Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

Calculate the equilibrium constant for the reaction between silver nitrate and metallic zinc.

Answer»

SOLUTION :Step 1 : Write the equation for the reaction
`2AG^(+)+Zn rArr Zn^(2+)+2Ag""E_("cell")^(@)=1.56V`
Step 2 : Substitute values in the NERNST equation at equilibrium
`kigK=(nE_("cell)^(@))/(0.591)`
`0=2xx1.56-0.03logK`
`-1.56xx2=-0.03logK`
`logK=(-1.56xx2)/(-0.03)=52.79`
`K=6.19xx10^(52)`
Discussion
2.

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»


Solution :`Ni^(4+)+2E^(-)toNi^(2+),2Cl^(-)toCl_(2)+2e^(-)thereforen=2`
Discussion
3.

Calculate the equilibrium constant for the reaction at 298K. 4Br^(-)+O_(2)+4H^(+)to2Br_(2)+2H_(2)O. Given that E_(cell)^(@)=0.16V.

Answer»


SOLUTION :`E_(cell)^(@)=(0.0591)/(N)"LOG "K_(c),0.16=(0.0591)/(4)" log "K_(c)` or `log" "K_(c)=10.8291` or `K_(c)=6.747xx10^(10)`
Discussion
4.

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

Answer»


ANSWER :`6.26xx10^(7)`
Discussion
5.

Calculate the equilibrium constant for the reaction, 2Fe^(3+)+3^(-)hArr2Fe^(2+)+I_(3)^(-). The standard reduction potentials in acidic condtitions are 0.77 and 0.54 V respectively for Fe^(3+)//Fe^(2+) and I_(3)^(-)//I^(-) couples.

Answer»


SOLUTION :CELL is `I^(-)|I_(3)^(-)||Fe^(3+)|Fe^(2+),E_(cell)^(@)=0.77-0.54=0.23V,n=2`
Discussion
6.

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»


Solution :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
7.

Calculate the equilibrium constant for the reactant at 298 K Cu (s) + 2 Ag^(+) (aq) to Cu^(2+) (aq) + 2 Ag (s) Given that E_((Ag^(+) |Ag))^(Theta) = 0.80 Vand E_((Cu^(2+) |Cu))^(Theta) = 0.34 V .

Answer»

Solution :`E_(cell)^(Theta)` is RELATED to equilibrum constant `K_(E)` as : 298 K as:
`E_(cell)^(Theta) = (0.059)/(n) log K_(c) ` or `log K_(c) = (n E_(cell)^(Theta))/(0.059)`
`E_(cell)^(Theta) = E_((Ag^(+) | Ag))^(Theta) - E_((Cu^(2+) |Cu))^(Theta)`
`= 0.80 - 0.34 = 0.46` V
or log `K_(c) = (2 xx 0.46)/(0.059) = 15.59`
`THEREFORE "" K_(c) = 3.92 xx 10^(15)` .
Discussion
8.

Calculate the equilibrium constant for the following reaction at 298 K: Cu(s) + Cl_(2) (g) to CuCl_(2) (aq) Given: R =8.314 J K^(-1) mol^(-1) , E_(Cu^(2+)//Cu)^(@) = 0.34 V, E_(1/2Cl_(2)//Cl^(-))^(@) =1.36 V, 1 F = 96500 C mol^(-1)

Answer»

Solution :`E_("cell")^(@) =E_(1/2Cl_(2)//Cl^(-))^(@) - E_(Cu^(2+)//Cu)^(@) = +1.36 V - 0.34 V = 1.02 V`
Applying the following relation and substituting the VALUES, we get
`LOG K = (nE^(@))/(0.0591) = (2 xx 1.02)/(0.0591) = 2.04/(0.0591) = 34.5177`
K = Antilog `34.5177 = 3.294 xx 10^(34)`
THUS, EQUILIBRIUM constant `=3.294 xx 10^(34)`
Discussion
9.

Calculate the equilibrium constant at 25^@C for the reaction:2Fe^(3+) + 2I^(-) = 2Fe^(2+) + I_2Given that E^@ Fe^(3+) , Fe^(2+) = 0.77 V, E^@ I_2, I^(-) = 0.536V

Answer»

SOLUTION :`8.29 XX 10^7`
Discussion
10.

Calculate the equilibrium constant for the cell obtained by connecting two electrode E_((Sn^(2+)|Sn))^(Theta)=0.14V and E_((Ni^(2+)|Ni))^(Theta)=-0.23V at 298K Temperature.

Answer»


ANSWER :`K_(C)=1.122xx10^(3)`
Discussion
11.

Calculate the entropy of a substance at 600 K using the following data. (i) Heat capacity of solid from 0 K to normal melting point 200 K C_("P.m.")(s)=0.035T" "JK^(-1)mol^(-1) (ii) Enthalpy of fusion = 7.5kJmol^(-1), (iii) Enthalpy of vaporisation = 30kJmol^(-1). (iv) Heat capacity of liquid from 200 K to normal boiling point 300KC_("P.m")(l)=60+0.016TJK^(-1)mol^(-1) (v) Heat capacity of gas from 300 K to 600 K at 1 atm C_("P.m")=50.0""JK^(-1)mol^(-1).

