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401.

The tendencies of the electrodes made up of Cu, Zn and Ag to release electrons when dipped in their respective salt solutions decrease in the order :A. `Zn gt Ag gt Cu`B. `Cu gt Zn gt Ag`C. `Zn gt Cu gt Ag`D. `Ag gt Cu gt Zn`

Answer» Correct Answer - C
Discussion
402.

The tendencies of the electrodes made up of Cu, Zn and Ag to release electrons when dipped in their respective salt solutions decrease in the orderA. ZngtAggtCuB. CugtZngtAgC. ZngtCugtAgD. AggtCugtZn

Answer» Correct Answer - C
In electrochemical series element are placed in order of their decreasing tendency to lose electrons. First Zn comes, then comes Cu and then comes Ag.
ZngtCugtAg
Discussion
403.

If hydrogen electrode dipped I 2 solution of pH=3 and pH=6 and salt bridge is connected the e.m.f. of resulting cell isA. 0.177VB. 0.3VC. 0.052VD. 0.104V

Answer» Correct Answer - A
`E_(cell)=-0.059log((10^(-6))/(10^(-3)))=-0.059log10^(-3)`
`=0.059xx(-3)=0.177V`.
Discussion
404.

A hydrogen electrode placed in a solution containing `CH_(3)COOK` and `CH_(3)COOH` in the ration `a:b` and `b:a` has electrode potential values of `-1.59` and `+1.0V,` respectively. Calculate `pK_(a)` of `CH_(3)COOH.`

Answer» Mixture of `CH_(3)COOHK(` salt of `W_(A)//S_(B))` and `CH_(3)COOH(W_(A))`
forms acidic buffer solution whose `pH` is
`pH_(aci dic buffer)=pK_(a)+log.([Sa l t])/([Aci d])`
`:. E_(1(H_(2)))=-0.059pH_(1)=-0.059[pK_(a)+log.(a)/(b)],,,(i)`
`E_(2(H_(2)))=-0.059pH_(2)=-0.059[Pk_(a)+log.(b)/(a)],,,(ii)`
Adding Eqs. `(i)` and `(ii)`, we get
`E_(1)+E_(2)=-0.059[pK_(a)+log.(a)/(b)]-0.059[pK_(a)+log.(b)/(a)]`
`=-2xx0.059pK_(a)`
`:. pK_(a)=(-(E_(1)+E_(2)))/(2xx0.059)=(-(-1.59+1.0))/(2xx0.059)`
`=(0.059)/(2xx0.059)=5`
`:. pK_(a)=5`
Discussion
405.

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» `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)` is an 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)` is a 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
406.

The `EMF` of the following cell is `0.180 V` at `30^(@)C`. Find `EMF_(cell)` when `a. 40 mL of 0.2M NaOH` is added to the negative terminal of the battery . `b.` `50 mL` of `0.2 M NaOH` is added to the negative terminal of the battery . `c.` `50mL` OF `0.2M NaOH` is added to `100 mL` of `H_(2)SO_(4)` at the positive terminal of the battery.

Answer» It is an electrolytic concentration cell.
`[H_(2)SO_(4)]0.05M=0.05xx(n` factor`)`
`=0.1N=10^(-1)N`
`pH_(c)=-log(10^(-1)N)=1`
`pH_(a)=?`
`E_(cell)=-0.06(pH_(c)-pH_(a))=-0.06(1-pH_(a))`
`0.180V=-0.06(1-Ph_(a))`
`pH_(a)=(0.180)/(0.06)+1=4`
`:. [H^(o+)]_(a)=10^(-4)M `
Since `HA` is a weak acid `(W_(A))`,
`:. pH_(a)=pH_(wA)=(1)/(2)(pK_(a)-logC)`
`4=(1)/(2)(pK_(a)-log10^(-1))`
`:. pK_(a)=7`
`a.` When` 40mL` of `0.2 M NaOH(=40xx0.2=8 mmol)` is added to a weak acid, acidic buffer is formed.
`HAA+NaOH rarrNaA+H_(2)O`


