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2301.

For the cell : `Zn|{:(Zn_(aq.)^(2+)),(1M):}||{:(Cu_(aq.)^(2+)),(2M):}|Cu` Calculate the values for , (a) cell reaction, (b) `E_(cell)^(@)` (c) `E_(cell)` (d) the minimum concentration of `Cu^(2+)` at which cell reaction is spontaneous if `Zn^(2+)` is `1M`, (e) does the displacement of `Cu^(2+)` goes almost to completion. Given : `E_(RP_(Cu^(2)//Cu))^(@) = +0.35V` `E_(RP_(Zn^(2)//Zn))^(@) = -0.76V`

Answer» `E_(OP)^(@)` for `Cu//Cu^(2+) = -0.35 V`
`E_(OP)^(@)` for `Zn//Zn^(2+) = +0.76 V`
More is `E_(OP)^(@)`, more is tendency to show oxidation and thus `Zn` will oxidize and `Cu^(2+)` will reduce.
Anode : `Zn rarr Zn^(2+)+2e`
Cathode : `Cu^(2+)+2e rarr Cu`
Cell reaction : `Zn+Cu^(2+) rarr Zn^(2+)+Cu`
(b) Also, `E_(cell)^(@) = E_(OP_(Zn//Zn^(2+)))^(@)+E_(RP_(Cu^(2+)//Cu))^(@)`
`= 0.76+0.35 = 1.11 V`
(c) Also, `E_(cell) = E_(Zn//Zn^(2+))^(@).^(+)E_(Cu^(2+)//Cu)`
`+E_(RP_(Cu//Cu^(2+)))^(@)+(0.059)/(2)log[Cu^(2+)]`
`E_(cell) = E_(OP_(Zn//Zn^(2+)))^(@)+E_(RP_(Cu//Cu^(2+)))^(@)+(0.059)/(2)log.([Cu^(2+)])/([Zn^(2+)])`
`= 0.76 + 0.35 + (0.059)/(2)log.(2)/(1)`
`E_(cell) = 1.109 V`
(d) Also,`E_(cell) = 1.1 + (0.059)/(2)log.([Cu^(2)])/([Zn^(2+)])`
Thus, if `E_(Cell)` is `+ve` for `[Zn^(2+)] = 1`,
so cell reaction is spontaneous.
`(0.059)/(2)log.([Cu^(2)])/([Zn^(2+)])gt -1.1`
or `log.([Cu^(2)])/([Zn^(2+)])gt -(1.1 xx 2)/(0.059) = -37.29`
or `log .([Cu^(2+)])/(1)gt -37.29`
or `[Cu^(2+)] gt 5.13 xx 10^(-38)`
(e) Yes, the displacement almost goes to completion.
Discussion
2302.

A button cell used in watched funcations as follwing `Zn(s)+Ag_(2)O(s)+H_(2)O(l)hArr2Ag(s)+Zn^(2+)(aq.)+2OH^(-)(aq)` If half cell potentials are `Zn^(2+)(aq.)+2e^(-) rarr Zn(s), E^(@) = -0.76 V` `Ag_(2)O(s)+H_(2)O(l)+2e^(-) rarr 2Ag(s)+2OH^(-)(aq.),, E^(@) = 0.34V` The cell potential will beA. 1.10VB. 0.42VC. 0.84VD. 1.3V

Answer» Correct Answer - A
For hy drogen electrode , reduction reaction is
`H^(+)+e=(1)/(2)H_(2)`
`E_(H^(+)//(1)/(2)H_(2))^(@)=E_(H^(+)//(1)/(2)H_(2))^(@)-(0.059)/(n)"log"(p^(1//2)H_(2))/([H^(+)])`
`=0-(0.059)/(n)"log"(1)/(10^(-10))=0.0591`
[pH=10means `|H^(+)| =10^(-1)M`]
`therefore` Oxidation potential `=0.591V`.
Discussion
2303.

For the cell reaction, `Cu_(C_(2))^(2+)(aq)+Zn(s) to Zn_(C_(1))^(2+)(aq)+Cu(s)` The change in free energy `(DeltaG)` at a given temperature is a function of :A. In `C_(1)`B. In `(c_(2)//c_(1))`C. In `(c_(1)+c_(2))`D. In `c_(2)`

Answer» Correct Answer - B
(b) `-DeltaG=nFE_(cell)=nFE_(cell)^(@)+RT" In "([Cu^(2+)])/([Zn^(2+)])`
`-DeltaG` is a function of In `(C_(2)//C_(1))`
Discussion
2304.

In the button cell, widely used in watches, the following reaction takes place `Zn(s)+Ag_(2)O(s)+H_(2)O to Zn^(2+)(aq)+2Ag(s)+2OH^(-)(aq)` Determine `E^(@)` and `DeltaG^(@)` for the reaction. (Given : `E_(Ag^(+)//Ag)^(@)=+0.80 V,E_(Zn^(2+)//Zn)^(@)=-0.76 V`

Answer» `"Anode: "Zn(s)toZn^(2+)(aq)+2e^(-)`
`underline(" ""Cathode: "2Ag^(+)(aq)+2e^(-)to2Ag(s)" ")`
`"Overall cell reaction :" " "Zn(s)+2Ag^(+)(aq)to Zn^(2+)(aq)+2Ag(s)`
`E_(cell)^(@)=(E_(cathode)^(@)-E_(anode)^(@))=(+0.80 V)-(-0.76 V)=+1.56 V`
`DeltaG^(@)=-"nF "E_(cell)^(@)=(-2)xx(965000 C)-(1.56 V)`
`=-301080" CV"=-301080" J "mol^(-1)=-301.08" kJ"mol^(-1)`.
Discussion
2305.

