

The electrode potential of hydrogen electrode at the pH=12 will be

Answer» Correct Answer - C
`E=E^(@)H_(2)+(0.059)/(n)"log"[H^(+)]^(2)`
` =0+(0.059)/(2)"log"(0^(-12))^(2)`
Discussion
2826.

For the electrochemical cell, `(M|M^(o+))||(X^(c-)|X),E^(c-)._((M^(o+)|M))=0.44V` and `E^(c-)._((X|X^(c-)))=0.334V`A. `M+Xrarr M^(o+)+X^(c-)` is a spontaneous reaction.B. `M^(o+)+X^(c-) rarr M+X` is the spontaneous reactionC. `E_(cell)=0.77V`D. `E_(cell)=-0.77`

Answer» Correct Answer - b
For reaction `:`
At anode `: X^(c-)rarr X+e^(c-),`
`E^(c-)._(red)=+0.334V(E^(c-)._(o x i d)=-0.034V)`
At cathode `: M^(o+)+e^(-) rarr M, " "E^(c-). (red)=0.44V`
`ulbar(M^(o+)+X^(c-)rarr M+X)`
`E^(c-)._(cell)=(E^(c-)._(red))_(c)-(E^(c-)._(red))_(a)`
`=0.44-0.334`
`0.106V`
Since `E^(c-)._(cell)` is positive, hence the reaction is spontaneous.
Discussion
2827.

Given `E^(@)` values as: `Ni^(2+)//Ni=0.25V,Cu^(2+)//Cu=0.34V,Ag^(+)//Ag=0.80V and Zn^(2+)//Zn=-0.76V` Which of the following reactions under standard conditions will not take place in the specified directions?A. `Ni^(2+)(aq)+Cu(s)toNi(s)+Cu^(2+)(aq)`B. `Cu(s)+2Ag^(+)(aq)toCu^(2+)(aq)+2Ag(s)`C. `Cu(s)+2H^(+)(aq)toCu^(2+)(aq)+H_(2)(g)`D. `Zn(s)+2H^(+)(aq)toZn^(2+)(aq)+H_(2)(g)`

Answer» Correct Answer - A::C
Cell reaction is spontaneous if its emf is +ve.
Discussion
2828.

Given`E^(c-)._(Ag^(o+)|Ag)=+0.80V,E^(c-)._(Co^(2)|Co)=-0.28V, E^(c-)._(Cu^(2+)|Cu)=+0.34V,E^(c-)._(Zn^(2+)|Zn)=-0.76V` Which metal will corrode fastest ?A. `Ag` will oxidize to `Ag^(o+)` and new `[Ag^(o+)]=1.36M`.B. `Cu`C. `Co`D. `Zn`

Answer» Correct Answer - d
Corrosion`:` Metal gets oxidized . `E^(c-)._(Zn^(2+)|Zn)` is most negative.
Discussion
2829.

The electrode potential measures theA. tendency of the electrode to gain or lose electronsB. tendency of a cell reaction to occurC. difference in the ionisation potential of electrode and metal ion.D. current carried by an electrode.

Answer» Correct Answer - A
Tendency of the electrode to get oxidised or reduced.
Discussion
2830.

The emf of the cell involving the reaction `2Ag^(+) (aq)+H_(2)(g) rarr 2Ag(s)+2H^(+) (aq)` is `0.080 V`. The standard oxidation potential od silver electrode isA. 0.80VB. `-0.80V`C. `0.40V`D. `0.20V`.

Answer» Correct Answer - B
Reduction potential of Ag=0.80V
Oxidation potential of `Ag= -0.80V`
Reduction potential and oxidation potential are having same value but opposite signs.
Discussion
2831.

Given : `Ag^(+)+//Agrarr+0.80V` `Co^(2+)//Cor rarr-.28V` `Cu^(2+)//Curarr+0.34V` `Zn^(2+)//Znrarr-0.76V`. The most reactive metal which displaces other metals from their salts in solution isA. AgB. CuC. CoD. Zn.

Answer» Correct Answer - D
The element with least value of reduction potential will displace other ions from the solution.
Discussion
2832.

Given : `Ag^(+)+//Agrarr+0.80V` `Co^(2+)//Cor rarr-.28V` `Cu^(2+)//Curarr+0.34V` `Zn^(2+)//Znrarr-0.76V`. The most reactive metal which displaces other metals from their salts in solution isA. AgB. CuC. CoD. Zn

Answer» Correct Answer - D
Zinc is more active than Ag , Cu , Co.
Discussion
2833.