Answer»

SOLUTION :`205.08JK^(-1)MOL^(-1)`
Discussion
12.

Calculate the entropy increase in the evaporation of 1 mole of a liquid when itboils at 100^@C having heat of vaporisation at 100^@C as 540 cals\gm.

Answer»

Solution :`DELTA S= 26.06 CAL K^(-1) MOL^(-1)`
Discussion
13.

Calculate the entropy change when 1 kg of water is heated from 27^(@)C to 200^(@)C forming super heated steam under constant pressure. Given specific heat of water = 4180 J/kg-K and specific heat of steam = 1670 + 0.49 T J/kg-K and latent heat of vaporization = 23 xx 10^(5) J/kg

Answer»


ANSWER :7522.50 J
Discussion
14.

Calculate the enthalpy of vaporisation of I_(2) if the sublimation energy and enthalpy of fusion of I_(2) is 57.3 kJ mol^(-1) and 15.5 kJ mol^(-1) respectively

Answer»

`-72.8" KJ "mol^(-1)`
`72.8" kJ "mol^(-1)`
`-41.8" kJ "mol^(-1)`
`+41.8" kJ "mol^(-1)`

Solution :(i) `I_(2)(s) to I_(2)(g),DeltaH_(1)=+57.3" kJ "mol^(-1)`
(ii) `I_(2)(s) to I_(2)(l),DeltaH_(2)=+15.5" kJ "mol^(-1)`
(III) `I_(2)(l)toI_(2)(g),DeltaH=?`
EQUATION (iii) can be achieved by =(i)-(ii)
`DeltaH=57.3-15.5=+41.8kJ" "mol^(-1)`
Discussion
15.

Calculate the enthalpy of formation of sulphuric acid (l) from the following data : (i) S(s)+O_(2)(g) to SO_(2)(g),DeltaH=-71.0"kcal" (ii) SO_(2)(g)+(1)/(2)O_(2)(g) to SO_(3)(g),DeltaH=-23.5"kcal" (iii) SO_(3)(g)+H_(2)O(l) to H_(2)SO_(4)(l), DeltaH=-31.2 "kcal" (iv) H_(2)(g)+(1)/(2)O_(2)(g) to H_(2)O(l),DeltaH=-68.5"kcal"

Answer»

Solution :From the above EQUATIONS, we have to calculate `DeltaH` for the equation
`H_(2)(G)+S(s)+2O_(2)(g) to H_(2)SO_(4)(l),DeltaH-=?`
Applying the INSPECTION METHOD,
[Eqn. (i)+Eqn. (ii) + Eqn. (III)+Eqn.(iv)], we get,
`S(s)+O_(2)(g)+SO_(2)+(1)/(2)O_(2)(g)+SO_(3)(g)+H_(2)O(l)+H_(2)(g) to (1)/(2)O_(2)(g) to SO_(2)(g)+SO_(3)(g)+H_(2)SO_(4)(l)+H_(2)O(l), DeltaH=(-71.0-23.5-31.2-68.5) kcal
or `H_(2)(g)+S(s)+2O_(2)(g) to H_(2)SO_(4)(l),DeltaH=-194.2"kcal"`
Discussion
16.

Calculate the enthalpy of formation of HBr (g) from the following data : SO_(2)(aq)+(1)/(2)O_(2)(g) to SO_(3)(aq), DeltaH=-63.7 kcal 2Br(g)+SO_(2)(aq)+H_(2)O(l) to 2HBr (aq)+SO_(3)(aq), DeltaH=-54 kcal H_(2)(g)+(1)/(2)O_(2)(g) to H_(2)O (l), DeltaH=-68.4 kcal HBr(g)+aq to HBr (aq), DeltaH=-20 kcal

Answer»


ANSWER :(-9.4 KCAL)
Discussion
17.

Calculate the enthalpy of formation of Ca(OH)_(2) (s) from the following data : H_(2)(g)+(1)/(2)O_(2)(g) to H_(2)O(l), DeltaH=-68.3 kcal CaO(s)+H_(2)O(l) to Ca(OH)_(2)(s), DeltaH=-15.3 kcal Ca(s)+(1)/(2)O_(2)(g) to CaO(s), DeltaH=-151.8 kcal

Answer»


ANSWER :(-235.4 KCAL)
Discussion
18.