A button cell used in watches function as following `Zn(s)+Ag_(2)O(s)+H_(2)O(l)hArr2Ag(s)+Zn^(2+)(aq)+2OH^(-)(aq)` if half cell potential are `Zn^(2+)(aq)+2e^(-)toZn(s),` `Ag_(2)O(s)+H_(2)O(l)+2e^(-)to2Ag(s)+2OH^(-)(aq),E^(@)=0.34V ` The cell potential will beA. 1.34VB. 1.10VC. 0.42VD. 0.84V

Answer» Correct Answer - B
]`Zn^(2+)(aq)+2e^(-)toZn(s),E^(o)=-0.76V`
`Ag_(2)O(s)+H_(2)O(l)+2e^(-)to2Ag(s)+2OH^(-)(aq),E^(o)=0.34V`
`Zn(s)+Ag_(2)O(s)+H_(2)O(l)+2e^(-)to2Ag(s)+2OH^(-)(aq)`
`+2OH^(-)(aq),E_(cell)=?`
`E_(cell)^(o)=(E_(R.P.)^(o))_("cathode")-(E_(R.P.)^(o))_("anode")`
`E_(cell)^(o)=0.34(-0.76)=1.10V`
`E_(cell)=E_(cell)^(o)=1.10V`
Discussion
2306.

Determine `DeltaG^(0)` for the button cell used in the watches. The cell reactions is `Zn_((s))+Ag _(2)O_((s))H_(2)O _((l))to Zn_((aq))^(2+)+2 Ag _((s))+2OH_((aq))^(-)` `E_(Ag^(+)//Ag)^(0) =+0.80V,E_(Zn^(2+)//Zn)^(0)=-0.76V.`

Answer» Given `E^(0) of Ag^(+)//Ag =0.80 V`
`E^(0) of Z n ^(+2)//Zn =-0.76V`
Cell representation `Zn //Zn^(+2)||Ag^(+)//Ag`
`EMF=E_(RHS)-E_(LHS) =0.80-(-0.76)=1.56V`
`DeltaG=-nFE^(0)=-2xx96500xx1.53=-301.08kJ//"mole"`
Discussion
2307.

What is Rate of a reaction

Answer» Rate of reaction : The [speed or velocity] of a reaction is the change in concentration of reactants or products per unit time.
Rate `(Deltax)/(Deltat)` Where, `Deltax` is the change in concentration,` Deltat` is the time interval.
The rate of reaction is always positive and it decreases as the reaction proceeds.
For the reaction,` R to P,` rate may be expressed as
a) Rate of disappearance of `R=("Decrease in concentration of R")/("Time taken for the decrease")=-(DeltaR)/(Deltat)`
b) Rate of appearance of `P=("Increase in concentration of R")/("Time taken for the increase")=+ (DeltaP )/(Deltat)`
Discussion
2308.

A button cell used in watches functions as following `Zn(s) + Ag _2 O(s) + H_2 O(l)` `2 A g(s) + Zn^(2+) (aq) = 2 OH^- (aq)` If half cell potentials are `Zn^(2+) (aq) +2e^- rarr Zn (s) , E^@ =- 0. 76 V` `Af_2 O(s) +H_2 O (l) +2e^- rarr` `2Ag(s) + 2OH^- (aq) , E^@ =0. 34 V`. The cell potential will be .A. ` 0.42 V`B. ` 0. 84 V`C. ` 1.10 V`D. ` 1.34 V`

Answer» Correct Answer - C
`E_(cell) = E_(OPZ//Zn^(2) ) + E_(PRAg+//Ag)`
`E_(OPZn)^@ + E_(PRAg)^@ + (0.059)/2 log. ([Ag^+]^2)/([Zn^(+2)])`
`=0.76 + 0. 34 + (0.095)/5 log.(0)/([Zn^(2+)])`
`E_(cell)^@ = 1.10 V`.
This is `E_(cell)^@ ` and nor `E_(cell)` since`[ Zn^(2+)]` is not given .
Discussion
2309.

If every quantity is expressed in SI units, then molar conductivity `(wedge_(m))`, conducitivy `(kappa)` and molarity (M) are related as____.

Answer» `wedge_(m)=(kappa)/(M)`
Discussion
2310.

The molar conductivity of `0.025mol L^(-1)` methanoic acid is `46.1 S cm^(2)mol ^(-1).` Calculate its degree of dissociation and dissociation constant. Given, `lamda^(0)(H^(+))=349.6S cm^(2) mol ^(-1)` `and lamda^(0)(HCOO^(-))=54.6S cm^(2)mol^(-1)`

Answer» Step I: Calculatio of degree of dissociatiation (a) `HCOOH`
`^^_(m)^(c)=46.1S cm^(2) mol ^(-1)`
`^^_(m(HCOOH))^(0)=^^_(m(HCOO^(-)))^(0) +lamda_(m(H^(+)))^(0)`
`=(54.6+349.6)S cm^(2) mol ^(-1)`
`=404.2 S cm ^(2)mol ^(-1)`
`alpha =(^^_(m)^(c))/(^^_(m)^(0))=((46.1)S cm^(2)mol ^(-1))/((404.2)S cm^(2) mol ^(-1))=0.1140`
Step II : Calculation of dissociation constant
`HCOOH_((aq))overset("water")hArrHCOO_((aq))^(-)+H_((aq))^(+)`
`{:("Initial conc.", C, 0, 0),("Equlibrium conc", C(1-alpha), C alpha, C alpha):}`
Dissociation constant, `K_(a)=([HCOO^(-)][H^(+)])/([HCOOH])=(CalphaxxCalpha)/(C(1-alpha))=(Calpha^(2))/((1-alpha))`
Placing values `K_(a) =((0.025 mol L^(-1))xx (0.114)^(2))/((1-0.114))`
`=((3.249xx10^(-4)mol L^(-1)))/((0.886))`
`=3.67 xx10^(-4) mol L^(-1)`
`alpha= 0.114`
`K_(a) =3.67 xx10^(-4)mol L^(-1)`
Discussion
2311.