The electolysis of acetate solution produces ethane according to reaction: `2CH_3COO^-)toC_6H6(g)+2CO_2(g)+2e^-` The current efficiency of the process is 80% . What volume of gases would be produced at `27^(@)` C and 740 torr, if the current of 0.5 amp is used though the solution for 96.45 min?A. 6.0LB. 0.60LC. 1.365LD. 0.91L

Answer» Correct Answer - D
Discussion
2834.

How many grams of silver could be plated out on a serving tray be electrolysis of solution containing silver in `+1` oxidation state for a period of `8.0` hour at a current of `8.46` ampere? What is the area of the tray if the thickness of the silver plating is `0.00254 cm`? Density of silver is `10.5 g//cm^(3)`.

Answer» Correct Answer - `300 cm^(2)`
Discussion
2835.

On the basis of the following `E^(@)` values, the stongest oxidizing agent is `[Fe(CN)_(6)]^(4-) rarr [Fe(CN)_(6)]^(3-)+e^(-), E^(@) = -0.35 V` `Fe^(2+) rarr Fe^(3+)+e^(-), E^(@) = -0.77 V`A. `[Fe(CN)_(6)]^(4-)`B. `Fe^(2+)`C. `Fe^(3+)`D. `[Fe(CN)_(6)]^(2-)`

Answer» Correct Answer - C
Oxidizing agent oxidizes the other substance and gets reduced. Thus, higher the reduction potential, stronger the reducing agent:
`[Fe(CN)_(6)]^(3-)+e^(-) rarr [Fe(CN)_(6)]^(4-),, E^(@) = 0.35 V`
`Fe^(3+)+e^(-) rarr Fe^(2+),, E^(@) = 0.77 V`
Since `E_(Fe^(3+)//Fe^(2+))^(@)gt E_([F^e(CN)_(6)]^(3-)//[Fe(CN)_(6)]^(4-)), Fe^(3+)` is the strongest oxidizing agent.
Discussion
2836.

On the basis of the following `E^(@)` values, the stongest oxidizing agent is `[Fe(CN)_(6)]^(4-) rarr [Fe(CN)_(6)]^(3-)+e^(-), E^(@) = -0.35 V` `Fe^(2+) rarr Fe^(3+)+e^(-), E^(@) = -0.77 V`A. `[Fe(CN)_(6)]^(4-)`B. `Fe^(2+)`C. `Fe^(3+)`D. `[Fe(CN)_(6)]^(3-)`

Answer» Correct Answer - C
( c) Substance which have higher reduction potential are stronger oxidising agent .
`[Fe(CN)_(6)]^(4-) to [Fe(CN)_(6)]^(3-)+e^(-),E^(@)=-0.35 V`
`Fe^(2+) to Fe^(3+)+e^(-), E^(@)=-0.77 V` ltbr ` because E_(oxi)^(@)=-E_(red)^(@)`
`:. [Fe(CN)_(6)]^(3-)+e^(-) to [Fe(CN)_(6)]^(4-),E^(@)=-0.35 V`
`Fe^(3+)+e^(-) to Fe^(2+), E^(@)=-0.77 V`
Hence , `Fe^(3+)` has maximum tendency to reduced, so it is the strongest oxidising agent.
Discussion
2837.

On the basis of the folwing `E^@` values, the strongest oxidizing agent is `[Fe(CN)_6 ]_4^(-) rarr [Fe(CN)_5]^(3-) +3^- , E^@ =- 0.35 V` `Fe^(2+) rarr Fe^(3+) + e^(- : E^@ =- 0. 77 V`.A. ` Fe^(2+)`B. ` Fe^(3+)`C. ` [Fe(CN)_6]^(3-)`D. ` [Fe(CN)_6]^(4-)`

Answer» Correct Answer - B
`Fe^(3+)` and ` Fe (CN)_(6)^(3)` are oxidants. Higher si `E_(RP)^(@)` stronger is oxidant .
` fe^(3+) = e rarr Fe^(2+) , E_(RP)^2 = 0.77 V`
`[ Fe (CN)_6 ]^(3-) + e rarr [Fe(CN)_6]^(4-) , E_(RP)^@ = 0.35 V`.
Discussion
2838.

In the electolysis of a `CuSO_4` solution, how many grams of Cu are plated out on the cathode in the time that it takes to liberate 5.6 litre of `O_2`(g) , measured at 1 atm and 273 K, at the node?A. 31.75B. 14.2C. 4.32D. None of these

Answer» Correct Answer - A
Discussion
2839.