Calculate the enthalpy of formation of acetic acid from the following data. CH_3COOH(I) +202 (g) — 2C02 (g) +2H20(I),"" Delta H= -207.9 kcal C(s) +O_2 (g) —C0_2 (g) . ..(ii)Delta H= -94.48 kcal H_(2)(g)+(1)/(2)O_(2)(g)to H_(2)O(l),""Delta H =-68.4k cal

Answer»

Solution :The REQUIRED equation is ,
`2C(s)+2H_(2)(G)+O_(2)(g)to CH_(3)COOH(l), "" Delta H=?`
This equation can be obtained by multiplying equation (ii) by 2 and also equation (iii) by 2 and ADDING both and finallysubtracting equation (i).
`2C(s)+2O_(2)(g)+2H_(2)(g)+O_(2)(g)-CH_(3)COOH(l)-2O_(2)(g)to 2CO_(2)(g)+2H_(2)O(l)-2CO_(2)(g)-2H_(2)O(l)`
`2C(s)+2H_(2)(g)+O_(2)(g)to CH_(3)COOH(l)`
`Delta_(f)H[CH_(3)COOH(l)=2xx(-94.48)+2xx(-68.4)-(-207.9)`
`=-188.96-136.8+207.9`
`=-325.76=-117.86` K cal.
Discussion
19.

Calculate the enthalpy of formation of acetic acid from the following data.CH_(3)COOH(l)+2O_(2)(g)rarr 2CO_(2)(g)+2H_(2)O(l),"" Delta H=-207.9 k cal

Answer»

Solution :The required equation is ,
`2C(s)+2H_(2)(g)+O_(2)(g)to CH_(3)COOH(l), "" Delta H=?`
This equation can be obtained by multiplying equation (II) by 2 and also equation (iii) by 2 and adding both and finallysubtracting equation (i).
`2C(s)+2O_(2)(g)+2H_(2)(g)+O_(2)(g)-CH_(3)COOH(l)-2O_(2)(g)to 2CO_(2)(g)+2H_(2)O(l)-2CO_(2)(g)-2H_(2)O(l)`
`2C(s)+2H_(2)(g)+O_(2)(g)to CH_(3)COOH(l)`
`Delta_(f)H[CH_(3)COOH(l)=2xx(-94.48)+2xx(-68.4)-(-207.9)`
`=-188.96-136.8+207.9`
`=-325.76=-117.86` k cal.
Discussion
20.

Calculate the enthalpy of combustion ofmethyl alcohol at 298 K from the following data {:("Bond",C-H,C-O,O-H,O=O,C=O),("Bond Enthalpy" (kJ mol^(-1)),414,351.5,464.5,494,711):} Resonance energy of CO_(2)=-143" kJ mol"^(-1) Latent heat of vaporisation of methyl alcohol =35.5" kJ mol"^(-1) Latent heat of vaporisation of water =40.6" kJ mol"^(-1).

Answer»


ANSWER :`-669.7" KJ MOL"^(-1)`
Discussion
21.

Calculate the enthalpy of combustion of benzene (l) on the basis of the following data: (i) Resonance energy of benzene (l) = -152 kJ/mol. (ii) Enthalpy of hydrogenation of cyclohexene (l) = -119 kJ/mol. (iii) DeltaH_(l)^(@) of H_(2)O (l) = -285.8 kJ/mol. (iv) DeltaH_(l)^(@) of H_(2)O (l) = -285.8 kJ/mol (v) DeltaH_(l)^(@) of CO_(2)(g) = -393.5 kJ/mol.

Answer»

SOLUTION :
`DeltaH_(1)^(@)` (benzene) =-156 + 357 = 201 kJ/mol.
RESONANCE energy `=DeltaH_(L)^(@) - DeltaH_(l)^(@)`
`DeltaH_(l)`(benzene) = -152 + 201 = 49 kJ/mol.
Actual + `15/2 O_(2)(l) to 6CO_(2)(G) + 3H_(2)O(l)`.
`DeltaH^(@) = 6DeltaH_(l)^(@) CO_(2)(g) + 3DeltaH_(l)^(@)H_(2)O(l) - DeltaH_(l)^(@)` (benzene)
`=6(-393.5) + 3(-285.8)-49`
`=-3267.4` kJ/mol.
Discussion
22.

Calculate the enthalpy change when infinitely solution of CaCl_(2) and Na_(2)CO_(3) are mixed Delta_(f)H^(@) for Ca^(2+)(aq, CO_(3)^(2-) (aq) and CaCO_(3) (s) are -129.80, -161.65, -288.5" kCal mol"^(-1) respectively.

Answer»


SOLUTION :`CA^(+2) + CO_(3)^(-2) rarr CaCO_(3)`
`Delta H = - 288.5 - [-129.8 - 161.65] = 2.95 KCAL`
Discussion
23.