Given the cell: `Cd(s)|Cd(OH)_2(s)|NaOH(aq,0.01M)|H_2(g,1"bar")|Pt(s)` with `E_(cell)=0.0V."if"E_(Cd^(2+)|Cd)^(@)=-0.39V,"then"K_(sp)"of"Cd(OH_2)`is:A. 0.1B. `10^(-13)`C. `10^(-15)`D. None of these

Answer» Correct Answer - C
Discussion
2312.

Diffine conductivity of a material. Give its SI units.

Answer» The reciprocal of specific resistance (or) resistivity is called conductivity. It is represented by (k)
or
The conductance of one unit cube of a conductor is also called conductivity
SI unit : `ohm^(-1)m^(-1) (or)Sm^(-1) S =` Siemen
Discussion
2313.

Calculate the `e.m.f.` of cell `{:(Pt_(H_(2))),(1atm):}|{:(CH_(3)COOH),(0.1M):}||{:(NH_(4)OH),(0.01M):}|{:(Pt_(H_(2))),(1atm):}` `(K_(a)` for `CH_(3)COOH = 1.8 xx 10^(-5)`, `K_(b)` for `NH_(4)OH = 1.8 xx 10^(-5)`)

Answer» Correct Answer - `-0.4575` volt ;
Discussion
2314.

The `E^(o)` values of the following reduction reactions are given `Fe_((aq))^(3+)+etoFe_((aq))^(2+),E^(o)=0.771V` `Fe_((aq))^(2+)to2etoFe_((s)),E^(o)=-0.447V` What will be the free energy change for the reaction `Fe_((aq))^(3+)+3e^(-)toFe_((s))" "(1F=96485C" mol"^(-1))`A. `+18.51kJ" "mol^(-1)`B. `+11.87kJ" "mol^(-1)`C. `-8.10kJ" "mol^(-1)`D. `-10.41kJ" "mol^(-1)`

Answer» Correct Answer - B
`DeltaG^(o)=-nFE_(cell)^(o)`
`Fe_((aq))^(3+)+e^(-)toFe_((aq))^(2+),DeltaG_(1)^(o)=-Fxx0.771`
`Fe_((aq))^(2+)+2e^(-)toFe_((s)),DeltaG_(2)^(o)=-2xxFxx(-0.447)`
`=0.894F`
`Fe_((aq))^(3+)+3e^(-)toFe_((s)),DeltaG_(3)^(o)=?`
`DeltaG_(3)^(o)=DeltaG_(1)^(o)+DeltaG_(2)^(o)`
`DeltaG_(3)^(o)=-0.771F+0.894F`
`DeltaG_(3)^(o)=0.123F=0.123xx96485`
`DeltaG_(3)^(o)=11867.6J" "mol^(-1)` or `11.87kJ" "mol^(-1)`.
Discussion
2315.

Consider the cell `Pt|H_(2(g,1atm))|H_((aq.1M))^(+)||Fe_((aq))^(3+),Fe_((aq))^(2+)||Pt_((s))` Given that `E_(Fe^(3+)|Fe^(2+))^(o)=0.771V` the ratio of conc. Of `Fe_((aq))^(2+)` to `Fe_((aq))^(3+)` is, when the cell potential is 0.830VA. 0.101B. 0.924C. 0.12D. None of these

Answer» Correct Answer - A
For oxidation-reduction in half-cell:
The half-cell reaction is `Fe_((aq))^(3+)+etoFe_((aq))^(2+)`
`Fe_(Fe^(3+)//Fe^(2+))=E_(Fe^(3+)//Fe^(2+))^(o)-(0.059)/(1)log(([Fe^(2+)])/([Fe^(3+)]))`
`0.83=0.771-(0.0591)/(1)log(([Fe^(2+)])/([Fe^(3+)]))`
`implies([Fe^(2+)])/([Fe^(+3)])=0.10039~0.1004`
Discussion
2316.

Calculate the emf of the following cell at 298K `Fe_((s))|Fe^(2+)(0.001M)||H^(+)(1M)|H_(2(g))(1" bar"),Pt_((s))` (Given `E_(cell)^(@)=+0.44V`)

Answer» Correct Answer - 0.53 V
`Fe+2H^(+)toFe^(2+)+H_(2)`,
`E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"([Fe^(2+)])/([H^(+)]^(2))=0.44-(0.0591)/(2)"log"(0.001)/((1)^(2))`
Discussion
2317.

Calculate the equilibrium constant for the reaction at 298K: `NiO_(2)+2Cl^(-)+4H^(+)toCl^(2)+Ni^(2+)+2H_(2)O` if `E_(cell)^(@)=0.320V`.

Answer» Correct Answer - `6.747xx10^(10)`
`Ni^(4+)+2e^(-)toNi^(2+),2Cl^(-)toCl_(2)+2e^(-)thereforen=2`
Discussion
2318.