Total volume `=100+40=140mL`
`[Sa l t] =[NaA]=(8mmol)/(140mL)`
`[W_(A)]_(l e f t)=[HA]_(l eft)=(2mmol)/(140mL)`
`pH_(a)=pH_(aci dic buffter)=pK_(a)+log .([Sa l t])/([Ac i d])`

`=7+log.((8//140)/(2//140))`
`=7+log2^(2)`
`=7+2xx0.3=7.6`
Thus,
`E_(cell)=-0.06(pH_(c)-pH_(a))`
`=-0.06(1-7.6)`
`=-0.06xx-6.6=0.396V`
`b.` When `50 mL` of `0.2M NaOH(=50xx0.2=10mmol)`
is added to weak acid `(HA)`, salt of `W_(A)//S_(B)(NaA)` is formed.



`:.[NaA]=(10mmol)/(150mL)=(1)/(15)M`
Thus, `pH` of a salt of `W_(A)//S_(B)` is given by `:`
`pH_(a)=pH_(sal tW_(A)//S_(B))=(1)/(2)(pK_(w)+pK_(a)+log C)`
`[pK_(a)` of `HA=7` from part `(a)` above `]`
`pH_(a)=(1)/(2)(14+7+log.(1)/(15))`
`=(1)/(2)(21+log1-log15)`
`=(1)/(2)(21+0-log3-log5)`
`=(1)/(2)(21-0.48-0.7)=9.91`
Thus,
`E_(cell)=-0.06(pH_(c)-pH_(a))`
`=-0.06xx(1-0.01)`
`-0.06xx(-8.91)=0.5346V`
`c.` When `50mL` of `0.2M NaOH(=50xx0.2=10mmol=10mEq)` is added to `100mL ` of `0.05 M H_(2)SO_(4)` to the positive terminal of battery , then
`mEq` of `NaOH=10`
`mEq` of `H_(2)SO_(4)=100mLxx0.05xx2(n` factor `)`
`=10`
When `10mEq` of `H_(2)SO_(4)` reacts with `10mEq` of `NaOH,` a salt `(Na_(2)SO_(4))` of `S_(A)//S_(B)` is formed, which does not hydrolyze . So `pH` of such solution is `=7`.
`pH_(c)=7`
`pH_(a)=?`
`HA` is a weak acid whose `pK_(a)=7(` determined in part `(a)` above `)` and `[HA]=0.1M=10^(-1)M`.
Thus, `pH_(wA)=(1)/(2)(pK_(a)-logC)`
`=(1)/(2)(7-log10^(-1))=4`
Hence,
`E_(cell)=-0.06(pH_(c)-pH_(a))`
`=-0.06(7-4)=-0.06xx3=-0.18V`
Discussion
407.

`Pt(H_(2))(p_(1))|H^(o+)(1M)|(H_(2))(p_(2)),Pt` cell reaction will be exergonic ifA. `p_(1) = p_(2)`B. `p_(1) gt p_(2)`C. `p_(2) gt p_(1)`D. `p_(1) = 1atm`

Answer» Correct Answer - B
Discussion
408.

`Pt(H_(2))(p_(1))|H^(o+)(1M)|(H_(2))(p_(2)),Pt` cell reaction will be exergonic ifA. `P_(1) gt P_(2)`B. `P_(1) gt P_(2)`C. `P_(2) gt P_(1)`D. `P_(1) = 1` atm

Answer» Correct Answer - B
`E_("cell")` for the given change
`E_("cell") = (0.059)/(2) "log" ([H^(+)]xxP_(1(H_(2))))/(P_(2H_(2))xx[H^(+)])`
Thus For `+ve E_("cell") = P_(1H_(2)) gt P_(2H_(2))`
Discussion
409.

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

Answer» 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
410.