Calculate the enthalpy change on freezing of 1.0 mole of water at 10.0^(@)C to ice at -10.0^(@)C [Delta_("fus")H=6.03kJmol^(-1) at 0^(@)C] C_(p)[H_(2)O(l)]=75.3Jmol^(-1)K^(-1) C_(p)[H_(2)O(s)]=36.8Jmol^(-1)K^(-1)

Answer»

`-753Jmol^(-1)`
`-368Jmol^(-1)`
`-7.151kJmol^(-1)`
`-6.03kJmol^(-1)`

SOLUTION :Enthalpy change for the conversion of 1 mole of LIQUID water at `10^(@)C` into 1 mole of liquid water at `0^(@)`C,
`DeltaH_(1)=C_(p)[H_(2)O(l)]xxDeltaT`
`=-75.3"J "mol^(-1)K^(-1)xx10K=-753J" "mol^(-1)`
Enthalpy of fusion,
`DeltaH_(2)=DeltaH_("freezing")=-DeltaH_("fusion")=-6.03"kJ "mol^(-1)`
Enthalpy change for the conversion of 1 mole of ice at `0^(@)C`
to 1 mole of ice at `10^(@)C`,
`DeltaH_(3)=C_(p)[H_(2)O(s)]xxDeltaT=-36.8J" "mol^(-1)K^(-1)xx10K`
`=-368"J "mol^(-1)`
`DeltaH_("total")=-(0.753+6.03+0.368)=-7.151" kJ "mol^(-1)`
Discussion
24.

Calculate the enthalpy change for the reaction Fe_(2)O_(3)+3CO to 2Fe+3CO_(2), from the following data : 2Fe+(3)/(2)O_(2) to Fe_(2)O_(3),DeltaH=-177.1 kcal C+(1)/(2)O_(2) to CO, DeltaH=-32.8 kcal C+O_(2) to CO_(2), DeltaH=-94.3 kcal

Answer»


ANSWER :(-8.4 KCAL)
Discussion
25.

Calculate the enthalpy change accompanying the conversion of 10 g of graphite into diamond if the heats of combustion of C (graphite) and C (diamond) are -94.05 and -94.50 kcal respectively.

Answer»

Solution :Given that,
(i) `C("GRAPHITE")+O_(2)(g) to CO_(2)(g),DeltaH=-94.05"kcal"`
(ii) `C("diamond")+O_(2)(g) to CO_(2)(g),DeltaH=-94.05"kcal"`
Thus, applying the inspection method,
[Eqn. (i)- Eqn. (ii)], we get,
`C("graphite")+O_(2)(g)-C("dimond")-O_(2)(g) to CO_(2)(g)-CO_(2)(g),DeltaH=-94.05-(-94.50)`
or `C("graphite") to C("dimond"),DeltaH=+0.45"kcal"`
SINCE this enthalpy CHANGE is only for conversion of 1 "mole", i.e., 12 g of `C("graphite")`to `C("diamond")`, therefore, for the conversion of 10 g of C ("graphite") to C("diamond")
`DeltaH=0.45xx(10)/(12)=0.375"kcal"`
Discussion
26.

Calculate the enthalpy change for the given reaction from data provided (kJ/mole) {:(HA(g),+,B(g),rarr,AHB(s)),("(weak acid)",,"(weak acid)",,"(Salt)"):} DeltaH_("neutralization") {HA(aq) ["at infinite dilution"]//B(aq)["at infinite dilution"]}=-40" kJ/mole" {:(DeltaH_("solution")[HA(g)],=-10" kJ/mole",{"at infinite dilution"}),(DeltaH_("solution")[B(g)],=-5" kJ/mole",{"at infinite dilution"}),(DeltaH_("solution")[HAB(s)],=+8" kJ/mole",{"at infinite dilution"}):}

Answer»

`-36`
`-63`
`-45`
`-37`

Solution :`{:(""HA(g),+,""B(g),overset(DeltaH)(RARR),""AHB(s)),(+nH_(2)O darrDeltaH_(1),,+nH_(2)O darr DeltaH_(2),,+nH_(2)O darr DeltaH_(4)),(""HA(aq),+,""B(aq),overset(DeltaH_(3))(rarr)," "A^(-) (aq)+B^(+)(aq)):}`
`DeltaH=DeltaH_(1)+DeltaH_(2)+DeltaH_(3)-DeltaH_(4)`
`=-10-5-8-40=-63" kJ/mole"`
Discussion
27.

Calculate the energy required to remove an electron completely from n = 2 orbit. What is the longest wave length of light that can be used to cause this transition?

Answer»

SOLUTION :`3647Å`
Discussion
28.

Calculate the energy required to excite 1 litre of H_2gas at 1 atm and 298 K to thefirst excited state of atomic hydrogen. The energy for the dissociation of H-H bonds is 436 kJ "mol"^(-1) . Rydberg constant for H = 109679 cm^(-1) , h = 6.626 xx 10^(-34) Js and C = 3 xx 10^8 m s^(-1) .

Answer»

Solution :Mole of H = 2 x mole `H_2 = 2 xx(PV)/(RT)`
Energy to EXCITE 1 H atom `=hv = (HC)/(lamda) = hcR ((1)/(n_1) - (1)/(n_2))`
Total energy = energy to break H-H bonds + energy to excite H atom
98.19kJ
Discussion
29.

Calculate the energy of an electron in the first Bohr orbit of hydrogen.