Calculate the equilibrium constant for the following reaction at 298K. `Cu(s)+Cl_(2)(g)toCuCl_(2)(aq)` `R=8.314JK^(-1)mol^(-1),E_(Cu^(2+)//Cu)^(@)=0.34V,E_(1//2" "Cl_(2)//Cl^(-))^(@)=1.36V,F=96500" C "mol^(-1)`

Answer» Correct Answer - `3.295xx10^(34)`.
Anode raction: `Cu(s)toCu^(2+)(aq)+2e^(-)`
Net cell reaction: `Cu(s)+Cl_(2)(g)toCuCl_(2)(aq)`
`E_(cell)^(@)=E_(cathode)^(@)-E_(anode)^(@)=1.36-0.34=1.02V`
`E_(cell)^(@)=(0.0591)/(n)log" "K_(c),i.e., 1.02=(0.0591)/(2)" log "K_(c)` or `log" "K_(c)=34.5178`
or `K_(c)=`antilog `34.5178=3.295xx10^(34)`.
Discussion
2319.

Calculate the half cell potential of a reaction `Ag_(2)S + 2e rarr 2Ag + S^(2-)` in a solution beffered at `pH = 3` and also saturated with `0.1 M H_(2)S. K_(1)` and `K_(2)` for `H_(2)S` are `10^(-8)` and `1.1 xx 10^(-8)` and `1.1 xx 10^(-13)` respectively. `(K_(SP_(Ag_(2)S)) = 2 xx 10^(-49), E_(Ag^(+)//Ag)^(@) = 0.8 V`)

Answer» Correct Answer - `-0.1658 V ;`
Discussion
2320.

Consider the cell `Ag|AgBr_((s)), Br^(-)||AgCl_((s)), Cl^(-)|Ag` at `25^(@)C`. The solubility product and `AgCl` and `AgBr` are `1 xx 10^(-10)` and `5 xx 10^(-13)` respectively. For what ratio of the concentrations of `Br^(-)` and `Cl^(-)` ions would the `e.m.f.` of the cell be zero ?

Answer» Correct Answer - `(1)/(200) ;`
Discussion
2321.

Consider the cell `AG|AgBr(s)|Br^(-)||AgCI(s)|CI^(-)|Ag` at `25^(@)C`. The solubility product constants of `AgBr & AgCI` are respectively `5 xx 10^(-13) & 1 xx 10^(-10)`. For what ratio of the concentration of `Br^(-) & CI^(-)` ions would the emf of the cell be zero?

Answer» Correct Answer - `[Br]: [CI] = 1:200`
`Ag|AgBr_((s)) |AgCI_((s)) |CI^(-)|Ag`
`E_(cell) =(0.0591)/(1)log. ([Ag^(+)]_(R.H.S))/([Ag^(+)]_(L.H.S))`
For `L.H.S`.
`K_(sp) of AgBr = 5 xx 10^(-13)`
`[Ag^(+)] [Br^(-)] = 5 xx 10^(-13)` ltbr? `[Ag^(+)]_(L.H.S) = (5 xx 10^(-13))/([Br^(-)])`
One for `R.H.S`
`[Ag^(+)] [CI^(-)] = K_(sp) of AgCI`
`[Ag^(+)]_(R.H.S) = (10^(-10))/([CI^(-1)])`
`10^(@) = (|Br^(-)|xx10^(10))/(|CI^(-)|xx5 xx 10^(-13))`
`(|Br^(-)|)/(|CI^(-)|) =(5)/(10^(3)) = (1)/(200)`
Discussion
2322.

Consider the cell `Ag(s)|AgBr(s)Br^(c-)(aq)||AgCl(s),Cl^(c-)(aq)|Ag(s)` at `298 K`. The `K_(sp)` of `AgBr` and `AgCl`, respectively are `5xx10^(-13)` and `1xx10^(-10)` . At what ratio of `[Br^(c-)]` and `[Cl^(c-)]` ions, `EMF_(cell)` would be zero ?A. `200:1`B. `1:200`C. `1:100`D. `1:500`

Answer» Correct Answer - a
Cell reaction `:`
`Ag(s) +Ag^(o+)(` from `AgCl) rarr Ag(s)+Ag^(o+)(` from `AgBr)`
`E_(cell)=E^(c-)._(cell)-(0.059)/(1)log .([Ag^(o+)]_(AgBr))/([Ag^(o+)]_(AgCl))`
`=0-0.059log .((K_(sp)(AgBr))/([Br^(c-)])xx([Cl^(c-)])/(K_(sp(AgCl))))`
For `E_(cell)=0`
`(K_(sp)(AgBr)xx[Cl^(c-)])/(K_(sp)(AgCl)xx[Br^(c-)])=1`
`([Cl^(c-)])/([Br^(c-)])xx(5xx10^(-13))/(1xx10^(-10))implies[(Cl^(c-))/[Br^(c-))]=(1)/(200)`
`[(Br^(c-))/(Cl^(c-))]=200:1`
Discussion
2323.

From the following information, calculate the solubility product of silver bromide ? `{:(AgBr(s)+e^(-)hArr Ag(s)+Br^(-)(aq.)E^(@)=0.07),(Ag^(+)(aq.)+e^(-)hArr Ag(s)E^(@)=0.80 V):}`A. `4 xx 10^(-7)`B. `4. xx 10^(-17)`C. `4 xx 10^(-13)`D. `4 xx 10^(-10)`