An electrolytic cell is constructed for preparing hudrogen. For an averge current of `1` amper in the circuit , the time required to produce `450 mL` of hydrogen at `NTP` is appr.A. ` 30` minB. `1` hourC. ` 2` hoursD. ` 5` hours

Answer» Correct Answer - B
` 22, ml H_2 at STP + 1 mol H_2`
`:. 450 mL H_2 at STP= ( 450 mL)/(22. 400mL) xx 1 "mole" [email protected]`
`= 0.02` mol
Electrode reaction :
` 2H^+ +2E^(-_ rarrH_2`
`2 F 1 mli`
So, the liberation of `0.02` mole of `H_2` required
`=2 xx 96, 500 xx 0.02 C = 3860 C`
Time required to produce `0.2` mole
`H_2 = Q/I = ( 3. 860)/(1 A) = 1.07 ~~1` hour .
Discussion
411.

The value of `(E_(H_(2)O)//H_(2)^(@))` (1atm) Pt at 298K would b eA. `+0.207`B. `-0.414V`C. `-0.207V`D. `+0.414V`

Answer» Correct Answer - B
For pure water `[H^(+)]=10^(-7)M`
`H^(+)(aq)+e^(-)rarr1//2H_(2)`
`E=E^(@)-(0.0591V)/(1)"log "(1)/([H^(+)])`
`=0V-(0.0591V)/(1)"log"(1)/(10^(-7))`
`=-0.0591Vxx"log"10^(7)`
`=-0. 0591xx7V=-0.414V`
Discussion
412.

`Pt(Cl_(2))(p_(1))|HCl(0.1M)|(Cl_(2))(p_(2)),Pt` cell reaction will be endergonic ifA. `p_(1)=p_(2)`B. `p_(1)gtp_(2)`C. `p_(2)gtp_(1)`D. `p_(1)=p_(2)=1atm`

Answer» Correct Answer - c
In case of `-ve` ion, cell is spontaneous if `[m_(1)^(c-)]_(a)gt[m_(2)^(c-)]_(c)`
or `(p_(2))_(c)gt(p_(1))_(a)` and vice versa.
This is only possible , when `p_(2)gtp_(1) or c_(1)gtc_(2) or M_(1)gtM_(2)`.
Discussion
413.

During the electrolysis of aqueous zinc nitrate.A. Zince plates out at the cathodeB. Zince plates out at the anodeC. Hydrogen gen `H_(2)` is evolved at the anode.D. Oxygen gas `O_(2)` is evolved at anode

Answer» Correct Answer - a,d
Anqueous `Zn(NO_(3))_(2):`
Cathode `: (sqrt()) Zn^(2+)(aq)+2e^(-) rarr Zn(s)`
`E^(c-)._(Zn^(2+)|Zn)=-0.76V`
`(X)2H_(2)O(l)+2e^(-) rarr H_(2)(g)+2overset(c-)(O)H(aq)`
`E^(c-)._(H_(2)O|H_(2))=-0.83V`
Anode `: (sqrt())2H_(2)O(l) rarr O_(2)(g)+4H^(o+)(aq)+4e^(-)`
`E^(c-)._(H_(2)O|O_(2))`
`(X)NO_(3)^(c-)rarr x`
Discussion
414.

`Pt(H_(2))(1atm)|H_(2)O`, electrode potential at `298K` isA. `-0.2364V`B. `-0.4137V`C. `0.4137V`D. `0.00V`

Answer» Correct Answer - c
Since oxidation occurs.
`H_(2) rarr 2H^(o+)+2e^(-)`
`Q=[H^(o+)]^(2)=10^(-14)M^(2) i n H_(2)O`
` E=E^(c-)-(0.0591)/(2)log10^(-14)`
`=0+0.0.591xx7=0.4137V`
Discussion
415.

A galvanic cell has electrical potenetial of 1.1V. If an opposing potnetial of 1.1 V is applid to this cell, what will happen to the cell reaction and current flowing through the cell?

Answer» when an oppsoing potential of 1.1 V is applied to a galvanic cell having electrical potential of 1.1 V then cell reaction stops completely and no current will flow through the cell.
Discussion
416.