Answer»

SOLUTION :For hydrogen, Z=1, n=1, we have,
`E_(1)= - (2pi^(2) e^(4)m)/(H^(2))` ….(EQN 2)
`=-(2 xx (3.14)^(2) xx (4.8xx 10^(-10))^(4) xx (9 xx 10^(-28)))/((6.63 xx 10^(-27))^(2))`
`= -2.18 xx 10^(-11)` erg
Discussion
30.

Calculate the energy of an electron in 3rd Bohr orbit.

Answer»


ANSWER :`-2.42xx10^(-19)J`
Discussion
31.

Calculate the energy in eV required to ionise 1 mole of hydrogen

Answer»

SOLUTION :`8.189 XX 10^(24)EV`
Discussion
32.

Calculate the energy in calories required to produce, from neutral He atoms, 1 mole of (a) He^(+) ions (b ) 'He^(+ +) ions using Bohr's equations

Answer»


SOLUTION :1,821000 CAL
Discussion
33.

Calculate the energy emitted when electrons of 1.0 g of hydrogen atoms undergoestransition giving the spectral line of lowest energy in the visible region of its atomic spectrum.R_H = 1.1 xx 10^7 m^(-1) , c = 3 xx 10^8 ms^(-1) , h = 6.62 xx 10^(-34) Js .

Answer»

SOLUTION :182.656kJ
Discussion
34.

Calculate the energy associated with the first orbit of He^(+) . What is the radius of this orbit?

Answer»

SOLUTION :`-54.4eV, 0.2645Å`
Discussion
35.

Calculate the empirical formula of a mineral having the following composition : CaO=48.0%, P_(2)O_(2)=41%,CaCl_(2)=10.7%

Answer»

SOLUTION :`9CaO.3P_(2)O_(5).CaCl_(2)`
Discussion
36.

Calculate the empirical formula of gold chloride which contains 35.1% of chlorine. At mass of Au = 197.

Answer»


Solution :`{:("CONSTITUENT",%"age","Mol. MASS","Moles of CONST."=(%)/("Mol. mass"),"SIMPLEST Ratio"),(CaO,48.0,56,48.0//56=0.857,9),(P_(2)O_(5),41.3,142,41.3//142=0.291,3),(CaCl_(2),10.7,111,10.7//111=0.096,1):}`
`therefore"Empirical Formula"=9CaO.3P_(2)O_(5).CaCl_(2).`
Discussion
37.

Calculate the empirical formula for a compound that contains 26.6% potassium, 35.4% chromium and 38.1% oxygen. [Given K=39.1,Cr=52,O=16]

Answer»

Solution :
Therefore, EMPIRICAL FORMULA is `K_(2)Cr_(2)O_(7)`.
Discussion
38.

Calculate the EMF of Zn//Zn^(2+)(0.1)"//"Cu^(2+)(0.1)//Cu E^(@) of Zn^(2+)//Zn = 0.762V , E^(@) of Cu^(2+)//Cu = +0.337V

Answer»


ANSWER :1.099V
Discussion
39.

Calculate the EMF of the following concentration cells a 30^(@)C and predict whether the cells are exergonic or endergonic. [ Assume Kw does not chage at 30^(@)C] a. Pt|H_(2)(g)(1 atm)|H^(o+)(10^(-6)M)||H^(o+)(10^(-4)M)|H_(2)(g)(1atm)|Pt b. Pt||Hg_(2)(1atm)|NaOH(10^(-4)M)||H^(o+)(10^(-5)M)|H_(2)(g),(1 atm)|Pt c.Pt|H_(2)(g)(1 atm)|H_(2)SO_(4)(0.05M)||KOH(10^(-3)M)|H_(2)(g)(1atm)|Pt d.Pt|H_(2)(g)(1 atm)|CH_(2)COOH(10^(-2)M)||CsOH(10^(-3)M)|H_(2)(g)(1atm)|Pt(pK_(a) of CH_(3)COOH=4.74) e. Pt|H_(2)(g)(1atm)|H_(2)O||HCl(10^(-3)M)|H_(2)(g)(1 atm)|Pt f.

Answer»