Answer» Correct Answer - C
We can even use reducation potentials to calculate the equilibrium constant of reactions that are not oxidation-reductions if they are the combination of two half-reactions listed in the `emf` series. Some examples of such reactions are the dissociation of complex ions intop the metal ion and its ligands, and the dissolutionof a sparingly soluble solid in water.
The reaction of interest that defines the `K_(eq.)` is
`AgBr(s)hArr Ag^(+)(aq.)+Br^(-)(aq.)`
One of the half-reactions in questions include solild silver bromide:
`AgBr(s)+e^(-)hArr Ag(s)+Br^(-)(aq.)E^(@)=0.07 V`
this half-reaction is the reduction of silver form the `+1` to the `0` oxidation state. to obtain an overall reaction, we must combine. It with another half-reaction that includes the same change in oxidation state. We find in question the half-reaction.
`Ag^(+)(aq.)+e^(-)hArr Ag(s)E^(@)=0.80V`
The overall reaction that we are tryping to form must have `Ag^(+)` on the right. This half reaction has `Ag^(+)` on the left, so it is the one that we must reaverse. Therefore, we change the sign of the `E^(@)` of this half-reaction. In this tabular form:
`{:(AgBr(s)+e^(-)hArrAg(s)+Br^(-)(aq.)" "E^(@)=0.07 V),(Ag(s)hArr Ag^(+)(aq.)+e^(-)" "E^(@)=-0.80 V),(bar(AgBr(s) hArr Ag^(+)(aq.).Br^(-)(aq.)E^(@)=0.73 V)):}`
The number of moles of electrons transferred, `n`, is `1`, the number of electrons that cancel when the half-reaction are combind. We calulate `K_(sp)` by using Equation 3.
`E_("cell")^(@) = (0.0592 V)/(n) log K`
`0.73 V = (0.592 V)/(1) log K_(sp)`
Solving for `log K_(sp)` we find
`log K_(sp) = - 12.4`
Now take the antilog of both sides
`K_(sp) = "antilog" (-12.4)`
`= 4 xx 10^(-13)`
The small equilibrium constant is consistant with the negative `E^(@)`.
Discussion
2324.

The molar conductivity of acetic acid solution at infinite dilution is 390.7 `Omega^(-1)cm^(2)mol^(-1)`. Calculate the moalr conductivity of 0.01M acetic acid solution, given that the dissociation of acetic acid is `1.8xx10^(-5)`

Answer» `underset(c(1-alpha))(CH_(3)COOH)hArrunderset(calpha)(CH_(3)COO^(-))+underset(calpha)(H^(+))`
`K_(eq)=calpha^(2)" or "alpha=sqrt((K_(eq))/(c))=(wedge_(m)^(c))/(wedge_(m)^(oo))`
`therefore(wedge_(m)^(c))/(390.7)=sqrt((1.8xx10^(-5))/(0.01))=sqrt(18xx10^(-4))=4.243xx10^(-2)` or `wedge_(m)^(c)=16.57Omega^(-1)cm^(2)`.
Discussion
2325.

A buffer solution is prepared in which the concentration of `NH_(3)` is 0.30M and the concentration of `NH_(4)^(+)` is 0.20M . If the equilibrium constant `k_(b)` for `NH_(3)` equals `1.8xx10^(-3)` what is the `pH` of this solution?A. 11.72B. 8.73C. 9.08D. 9.43.

Answer» Correct Answer - D
`[NH_(3)]=0.3M,[NH_(4)^(+)]=0.2M`
`k_(b)=1.8xx10^(-5)`
`NH_(3)+H_(2)OhArrNH_(4)^(+)+OH^(-)`
`K_(b)=([NH_(4)^(+)][OH^(-)])/([NH_(3)])rArr[OH^(-)]=(k_(b)xx[NH_(3)])/([NH_(4)^(+)])`
`[OH^(-)]=(1.8xx10^(-5)xx0.3)/(0.2)=2.7xx10^(-5)`
`pOH=-"log"[OH^(-)]=-"log"(2.7xx10^(-5))`
`=5-"log"2.7=5-0.43=4.57`
`therefore pH=14-pOH=14-4.57=9.43`
Discussion
2326.

`Cu^(2+)(aq.)` is unstable in solution and under goes simultaneous oxidation and reduction according to the reaction `2Cu^(+)(aq.)hArr Cu^(2+)(aq.)+Cu(s)` Choose the correct `E^(@)` for the above reaction if `E_(Cu^(2+))^(@)//Cu = 0.34 V and E_(Cu^(2+))^(@)//Cu^(+) = 0.15 V`A. `+0.49V`B. `+0.38V`C. `-0.19V`D. `-0.38V`

Answer» Correct Answer - B
See A-Level information `E_(Cu^(+)//Cu)^(@)=2E_(Cu^(2+)Cu^(2+)//Cu)-E_(Cu^(2+)//C u)=2xx0.337V-0.153V`
`=0.674V-0.153V=0.521V`
Discussion
2327.

Consider the following cell reaction. `2Fe(s)+O_(2)(g)+4H^(+)(aq)rarr2Fe^(2+)(aq)+2H_(2)O(l),` `E^(@)=1.67V` At `[Fe^(2+)]=10^(-3)M,P(O_(2))=0.1` atm and pH=3, the cell potential at `25^(@)C` isA. 1.47VB. 1.77VC. 1.87VD. 1.57V

Answer» Correct Answer - D
`pH=3rArr[H^(+)]=10^(-3)M`
`E_("cell")=E_("cell")^(@)-(0.059)/(4)"log"([Fe^(2+)]^(2))/([H^(+)]^(4)p(O_(2)))`
`=1.67-(0.059)/( 4)"log"(10^(-6))/((10^(-3))^(4)xx0.1)`
`=1. 67-(0.059)/(4)"log"(10^(-6+13)`
`=1.67-(0.059xx7)/(4)=1.67-0.1`
`=1.57V`
Discussion
2328.