On the basis of the following `E^(@)` values, the stongest oxidizing agent is `[Fe(CN)_(6)]^(4-) rarr [Fe(CN)_(6)]^(3-)+e^(-), E^(@) = -0.35 V` `Fe^(2+) rarr Fe^(3+)+e^(-), E^(@) = -0.77 V`A. `[Fe(CN)_(4)]^(-)`B. `Fe^(2+)`C. `Fe^(3+)`D. `[Fe(CN)_(6)]^(3-)`

Answer» Correct Answer - C
Strongest oxidizing agent is the substance which is reduced most easily i.e., has the highest reduction potential Reduction potential s of `[Fe(CN)_(6)]^(3-)` and `Fe^(3+)` are `+0.35V` and `+0.77V` Thus `Fe^(3+)` is reduced most easily and hence is the stronger oxidising agent.
Discussion
2840.

A current of `2` A was passed for `1 h` through a solution of `CuSO_4. 0.237g` of `Cu^(2+)` ions was discharged at cathode . The current efficiency is .A. ` 42. 2%`B. ` 26 .1%`C. ` 10%`D. ` 40.01 %`

Answer» Correct Answer - C
` W= ( 63.5)/(2xx 96500) xx 2 xx 60 xx 60= 2 37 g`
`%` of efficlency `= (0. 237)/(2.37) xx 10%`.
Discussion
2841.

Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time required to liberate 0.01 mol of `H_(2)` gas at the cathode is (1 Faraday=96500 C `"mol"^(-1)`)A. `9.65xx10^(4)`secB. `19.3xx10^(4)` secC. `28.95xx10^(4)` secD. `38.6xx10^(4)` sec.

Answer» Correct Answer - B
`NaCl+aqrarrNa++Cl^(-)`
`H_(2)OhArrH^(+)+OH^(-)`
`H_(2)^(+)+e^(-)rarr+1//2H_(2)`
`therefore` 0.5 mole of `H_(2)` is liberated by 1F= 96500C
0.01 mole of `H_(2)` will be liberated by charge
`=(96 500)/(0.5)xx0.01=1930C`
`Q=1xxt` or `t=(Q)/(I)=(1930)/(10xx10^(-3))A`
`=193000"sec" =19.3xx10^(4)` sec.
Discussion
2842.

The cell , `Zn | Zn^(2+) (1M) || Cu^(2+) (1M) Cu (E_("cell")^@ = 1. 10 V)`, Was allowed to be completely discharfed at `298 K `. The relative concentration of `2+` to `Cu^(2+) [(Zn^(2=))/(Cu^(2+))]` is :A. `9.65xx10^(4)`B. antilog 24.08C. 37.3D. `10^(37.3)`

Answer» Correct Answer - D
The cell reaction is
`Zn+Cu^(2+)rarrZn^(2+)+Cu`
When the cell is completely discharged.
`E_("cell")=0`
Hence
` E_("cell")^(@)=(0.059)/( 2)"log"( [Zn^(2+)])/([Cu^(2+)])`
` 1.10=(0.0591)/(2)"log"([Zn^(2+)])/([C u^(2+)])`
or log `([Zn^(2+)])/(Cu^(2+)]=37.3`
or `([Zn^(2+)])/([Cu^(2+)])= 10^(37.3)`
Discussion
2843.

The cell , `Zn | Zn^(2+) (1M) || Cu^(2+) (1M) Cu (E_("cell")^@ = 1. 10 V)`, Was allowed to be completely discharfed at `298 K `. The relative concentration of `2+` to `Cu^(2+) [(Zn^(2=))/(Cu^(2+))]` is :A. `9.65xx10^(4)`B. Antilog 24.08C. 37.7D. `10^(37.3)`

Answer» Correct Answer - D
(d) `E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"[(Zn^(2+))/(Cu^(2+))]`
When cell is completely discharged, `E_(cell)=0`
`:. " " 0=E_(cell)^(@)-(0.0591)/(2)"log"([Zn^(2+)])/([Cu^(2+)])`
`0=1.1-(0.0591)/(2)"log"([Zn^(2+)])/([Cu^(2+)])`
`"log"([Zn^(2+)])/([Cu^(2+)])=(1.1xx2)/(0.0591)=37.3`
`([Zn^2+])/([Cu^(2+)])=10^(37.3)`
Discussion
2844.