Which of the following changes will increase the `EMF` of the cell `:` `Co(s)|CoCl_(2)(M_(2))||HCl(M_(2))||(H_(2),g)Pt`A. Increase the volume of `CoCl_(2)` from `100mL` to `200mL`B. Increase `M_(2)` from `0.1 M` to `0.50M`.C. Increase the pressure of the `H_(2)(g)` from `1.0` to `2.0 atm`.D. Increase `M_(1)` from `0.01M` to `0.50M`.

Answer» Correct Answer - a,b
Cell reaction `:` `Co(s)+2H^(o+)(M_(2))rarrCo^(2+)(M_(1))+H_(2)(g)`
`E_(cell)=E^(c-)._(cell)_-(0.059)/(2)log.([Co^(2+)])/([H^(o+)])(p_(H_(2))-=Q_(cell))`
`a.` Increase in the volume `(i.e.,` dilution `)` of `CoCl_(2)` solution `(i.e.,` decrease in `[Co^(2+)])` will cause `Q_(cell)` to decrease causing `E_(cell)` to increase.
`b.` Increaseing `M_(2)` will cause a reduction in `Q` thus increasing `E_(cell)`
`c.` Increasing `p_(H_(2))` will increase `Q` thus reducing `E_(cell)`.
`d.` Increasing `M_(1)` will increase `Q` thus reducting `E_(cell)`.
Discussion
417.

Which of the following changes will increase the `EMF` of the cell `:` `Co(s)|CoCl_(2)(M_(2))||HCl(M_(2))||(H_(2),g)Pt`A. Increae in the volume of `CoCl_(2)` solution from `100mL` to `200mL`.B. Increase `M_(2)` from `0.1 M` to `0.50M`.C. Increase the pressure of the `H_(2)(g)` from `1.0` to `2.0 atm`.D. Increase `M_(1)` from `0.01M` to `0.50M`.

Answer» Correct Answer - b
`Co+2H^(o+) rarr Co^(2+)+H_(2)`
`Q=([Co^(2+)])/([H^(o+)]^(2))`
`E=E^(c-)-(0.0591)/(2) log .([Co^(2+)])/([H^(o+)]^(2))`
`E` will increase when `[H^(o+)]` increases of `[Co^(2+)]` decreases.
Discussion
418.

`underset(1 L solution)(Ag|Ag^(o+)(1M))||underset(1 L solution)(Ag^(o+)||Ag)` `0.5F` electricity in the `LHS(` anode `)` and `1F` of electricity in the `RHS(` cathode`)` is first passed making them independent electrolytic cells at `298K`. `EMF` of the cell after electrolysis will beA. IncreasedB. DecreasedC. No changeD. Time is also required

Answer» Correct Answer - c
`E=E^(c-)+(0.0591)/(1)log.([Ag^(o+)]_(R))/([Ag^(o+)]_(L))`
`=0+0.0591log 2=0.0591xx0.301V`
After current is passed `[Ag^(o+)]_(R)=1M`
`[Ag^(o+)]_(L)=0.5M`
Hence, no change in `EMF`.
Discussion
419.

`9.65` amp of current was passed for one hour through Daniel cell. The loss of mass of zinc anode isA. 11.77 g , 11.43 gB. 11.43 g , 11.77 gC. 23.54 g , 22.84 gD. 22.86 g 23.54 g

Answer» Correct Answer - A
`Q = 9.65 xx 60 xx 60 = 34740` C
At anode ` : Zn to Zn^(2+) + 2e^(-)`
Loss mass of Zn by `2 xx 96500 C = 65.4` g
`therefore ` Loss in mass by 34740 C = 11.77 g
At cathode `: Cu^(2+) + 2e^(-) to Cu` .
Gain in mass of Cu by `2 xx 96500 C = 63.5`g
`therefore` Gain in mass of Cu by 34740 C
= `(63.5)/(2 xx 96500) xx 3474 = 11 .43`g
Discussion
420.