Consider the following cell reaction `2Fe(s)+O_(2)(g)+4H^(+)(aq)rarr2Fe^(2+)(aq),2H_(2)O(l),E^(@)=1.67V` `At|Fe^(2+)| =10^(-3) M,P(O_(2))=0.1` atm pH =3, the cell potential at `25^(@)C` isA. 1.47VB. 1.77VC. 1.87VD. 1.57V

Answer» Correct Answer - D
Applying Nernst equation of the given reaction
`E_("cell")=E_("cell")^(@)-(0.0591)/(n)"log"([Fe ^(+2)]^ (2))/(P_(O_(2))xx[H^(+)]^(4))`
For the given reaction, n=4 pH=3 means
`[H^(+)]=10^(-3)M`
`therefore E_("cell")=1.67-(0.059)/(4)"log"((10^(-3))^(2))/(0.1xx(10^(-3))^(4))`
`=1.67-(0.059)/(4)"log"((10^(-1))^(2))/(0.1xx(10^(-3))^(4))`
`=1.69-(0.059)/(4)"log"10^(7)`
`=1.67-0.10=1.57V`
Discussion
2329.

Consider the following cell reation `:` `2Fe(s)+O_(2)(g)+4H^(o+)(aq) rarr 2Fe^(2+)(aq)+2H_(2)O(l)" "E^(c-)=1.67V` `At[Fe^(2+)]=10^(-3)M,p(O_(2))=0.1atm` and `pH=3` . The cell potential at `25^(@)C` isA. 1.47 VB. 1.77 VC. 1.87 VD. 1.57 V

Answer» Correct Answer - D
For the given cell reaction, `n=4, E^(@)=1.67 V`
`Q=([Fe^(2+)]^(2))/(p_(O_(2)xx[H^(+)]^(4)))=((10^(-3))^(2))/(0.1xx(10^(-3))^(4))=10^(-6)/10^(-13)=10^(7)`
`E=E^(@)-0.059/n log_(10) Q`
`=+1.67-0.059/4 log10^(7) =+1.67-0.10=1.57 V`
Discussion
2330.

which of the following will increase the voltage of the cell with following cell reaction? `Sn(s)+2Ag^(+)+(aq)rarrSn^(+2)(aq)+2Ag(s)`A. Increase in the size of silver rodB. Increase in the conc. Of `Sn^(2+)` ionsC. Increase in the concentration of `Ag^(+)` ionsD. Increase in the concentration of `Ag^(+)` ions

Answer» Correct Answer - C
Since reduction occurs at Ag electrode. Hence increase in the conc. Of `Ag^(+)` or decrease in conc. Of `Sn^(2+)` wil increase the voltage.
Discussion
2331.

Select the spontaneous rreactions from the changes given below `a. Sn^(4+)+2Fe^(2+)rarrSn^(2+)+2Fe^(3+)` `b. 2Fe^(2+)=I_(2)rarr2Fe^(3+)+2I^(c-)` `c. Sn^(4+)+2I^(c-)rarr Sn^(2+)+I_(2)` `d. Sn^(2+)+I_(2) rarr Sn^(4+)+2I^(c-)`

Answer» For `(a) Sn^(4+)` reduces to `Sn^(2+)` and `Fe^(2+)` oxidizes to `Fe^(3+)` .
`:. E^(c-)._(cell)=E^(c-)._(cell)+E^(c-)._(o x i d)`
`=E^(c-)._(Sn^(4+)|Sn^(2+))+E^(c-)._(Fe^(2+)|Fe^(3+))`
`=0.1 V-0.77V=-0.67V`
Hence, `E^(c-)._(cell)` is negative and the cell is non`-` spontaneous.
For `(b), I_(2)` reduces to `2I^(c-)` and `Fe^(2+)` oxidizes to `Fe^(3+)`
`:. E^(c-)._(cell)=E^(c-)._(red)+E^(c-)._(o x i d)`
`=E^(c-)._(I_(2)|2I^(c-))+E^(c-)._(Fe^(2+)|Fe^(3+))`
`=0.54V-0.77V=-0.23V`
Hence, `E^(c-)._(cell)` is negative and the cell is non`-` spontaneous.
For `(c),Sn^(4+)` reduces to `Sn^(2+)` and `EI^(c-)` oxides to `I_(2)`.
`:. E^(c-)._(cell)=E^(c-)._(red)+E^(c-)._(o x i d)`
`=E^(c-)._(Sn^(4+)|Sn^(2+))+E^(c-)._(2I^(c-)|I_(2))`
`=0.1V-0.54V=-0.44V`
Hence , `E^(c-)._(cell)` is negative and the cell is non`-` spontaneous.
For `(d),I_(2)` reduces to `2I^(c-)` and `Sn^(2+)` oxides to `Sn^(4+)`.
`:. E^(c-)._(cell)=E^(c-)._(red)+E^(c-)._(o x i d)`
`=E^(c-)._(I_(2)|2I^(c-))+E^(c-)._(Sn^(2+)|Sn^(4+))`
`=0.54V-0.1V=0.44V`
Hence,` E^(c-)._(cell)` is positive and the cell is spontaneous.
Discussion
2332.

In the electrochemical reaction, `2Fe^(3+)+Zn rarr Zn^(2+) + 2Fe^(2+)` increasing the concentration of `Fe^(2+)`:A. increasing cell emfB. increasing the current flowC. decreases the cell emfD. alters the pH of the solution .

Answer» Correct Answer - C
`E_("cell") = E_("cell")^(@) - (RT)/(nF) "ln" ([Fe^(2+)]^(2))/([Fe^(3+)]^(2))` .
Increasing `[Fe^(2+)]` decreases `E_("cell")` .
Discussion
2333.

which of the following will increase the voltage of the cell with following cell reaction? `Sn(s)+2Ag^(+)+(aq)rarrSn^(+2)(aq)+2Ag(s)`A. Increase in the size of the silver rodB. Increase in the concentration of `Sn^(2+)` ionsC. Increase in the concentration of `Ag^(+)` ionsD. None of the above .