A lead storage cell is discharged which causes the `H_(2)SO_(4)` electrolyte to change from a concentration of `34.6%` by weight (density `1.261 gmL^(-1)` at `25^(@)C`) to one of `27%` by weight. The original volume of electrolyte is one litre. How many faraday have left the anode of battry ? Note the water is produced by the cell reaction as `H_(2)SO_(4)` is used up. Over all reaction is : `Pb_(s) + PbO_(2) + 2H_(2)SO_(4(l)) rarr 2PbSO_(4(s)) + 2H_(2)O`

Answer» Correct Answer - `1.255` faraday
Discussion
2845.

Why on dilution the `Lambda_(m)` of `CH_(3)COOH` increases drastically, while that of `CH_(3)COONa` increases gradually?

Answer» `Ch_(3)COOH` is a weak electrolyte and is dissociated to small extent (`alpha` is quite small). With dilution, its degree of dissociation increases i.e. more ions are released in solution. Therefore, `Lambda_(m)` of `CH_(3)COOH` increases drastically. On the other hand , `CH_(3)COONa` is a strong electrolyte and is almost completely and is almost completely dissociated in aqueous solution (`alpha` is nearly one). With dilution, only interionic forces of attraction increase resulting in only small increase in the value of `Lambda_(m)`. For more details, consult section 3.19
Discussion
2846.

How much charge is required for the following reductions? (i) 1 mol of Al3+ to Al (ii) 1 mol of Cu2+ to Cu (iii) 1 mol of MnO4- to Mn2+

Answer»

(i) Al3+ + 3e → Al 

charge required = 3F 

(ii) Cu2+ + 2e → Cu 

charge required = 2F 

(iii) MnO4- + 8H+ + Se- → Mn2+ + H2O 

charge required = 5F
Discussion
2847.

How much Charge is required for the following reductions `:` `a. 1 mol of Al^(3+) to Al` `b.` ` 1 mol of Cu^(2+) to Cu` `c.` `1 mol of MnO_(4)^(-)` to `Mn^(2+)?`

Answer» (i) `" " Al^(3+)(aq)+underset(3" Faraday")(3e^(-))toAl(s)`
`"Charge" =3xxF=3xx96500" C"= 2.895xx10^(5)C`
(ii) `" " Cu^(2+)(aq)+ underset(2 " Faraday")(2e^(-))toCu(s)`
`"Charge"=2xxF=2xx96500" C "=1.93xx10^(5) C`.
(iii) `" " underset(O.N. of Mn=+7)(MnO_(4)^(-))+5e^(-) to underset(O.N. of Mn=+2)(Mn^(2+))`
`"Charge"=5xxF=5xx96500" C"=4.825xx10^(5)" C"`.
Discussion
2848.

Conductivity of 0.00241 M acetic acid solution is `7.896xx10^(-5)" S "cm^(-1)`. Calculate its molar conductivity in this solution. If `wedge_(m)^(@)` for acetic acid be 390.5 S `cm^(2)mol^(-1)`, what would be its dissociation constant?

Answer» Correct Answer - 32.76 S `cm^(2)mol^(-1).1.85xx10^(-5)`
`wedge_(m)^(c)=(kappaxx1000)/("Molarity")=(7.896xx10^(-5)" S "cm^(-1)xx1000cm^(3)L^(-1))/(0.00241" mol "L^(-1))=32.76" S "cm^(2)mol^(-1)`
`alpha=(32.76)/(390.5)=0.084,K=(calpha^(2))/(1-alpha)=(0.00241xx(0.084)^(2))/(1-0.084)=1.85xx10^(-5)`.
Discussion
2849.

Calculate `A^(oo)` (in `Scm^(2)"mol"^(-1)`) for `NH_(4)OH`, given that value of `A^(oo)` for `Ba(OH)_(2),BaCl_(2)` and `NH_(4)Cl` as 523.28.280.0 and 1298 `Scm^(2)"mol"^(-1)` respectively.A. 373.8B. 673.48C. 543.68D. 251.44

Answer» Correct Answer - d
Discussion
2850.

Calculate `lambda_(m)^(@)` for `NH_(4)OH` given that values of `lambda_(m)^(@)` for `Ba(OH)_(2) BaCI_(2) and NH_(4)CI as 523.28 280.0and 129.8 S cm^(2) mol^(-1)` respectively

Answer» `lambda_(m)^(@)NH_(4)OH=lambda_(m)^(@)+lambda_(m)^(@)(OH^(-))`
Now
`wedge_(m)^(@)NH_(4)OH=lambda_(m)^(@)NH_(4)^(@)(OH^(-))+1/2(Ba^(2+))-1/2lambda_(m)^(@)(CI^(-))-lambda_(m)^(@)(CI^(-))`
`=129.8 +1/2(523.28-280.0)=251.44 S cm^(2) mol^(-1)`
Discussion