`a.` In the following reactions, what weight of substance would be liberated if `1F` of electricity were passed through the cell `:` `i. Cu^(2+)+2e^(-) rarr Cu` `ii. Al^(3+)+3e^(-) rarr Al` `iii. 2Cl^(c-) rarr Cl_(2)+2e^(-)` `iv. Ag^(o+)+e^(-) rarr Ag` `b.` In the series of cathodes `//` anodes given above, how many coulombs are needed to produce `1 g` of each ofA. `Cu`B. `Al`C. `Cl_(2)`D. `Ag`

Answer» Correct Answer - `a. i. 31.75g" "ii. 9g" "iii. 35.5g" "iv. 108g `
`b.` `i. 3039C" "ii. 10722.2C" "iii. 2718.3C" "iv. 893.5C`
`i. Cu^(2+)+2e^(-) rarr Cu`
`[{:(2e^(-),2F-=1 mol of Cu,,=,63.5g),(1F,(63.5)/(2)=31.75g,,,):}]`
`ii. Al^(3+)+3e^(-) rarr Al `
`[{:(3e^(-),3F-=1 mol of Al,,=,27g),(1F,(27)/(3)=9.0g,,,):}]`
`iii. 2Cl^(c-) rarr Cl_(2)+2e^(-)`
`[{:(2e^(-),2F-=1 mol of Cl_(2),,=,71.0g),(1F,(71.0)/(2)=35.5g,,,):}]`
`iv. Ag^(o+)+e^(-) rarr Ag`
`[1e^(-)=1F=1mol` of `Ag =108g]`
`b.` `i. Cuimplies((1)/(63.5))xx(96500xx2C)=3039.5C`
`ii. Alimplies((1)/(27))xx(96500xx3C)=10722.2C`
`iii. Cl_(2)implies((1)/(71))xx(96500xx2C)=2718.3C`
`iv. Agimplies((1)/(108))xx(96500xx1C)=893.5C`
Discussion
421.

How many electrons flow through a metallic wire if a current of 0.5 A is passed for 2 hours? (Given 1F=96,500 C `mol^(-1)`)

Answer» Charge flowing through the metallic wire`=Ixxt=0.5xx2xx60xx60C=3600C`
But charge flowing=no. of electrons flowing `xx`charge on each electron
`therefore3600=nxx1.60xx10^(-19)`
or `n=(3600)/(1.60xx10^(-19))=2.25xx10^(22)`.
Discussion
422.

The questions consist of two atatements each, printed as Assertion and Reason. While answering these questions you are required to choose any one of the following four responses : The electrical resistance of a column of `0.5 M NaOH` solution of diameter `1 cm` and length `50 cm` is ` 5.55 xx 10^3 "ohm"`. Its resistivity is equal to `75.234 Omega cm`.A. If both the assertion and reason are true but the reason is ont the correct explanation of assertionB. If both the assertion and reason are true but the reason is not the correct explanation of assertion.C. If the assertion is true but reason is false.D. If assertion is false but reason is rue

Answer» Correct Answer - c
` R = rho xx 1/A (A=pir^2)`
`:. rho = (Rxx A)/(l)= ( 5.55 xx 10^3 (0.5)^2xx 3.14)/(50) = 87 . 13` ohm.
Discussion
423.

Loss of electrons is oxidation. The process at anode isA. OxidationB. ReductionC. BothD. None.

Answer» Correct Answer - A
At anode, oxidation i.e., loss of electron occurs.
Discussion
424.

A current of 9.65 amperes is passed through excess of fused `AlCl_(3)` for 5 hours. How many litres of chlorine will be liberated at S.T.P. ? (1F=96500C)A. `2.016" L "`B. 1.008LC. 11.2 LD. 20.16 L

Answer» Correct Answer - D
(d) The anodic reaction is :
`2Cl^(-) to underset(22.4L)(Cl_(2))+underset(2F)(2e^(-))`
The quantity of charge (Q) passed`=ixxt`
No. of Faradays of `=(9.65xx5xx60(C ))/(96500(C ))`
charge passed =1.8 F
1F of charge evolves `Cl_(2)=11.2 L`
1.8 F of charge evolves `Cl_(2)=(11.2L)/((1F))xx(1.8F)=20.16" L"`
Discussion
425.