Answer» Correct Answer - C
`E_("cell") = E_("cell")^(@) - (RT)/(nF) "ln" ([Sn^(2+)])/([Ag^(+)]^(2))` .
Discussion
2334.

If `E_(1),E_(2)` and `E_(3)` are the emf values of the three galvanic cells respectivley (i)`Zn|Zn^(2+)(1M)||Cu^(2+)(0.1M)|Cu` (ii) `Zn|Zn^(2+)(1M)||Cu^(2+)(1M)|Cu` (iii) `Zn|Zn^(2+)(0.1)||Cu^(2+)(1M)|Cu`. Which one of the following is true.A. `E_(2)gtE_(3)gtE_(1)`B. `E_(3)gtE_(2)gtE_(1)`C. `E_(1)gtE_(2)gtE_(3)`D. `E_(1)gtE_(3)gtE_(2)`.

Answer» Correct Answer - B
`Zn+Cu^(2+)rarrZn^(2+)+Cu`
`E_("cell")=E_("cell")^(@)-(0.0591)/(2)"log"([Zn^(2+)])/([Cu^(2+)])`
Lesser the `[Zn^(2+)]//[Cu^(2+)]` value more will be `E_("cell")`
for (i) `([Zn^(2+)])/([Cu^(2+)])=(1)/(01)=10`
for (ii) `([Zn^(2+)])/([Cu^(2+)])=(1)/(1)=1` and
for (iii) `([Zn^(2+)])/([Cu^(2+)])=(01)/(1)=0.1`
Hence `E_(3)gtE_(2)gtE_(1)`
Discussion
2335.

If `E_(1),E_(2)` and `E_(3)` are the emf values of the three galvanic cells respectivley (i)`Zn|Zn^(2+)(1M)||Cu^(2+)(0.1M)|Cu` (ii) `Zn|Zn^(2+)(1M)||Cu^(2+)(1M)|Cu` (iii) `Zn|Zn^(2+)(0.1)||Cu^(2+)(1M)|Cu`. Which one of the following is true.A. `E_(1) gt E_(2) gt E_(3)`B. `E_(3) gt E_(2) gt E_(1)`C. `E_(3) gt E_(1) gt E_(2)`D. `E_(2) gt E_(1) gt E_(3)`

Answer» Correct Answer - D
Cell reaction is `Zn + Cu^(2+) to Zn^(2+) + Cu`.
`E_("cell") = E_("cell")^(@) - (RT)/(nF) "ln" ([Zn^(2+)])/([Cu^(2+)])`
Greater the factor `[Zn^(2+)] | [Cu^(2+)]` , less in the EMF .
Discussion
2336.

The emf of the following three galvanic cells :`1. Zn//Zn^(2+) ( 1M) || Cu^(2+) (1 M) //Cu` `2. Zn//ZN^(2+ ) (1 M)|| Cu^(2+) //cu `3. Zn//Zn^(2+) (1 M) || Cu^(2+) (0.1 M) Cu` are repersented by `E_1, E_2, E_3` which of the following statement is true ?A. ` E_1 gt E_2 gt E_3`B. ` E-3 gtE_2 gt E_1`C. ` E_3 gt E_1 gt E_2`D. ` E_2 gt E_1 E_3`

Answer» Correct Answer - D
` E_("cell") = E_("cell")^2 - ( 2.303RT)/(nT) log. ([Zn^(2+)))/([Cu^(2+)])`
`E_1 = E_(cell)^@ - ( 2.303 RT)/(2T) log. (1)/(1)`
`E_("cell")^2`
`E_2 = E_(cell)^2 - ( 2.303 RT)/(2 F) log.(0.1)/(1.0)`
`E_3 = E_(cell)^@ - ( 2.303)/(2 F) 1/(0.1)`
` :. E_2 gt E_1 gt E_3`.
Discussion
2337.

`E_(1),E_(2)` and `E_(3)` are the emf values of the three galvanic cells respectively (i) `Zn|Zn^(2+)(1M)||Cu^(2+)(0.1M)|Cu` (ii) `Zn|Zn^(2+)(1M)||Cu^(2+)(1M)|Cu` (iii)` Zn|Zn^(2+)(0.1M)||Cu^(2+)(1M)|Cu` Which one of the following is true?A. `E_(2) gt E_(3) gt E_(1)`B. `E_(3) gt E_(2) gt E_(1)`C. `E_(1) gt E_(2) gt E_(3)`D. `E_(1) gt E_(3) gt E_(2)`.

Answer» Correct Answer - A
`Zn+Cu^(2+)toZn^(2+)+Cu`
`E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"([Zn^(2+)])/([Cu^(2+)])`
Thus, higher the value of `[Zn^(2+)]//[Cu^(2+)]`, lower is the emf of the cell, `([Zn^(2+)])/([Cu^(2+)])` for `(i)=(1)/(0.1)=10`,
for (ii)`=(1)/(1)=1`, for `(iii)=(0.1)/(1)=0.1`
Hence, `E_(3) gt E_(2) gt E_(1)`.
Discussion
2338.