Solution :`a.""[H^(o+)]_(cathode )gt[H^(o+)]_(anode)or pH_(cathode)ltpH_(anode )`
So `EMF` of the cell will be positive and the cell will be spontaneous or exergonix `(i.e., DeltaG=-ve).`
`(pH)_(c)=4,(pH)_(a)=6`
`:.E_(cell)=-0.060(pH_(c)-pH_(a))=-0.060(4-6)=0.12V`
`b. (pH)_(c)=5,(pOH)_(a)=4`,or `(pH)_(a)=14-4=10`
SINCE `(pH)_(c)lt(pH)_(a)`, so `EMF` of the cell will be positive and the cell will be spontaneous or exergonic.
`:. E_(cell) =-0.06(5-10)=0.30V`
`c.` Since`[H_(2)SO_(4)]=[H^(o+)]_(a)=0.05xx2(n` factor `)`
`=0.1N=10^(-1)N`
`pH_(a)=1,(pOH)_(c)=3,pH_(c)=14-3=11`
Since `pH_(c)gtpH_(a)` so `EMF` of cell will be negative and the cell will not be feasible or non`-` spontaneous or endergonic `(i.e., Delta G=+ve)`.
`E_(cell)=-0.06(pH_(c)-pH_(a))=-0.06(11-1)=-0.6V.`
`d. CH_(3)COOH`is a weak acid and `CsOH` is a strong base.
THUS, `pH_(wA)=(1)/(2)(pK_(a)-log c)`
`=(1)/(2)(4.74-log 10^(-2))`
`=(1)/(2)(4.74+2)=3.37=pH_(a)`
`(pOH)_(c)=3,impliespH_(c)=14-3=11`
Since `pH_(c)gtpH_(a),` so `EMF` of cell will be negative and is endergonic.
`E_(cell)=-0.06(pH_(c)-pH_(a))=-0.06(11-3.37)`
`=-0.06xx7.63`
`=-0.457V`
`e.""pH` of `H_(2)O=7=pHa`
`pH_(c)=3 `
Since `pH_(c)ltpH_(a),EMF_(cell)=+ve,` cell is exergonic.
`E_(cell)=-0.06(pH_(c)-pH_(a))=-0.06(3-7)=0.24V`
`f.` `NH_(4)OH` is a weak base and `RbOH` is a strong base.
Thus, `pOH_(wB)=(1)/(2)(pK_(b)-log C)=(1)/(2)(4.74-log10^(-2))=3.37`
`pH_(a)=14-3.37=10.63`
`(pOH)=3,impliespH_(c)=14-3=11`
Since `pH_(c)gtpH_(a)`, therefore, `EMF_(cell)=-ve,` cell will be endergonic.
`E_(cell)=-0.06(pH_(c)-pH_(a))=-0.06(11-10.63)`
`=-0.022V`
`g.` Mixture of `CH_(3)COOH(W_(A))` and salt of `W_(A)//S_(B)`
`(CH_(3)COONa)` isan acidic buffer. Thus,
`:. pH_(ACI d b uffer )=pK_(a)+lo.([Sal t])/([Aci d])`
`=4.74+log.(10^(-1)M)/(10^(-2)M)`
`=4.74+1=5.74`
Mixture of `NH_(4)OH(W_(B))` and salt of `W_(B)//S_(A)(NH_(4)Cl)` isa basic buffer.
`:. pOH_(basic buffer)=pK_(a)+lo.([Sa l t])/([Base])`
`=4.74+log `(0.4M)/(0.2M)`
`=4.74+log 2 `
`=4.74+0.3 = 5.04`
`:. pH_9c)=14-5.04=8.96`
Since `pH_(c)gtpH_(a)`
`:. EMF_(cell)=-ve,` cell is endergonic.
`E_(cell)=-0.06(pH_(c)-pH_(a))=-0.06(8.96-5.4)`
`=-0.06xx3.22`
`=-0.1932V`
Discussion
40.

Calculate the EMF of the following concentration cells at 30^(@)C and predict whether the cells are exergonic or endergonic.

Answer»