A current of strength of 9.65 ampere is passed through excess fused AlCl3 for 5 hours. How many liters of chlorine will be liberated at STP.

Answer»

20.16 liters of chlorine will be liberated at STP.
Discussion
426.

The questions consist of two atatements each, printed as Assertion and Reason. While answering these questions you are required to choose any one of the following four responses :Assertion : The cell potential of mercury cell is `1.35 V`, which remains constant . Reason : In mercury cell, the electrolute is a paste of KOH and ZnO.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 incorrectD. If (A) is incorrect, but (R) is correct.

Answer» Correct Answer - B
Discussion
427.

1 faraday of electricity will liberate 1 gram atom of the metal from the solution ofA. Copper sulphateB. Calcium chlorideC. Gold III chlorideD. silver I chloride

Answer» Correct Answer - D
`Ag^(+)+e^(-)rarrAg(s)`
1mole 1 gm atom
1F 1 mole
Discussion
428.

1 faraday of electricity will liberate 1 gram atom of the metal from the solution ofA. `AuCl_(3)`B. `AgNO_(3)`C. `CaCl_(2)`D. `CuSO_(4)`

Answer» Correct Answer - B
Ag is monovalent ,
Thus 1 g atom = 1 g equivalent,
1 Faraday liberates 1 g equivalent .
Discussion
429.

Ione faraday of electricity will liberate one gram atom of a metal from a solution ofA. `AuCl_(3)`B. `CuSO_(4)`C. `BaCl_(2)`D. `KCl`.

Answer» Correct Answer - D
`K+e^(-)rarrK`
1mole 1 mole
1 faraday 39gm
1 gram atom
Discussion
430.

(A) The charge of one mole of electron is one faraday. (R) The quantity of current required to deposite one mol of Mg from `Mg^(2+)` electrolyte solution is two faradays.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 incorrectD. If (A) is incorrect, but (R) is correct.

Answer» Correct Answer - B
Discussion
431.

One faraday is equal toA. 9650 coulombB. 10,000 coulombC. 19640 coulombD. 96500 coulomb

Answer» Correct Answer - D
Discussion
432.

On passing one faraday of electricity through a dilute solution of an acid, the volume of hydrogen obtained at NTP is:A. 22400 mLB. 1120 mLC. 2240 mLD. 11200 mL

Answer» Correct Answer - D
Discussion
433.

When one faraday of electric current is passed, the mass deposited is equal to :A. one gram equivalentB. one gram moleC. electrochemical equivalentD. half gram equivalent

Answer» Correct Answer - A
Discussion
434.

Which one of the following metals could not be obtained on electrolysis of aqueous solution of its saltsA. `Ag`B. `Mg`C. `Cu`D. `Cr`

Answer» Correct Answer - B
The reduction potential of Mgis less than that of water `(E^(o)=-0.83V)`. Hence their ions in the aqueous solution cannot be reduced instead water will be reduced.
`2H_(2)O+2e^(-)toH_(2)+2OH^(-)`.
Discussion
435.

Which of the given statements for mercury cell are incorrect? (i)Mercury cell is suitable for low current devices like hearing aids, watches, etc. (ii)It consists of zinc-mercury amalgam as anode and a paste of HgO and carbon as the cathode. (iii)The electrolyte is a paste of `Zn(OH)_2` and `KO_2`. (iv)The electrolyte reactions for the cell are At anode : `Zn(Hg)+ H_2O to ZnO_((s)) + 2OH^(-) + 2e^(-)` At cathode : `HgO + H_2O + 2e^(-) to Hg_((l)) + 2OH^-`A. (ii) and (iii) onlyB. (i) and (ii) onlyC. (i),(iii) and (iv) onlyD. (iii) and (iv) only

Answer» Correct Answer - D
The electrolyte in a mercury cell is a paste of KOH and ZnO. The electrode reactions for the cell are `{:("At anode",Zn(Hg)+2OH^(-) to ZnO_((s))+H_2O+2e^(-)),("At cathode",HgO+H_2O + 2e^(-) to Hg_((l))+2OH^(-)):}`
Discussion
436.