The oxidation potential of hydrogen half-cell will be negative if:A. `p(H_(2))=1" atm and" [H^(+)]=1 M`B. `p(H_(2))=1" atm and" [H^(+)]=2 M`C. `p(H_(2))=0.2" atm and" [H^(+)]=1 M`D. `p(H_(2))=0.2" atm and" [H^(+)]=2 M`

Answer» Correct Answer - B::C
(b c ) The Nernst equation for the oxidation reaction :
`H_(2) to 2H^(+)+2e^(-)` is :
`E_("ox")=E_("ox")^(@)-(0.0591)/(n)"log"([H^(+)]^(2))/(pH_(2))`
`E_("ox")=-(0.0591)/(2)"log"([H^(+)])/(pH_(2))`
n=2 , `E_(ox)^(@)`
Let us determine `E_("ox")` in different cases : (a)`p_((H_(2)))=1" atm", [H^(+)]=1" M"`
`E_("ox")=-(0.0591)/(2)"log"((1)^(2))/(1)=0`
(b)`p_((H_(2)))=1" atm", [H^(+)]=1" M"`
`E_("ox")=-(0.0591)/(2)"log"((2)^(2))/(1)`
`=-(0.0591)/(2)xx2 log2=-0.178" V"`
(c )`p_((H_(2)))=0.2" atm", [H^(+)]=1" M"`
`E_("ox")=-(0.0591)/(2)"log"((1)^(2))/(0.2)=-(0.0591)/(2)log5`
`=-0.027" V"`
(d)`p_((H_(2)))=0.2" atm", [H^(+)]=0.2" M"`
`E_("ox")=-(0.0591)/(2)"log"((0.2)^(2))/(0.2)=-(0.0591)/(2)log0.2`
`=-(0.0591)/(2)xx(-0.6989)=+0.0207" V"`
Discussion
2339.

The emf of the following three galvanic cells :`1. Zn//Zn^(2+) ( 1M) || Cu^(2+) (1 M) //Cu` `2. Zn//ZN^(2+ ) (1 M)|| Cu^(2+) //cu `3. Zn//Zn^(2+) (1 M) || Cu^(2+) (0.1 M) Cu` are repersented by `E_1, E_2, E_3` which of the following statement is true ?A. `E_(1) gt E_(2) gt E_(3)`B. `E_(3) gt E_(2) gt E_(1)`C. `E_(3) gt E_(1) gt E_(2)`D. `E_(2) gt E_(1) gt E_(3)`.

Answer» Correct Answer - D
(d) According to Nernst equation
`E_(cell)=E_(cell)^(@)-(0.0591)/(n)"log"([Zn^(2+)])/([Cu^(2+)])`
In first case :`([Zn^(2+)])/([Cu^(2+)])=1`
`:. E_(cell)(E_(1))=E_(cell)^(@)`
In third case , `([Zn^(2+)])/([Cu^(2+)])=((1M))/((0.1 M))=10`
`E_(cell)(E_(3))=E_(cell)^(@)-(0.0591)/(2)`
`=E_(cell)^(@)-0.02955`
Thus,`E_(2) gt E_(1) gt E_(3)`.
Discussion
2340.

Sodium is made by the electrolysis of a molten mixture of about `40%` NaCl and `60%` CaCl_(2)` because

Answer» Correct Answer - B
Discussion
2341.

How long (in hours) must a current of 5 ampere be maintained to electroplate 60 g of calcium from molten `CaCl_(2)` ?A. 27 hoursB. 8.3 hoursC. 11 hoursD. 16 hours

Answer» Correct Answer - D
Discussion
2342.

For strong electrolytes, the plot or molar conductance versus `sqrt(C)` is:A. parabolicB. linearC. sinusoidalD. circular

Answer» Correct Answer - B
Discussion
2343.

The plot of molar conductance vs `sqrt(C)` in strong electrolyte , isA. circularB. linearC. parabolicD. sinusoidal

Answer» Correct Answer - B
The plot of molar conductance vs `sqrt(C)` in strong electrolyte is linear
Discussion
2344.

The variation of equivalent conductance verses concentration of a strong electrolyte is correctly given in the plotA. B. C. D.

Answer» Correct Answer - C
The value of equivalent conductance of a strong e,ectrp,ute decreases slighlty with increase in concentration.
Discussion
2345.

Write equation for the reaction between free energy change and standard cell potential.

Answer» `DeltaG^(@)=-nFE_(cell)^(2)`
Discussion
2346.

What is electrode potential?

Answer»

The potential difference between electrolyte Solution and electrode is called electrode potential. 
Discussion
2347.

Consider a cell given below :Cu | Cu2+ || Cl- | Cl2 , Pt(s)Write the reactions that occur at anode and cathode

Answer»

Cu → Cu2+ + 2e  At anode Cl2 + 2e → 2 Cl-  At cathode
Discussion
2348.

Depict the galvanic cell in which the reactionZn(s) +2Ag+ → (aq) Zn2+ (aq) +2Ag(s)takes place. Further show:(i) which of the electrode is negatively charged? (ii) the carriers of the current in the cell.(iii) Individual reaction of each electrode.

Answer»

Zn(s)|Zn2+(aq) || Ag+ (aq) | Ag (s)

(I) Zn- electrode

(ii) electrons flow from anode to cathode, current flows from cathode to anode

(iii) Reaction at Anode

 Zn (s)  → Zn 2+(aq) + 2e-

Reaction at Cathode:

Ag+ (aq) + e  → Ag(S
Discussion
2349.

Represent the cell in which the following reaction takes placeMg(s) + 2Ag+ (0.0001M) → Mg2+ (0.130 M) + 2Ag(s) Calculate its Ecell if Eo cell = 3.17 V . 

Answer»

Ecell = Eocell – (RT /2F) x ln{[Mg2+]/[Ag+]2

= 3.17 V - 0.059/2 x log 0.130/(0.0001)2

= 2.96 V
Discussion
2350.

What is the function of salt bridge?

Answer»

(i) To maintain the electrical neutrality of the electrolytes

(ii) To complete the internal circuit of the cell.
Discussion