Solution :Note that at `30^(@)C` , the value of `2.303(RT)/(F)=0.06`.
`a. CH_(3)COONa` is a salt of `W_(A)//S_(B).` Thus
`pH_(a)=(1)/(2)(pK_(w)+pK_(a)=log c)`
`(1)/(2)(14+4.74=log10^(-2))`
`=(1)/(2)(14+4.74-2)=8.37`
`gt.Ph_(a)=8.37`
`NH_(4)NOH_(3)` is a salt of`W_(B)//S_(A)`. Thus,
`pH_(c)=(1)/(2)(pK_(w)-pK_(b)-log c )`
`=(1)/(2)(14-4.74-log 0.2)`
`=(1)/(2)[14-4.74-log 2xx10^(-1)]`
`=(1)/(2)(14-4.74-0.3+1)=4.98.`
Since `pH_(c)ltpH_(a)`, therefore,`EMF_(cell)=+ve` and the cell will be exergonic.
`E_(cell)=-0.06(pH_(c)-pH_(a))=-0.06(4.98-8.37)`
`=-0.06xx-3.39`
`=0.2034V`
`b. CH_(3)COONH_(4)` is salt of `W_(A)//W_(B)` and its `PH` is independent of the concentration.
`:. pH_(a)=(1)/(2)(pK_(w)+pK_(a)-pK_(b))`
`=(1)/(2)(14+4.74-4.74)=7`
`CH_(3)COONa` is a salt fo `W_(A)//S_(B).`
`:. pH_(c)=(1)/(2)(pK_(w)+pK_(a)+logc)`
`=(1)/(2)(14+4.74+log 10^(-3))=7.87`
Since `pH_(c)gtpH_(a)`
`:. EMF_(cell)=-ve` and the cell will be endergonic
`:. E_(cell)=-0.06(pH_(c)-pH_(a))=-0.06(7.87-7)`
`=-0.06xx0.87`
`=0.05522V`
`c. CH_(3)COH`is a weak ACID `(W_(A)).`
Thus, `pH_(WA)=(1)/(2)(pK_(a)-log c)`
`=(1)/(2)(4.74- log 10^(-1))`
`=2.87`
`NH_(4)OH`is a weak base `(W_(B))`.
Thus, `pOH_(wB)=(POH)_(c)=(1)/(2)(pK_(b)-logc)`
`=(1)/(2)(4.74-log 10^(-2))`
`=3.37`
Thus, `pOH_(wB)=pH_(c)=14-3.33=10.63`
Since `pH_(c)gtpH_(a)`, therefore, `EMF_(cell)=-ve`, and the cell will be endergonic.
`E_(cell)=-0.06(pH_(c)-pH_(a))=-0.06(10.63-2.87)`
`=-0.06xx7.67`
`-0.4656V`
`d.` Anode reaction`: cancer(Ag(s)) rarrAg^(o+)Ag^(o+)(0.1M)+cancel(e^(-))`
Cathode reaction `:`
`Ag^(o+)(1M)+cancel(e^(-)) rarr Ag(s)`
Cell reaction `:` `ulbar(Ag^(o+)(1M)rarrAg^(o+)(0.1M))`
Since `[Ag^(o+)]_(c)gt[Ag^(o+)]_(a),` therefore , `EMF_(cell)=+ve` and the cell will be exergonic.
`E_(cell)=E^(c-)._(cell)-(0.06)/(n_(cell))log.([Ag^(o+)])/([Ag^(o+)])`
`=0-(0.06)/(1)log.(0.1M)/(1M)`
`E_(cell)=-0.06[log 10^(-1)]=-0.06 xx -1=0.06V`
`e.` Anode reaction `:`
`2Cl^(c-)(10^(-3)M)rarr cancel(Cl_(2)(g)(1atm))+cancel(2e^(-)) `
Cathode reaction `:`
`cancel (2Cl(g)(1atm))+cancel(2e^(-))rarr2Cl^(c-)(10^(-2)M)`
`ulbar(2Cl^(c-)(10^(-3)M) rarr 2Cl^(c-)(10^(-2)M)`
Since `[Cl^(c-)]_(c)gt[Cl^(c-)]_(a)`
Therefore, `EMF_(cell)=-ve` and the cell will be endergonic.
`E_(cell)=E^(c-)._(cell)-(0.06)/(n_(cell))log.([Cl^(c-)]_(c)^(2))/([Cl_(a)^(2)])`
`=0-(0.06)/(2)log.((10^(-2))^(2))/((10^(-3))^(2))`
`=-(0.06)/(2)xx2 log 10=-0.06V`

`f.` Anode reaction `:`
`2Cl^(c-)(10^(-3)M) rarr Cl_(2)(g)(2atm)+cancel(2e^(-)) `
Cathode reaction `:` `Cl_(2)(g)(1 ATM)+cancel(2e^(-)) rarr 2 Cl_(2)(g)(10^(-2)M)`
` Cell reaction `:``ul bar(2Cl^(c-)(10^(-3)M)+Cl_(2)(g)(1atm)rarr2Cl^(c-)(10^(-2)M)+Cl_(2)(g)(2atm))`
`E_(cell)=E^(c-)._(cell)-(0.06)/(n_(cell))log.([Cl^(c-)]_(c)^(2)[P_(Cl_(2))]_(a))/([Cl^(c-)]_(a)^(2)[p_(Cl_(2))]_(c)).=0-(0.06)/(2)log.((10^(-2))^(2)xx2atm)/((10^(-3))^(2)xx1atm)`
`=-(0.06)/(2)[log 10^(2)xx2]`
`=-(0.06)/(2)[2 log 10+log 2 ]`
`=-0.03[2+0.3]=-0.03xx2.3=-0.069V`
Therefore, `EMF_(cell)` is negative and the cell will be endergonic.
`EMF_(cell)` is positive , if `[Cl^(c-)]_(a)gt[Cl^(c-)]_(c)` and `(p_(Cl_(2)))_(c)gt(p_(Cl_(2)))_(a)`
Discussion
41.

Calculate the E.M.F. of the zinc - silver cell at 25^(@)C when [Zn^(2+)]=0.10M and [Ag^(+)]=10M.[E^(@) " cell at "25^(@)C="1.56 volt"]

Answer»

Solution :The CELL REACTION in the zinc - silver cell would be
`2Ag^(+)+Zn hArr 2Ag+Zn^(2+)`
The Nernst equation for the above all reaction MAY be written as :
`E_("cell")=E_("cell")^(@)-(RT)/(nF)ln([Ag]^(2)[Zn^(2+)])/([Ag^(+)]^(2)[Zn])`
(SINCE concentrations of solids are taken as unity)
`=E_("cell")^(@)-ln.([Zn^(2+)])/([Ag^(+)]^(2))`
Substituting the various values in Nernst equation, we have
`E_("cell")=1.56-(2.303xx8.314xx298)/(2xx96495)log""(0.1)/((10)^(2))`
`="1.648 VOLTS."`
Discussion
42.