Zinc/silver oxide cell is used in hearing aids and electric watches. The following reactions occur : `Zn(s) to Zn^(2+)(aq)+2e^(-) , E_(Zn^(2+)//Zn)^(@)=-0.76V` `Ag_(2)O+H_(2)O+2e^(-) to2Ag+2OH^(-),E_(Ag^(+)//Ag)^(@)=0.344" V"` Calculate (i) Standard potential of the cell (ii) Standard Gibbs energy.

Answer» In this reaction, Zn is oxidised and `Ag_(2)O` is reduced
(i)`" "E_(cell)^(@)=E_(cathode)^(@)-E_(anode)^(@)`
`=(0.344)-(-0.76)=1.104" V"`
(ii) `" " DeltaG^(@)=-nFE_(cell)^(@)`
`=-2xx(96500" C mol"^(-1))xx(1.104" V")`
`=-2.13xx10^(5)" CV "mol^(-1)`
`=-2.13xx10^(5)" J mol"^(-1)`
Discussion
437.

The zinc/silver oxide cell is used in hearing aids and electric watches. The following reactions take place: `ZntoZn^(2+)+2e^(-),E^(@)=0.76V` `Ag_(2)O+H_(2)O+2e^(-)to2Ag+2OH^(-),E^(@)=0.344V` (a) What is oxidized and reduced? (b) Find `E^(@)` of the cell and `DeltaG` in joules.

Answer» (a) Zn is oxidized and `Ag_(2)O` is reduced (as `Ag^(+)` ions change into Ag)
(b) `E_(cell)^(@)=E_(Ag_(2)O//Ag)^(@)(Red)+E_(Zn//Zn^(2+))^(@)(Ox)=0.344+0.76=1.104V`
`DeltaG=-nFE_(cell)^(@)=-2xx96500xx1.104J=-2.13xx10^(5)J`.
Discussion
438.

A silver oxide-zinc cell maintains a fairly constant voltage during discharge (1.50V). The button form of the cell is used in hearing aids. The half-reactions involved are : `Zn(s)+2OH^(-)(aq) to Zn(OH)_(2)(s)+2e^(-)` `Ag_(2)O(s)+H_(2)O(l)+2e^(-) to 2Ag(s)+2OH^(-)(aq)` Will change in `[OH^(-)]` affect `E_(cell)` ?

Answer» Let us add the two half cell reactions :
`Zn(s) +2OH^(-)(aq) to Zn (OH)_(2)(s)+2e^(-)`
`(Ag_(2)O(s)+H_(2)O(l)+2e^(-) to 2Ag(s)+2OH^(-)(aq))/(Zn(s)+Ag_(2)O(s)+H_(2)O(l) to Zn(OH)_(2)(s)+2Ag(s)`
As the `OH^(-)` ions are not involved in the final equation, any change in [OH^(-)] will not affect `E_(cell)`.
Discussion
439.

(a) In a cell reaction, equilibrium constant L is less than one. Is `E^(@)` for the cell positive or negative ? (b) What will be the value of K if `E_(cell)^(@)=0` ?

Answer» (a) For a cell , `E^(@)=(0.0591)/(n)logK`
Let the value of K be 0.01 (less than one)
`E^(@)=(0.0591)/(n)log0.01=(0.0591)/(n)log(10^(-2))=(-(2)xx(0.0591))/(n)=-"ve"`.
(b) If `E_(cell)^(@)=0, "then" 0=(0.0591)/(n)log K" or " log K=0,K="Antilog"(0)=1`.
Discussion
440.