Calculate the EMF of the following concentration cell at 298K Zn|ZnSO_(4)(0.05M)||ZnSO_(4)(0.5M)|Zn

Answer»

SOLUTION :`E_(Cell)=(0.0591)/(n)"LOG"(C_(2))/(C_(1))=(0.0591)/(2)"log"(0.5)/(0.05)=0.02955V`
Discussion
43.

Calculate the emf of the following concentration cell at 25^(@)C: Ag(s)|AgNO_(3) (0.01 M)||AgNO_(3) (0.05 M)|Ag (s)

Answer»

0.828V
0.0413V
`-0.0413` V
`-0.828` V

Solution :`E=E^@-0.0591/n` log Q
`E^@`=0 for all CONCENTRATION CELLS
=`0-0.0591/1 "log"(0.01/0.05)`=0.0413 V
Discussion
44.

Calculate the emf of the following cells, find their cells reactions using E^@ values from the table? Zn|ZnO_2^(2-) (0.1M), OH^(-) (1M) |HgO|Hg In each case, is the reaction as written spontaneous or not?

Answer»

SOLUTION :`HgO+Zn+2OH^(-) (1.0)=Hg+ZnO_2^(2-) (0.1)+H_2O, 1.343` VOLTS
Discussion
45.

Calculate the emf of the following cells, find their cells reactions using E^@ values from the table? Pt|Fe^(2+) (1M), Fe^(3+) (0.1M) ||Cl^(-)(0.001M) |AgCl|Ag In each case, is the reaction as written spontaneous or not?

Answer»

SOLUTION :`AgCl+FE^(2+)(1.0) =Ag+Cl^(-) (0.001) +Fe^(3+) (0.1),-0.313` VOLT
Discussion
46.

calculate the emf of the following cell: Pt(H_(2)" 1 atm")|CH_(3)CH_(2)COOH(0.15 M) ||0.01 M NH_(4)OH|H_(2)" (1 atm)"Pt K_(a) for CH_(3)CH_(2)COOH=1.4xx10^(-5) K_(b) for NH_(4)OH =1.8xx10^(-5)

Answer»


Solution :`[H^(+)]` in `CH_(3)CH_(2)COOH=sqrt(CxxK_(a))=sqrt(0.15xx1.4xx10^(-5))`
`=1.449xx10^(-3)`
`[OH^(-)]` in `NH_(4)OH=sqrt(C xxK_(b))=sqrt(0.01xx1.8xx10^(-5))=0.4242xx10^(-3)`
`[H^(+)]` in `NH_(4)OH=10^(-14)/(0.4242 XX 10^(-3))=2.3573xx10^(-11)`
`E_(cell)=0.0591 "log"([H^(+)]_(RHS))/([H^(+)]_(LHS))`
Discussion
47.

Calculate the emf of the following cells, find their cells reactions using E^@ values from the table? Ag|Ag^+ (0.01M) ||Zn^(2+) (0.1M)|Zn In each case, is the reaction as written spontaneous or not?

Answer»

SOLUTION :`ZN^(2+)(0.1)+2Ag=Zn+2Ag^+ (0.01), -1.473` VOLT
Discussion
48.

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»


SOLUTION :`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
49.

Calculate the emf of the following cell at 298K, Fe(s)Fe^(2+)(0.001M)||H^+(1M)|H_2(g)(1ba r), Pt(s)

Answer»

Solution :The cell REACTION:
`Fe(s)+2H^+(aq)to Fe^(2+)(aq)+H_2(g) `
`E_(cell)^@=0.44V`
NERNST equation
`E_(cell)=E_(cell)^@-0.059/2log""([Fe^(2+)])/([H^+]^2)`
`E_(cell)=0.44V-0.059/2log""((0.001M))/(1M)^2`
`=0.44V-0.059/2log(10^-3)`
`=0.44V+0.0885 V`
`=0.5285V`
Discussion
50.

Calculate the e.m.f. of the following cell at 298K: 2Cr(s)+3Fe^(2+)(0.1M)to2Cr^(3+)(0.01M)+3Fe(s) Given: E_((Cr^(3+)//Cr))^(@)=-0.74V,E_((Fe^(2+)//Fe))^(@)=-0.44V.

Answer»


SOLUTION :`E_(cell)^(@)=E_("cathode")^(@)-E_("anode")^(@)=0.44-(-0.74)=0.30V`
`E_(cell)=E_(cell)^(@)-(2.303RT)/(NF)"LOG"([Cr^(3+)]^(2))/([Fe^(2+)]^(3))(n=6)`
Discussion