For the cell `Tl|Tl^(+)(0.001M)||Cu^(2+)(0.01M)|Cu.E_("cell")` at `25^(@)C` is 0.83V, which can be increased:A. by increasing `[Cu^(2+)]`B. by increasing `[Tl^(+)]`C. by decreasing `[Cu^(2+)]`D. none

Answer» Correct Answer - A
`E_("cell") = E_("cell")^(@) + (0.0592)/(2) log _(10) ([Cu^(2+)])/([Tl^(2+)])`
`therefore E_(cell) prop [Cu^(2+)]`
Discussion
441.

For the cell, `TI|TI^(+)(0.001M)|Cu^(2+)(0.1M)|Cu,E_(cell)` at `25^(@)C` is 0.83 V. this can be increasedA. by increasing `[Cu^(2+)]`B. by increasing `[TI^(+)]`C. by decreasing `[Cu^(2+)]`D. by decreasing `[TI^(+)]`

Answer» Correct Answer - A::D
`2TI+Cu^(2+)to2TI^(+)+Cu,E_(cell)-E_(cell)^(@)-(0.0591)/(2)"log"([TI^(+)]^(2))/([Cu^(2+)])`
Discussion
442.

For the cell, `TI|TI^(+)(0.001 M)||Cu^(2+)(0.1 M)|Cu(s),E_(cell)^(@)` at `25^(@)C` is `0.83" V"`. It can be increased by :A. increasing `[Cu^(2+)]`B. increasing `[TI^(+)]`C. decreasing `[Cu^(2+)]`D. decreasing `[TI^(+)]`

Answer» Correct Answer - A::D
(a,d) are correct.
Discussion
443.

Calculate the voltage `E` of the cell at `25^(@)C` `Mn(s) |Mn(OH_(2))(s)|Mn^(2+)(x,M)OH^(-) (1.00 xx 10^(-4)M)||Cu^(2+) (0.675M)|Cu(s)` given that `K_(sp) = 1.9 xx 1-^(-13)` for `Mn(OH)_(2)(s) E^(@) (Mn^(2+)//Mn) =- 1.18 V, E^(@) (Cu^(+2)//C) = +0.34V`

Answer» Correct Answer - `1.66V`
`Mn rarr Mn^(2+) +2e^(-)`
`Cu^(2+) +2e^(-) rarr Cu K_(sp) = [Mn^(2+)] [OH^(-)]^(2)`
`E = (1.18 +0.34) - (0.0591)/(2) log. ([Mn^(2+)])/([Cu^(2+)])`
`E_(cell) = 1.66V`
Discussion
444.

Calculate the equilibrium constant for the reaction: `3Sn(s) +2Cr_(2)O_(7)^(2-) +28 H^(+) rarr 3Sn^(+4) +4Cr^(3+) +14H_(2)O` `E^(@)` for `Sn//Sn^(+2) = 0.136 V E^(@)` for4 `Sn^(2+)//Sn^(4+) =- 0.154V` `E^(@)` for `Cr_(2)O_(7)^(2-)//Cr^(3+) = 1.33V`

Answer» Correct Answer - `K = 10^(268)`
`Sn rarr Sn^(2+) +2e^(-) :. DeltaG_(1) =- 2 xx E_(1) xx F`
`Sn^(2+) rarr Sn^(4+) +2e^(-) :. DeltaG_(2) =- 2 xx E_(2) xx F`
`Sn rarr Sn^(4-) +4e^(-) :. DeltaG_(3) =- 4 xx E_(3) xx F`
`DeltaG_(3) = DeltaG_(1) +DeltaG_(2)`
`E_(3) = (2 xx 0 xx 136 +2 +(-0.154))/(4)`
`E_(3) rArr 0.009`
`E = E^(@)`
at equilibrium `E = 0`
`E^(@) = (0.0591)/(12) log K_(eq)`
`-0.009 +1.33 = (0.0591)/(12) xx log K_(eq)`
`1.321 =(0.0591)/(12) log K_(eq)`
`K_(eq) = 10^(268)`
Discussion
445.

The cathodic reaction in electrolysis of dilute `H_(2)SO_(4)` with platinum electrode is:A. oxidationB. reductionC. both oxidation and reductionD. neutralization

Answer» Correct Answer - D
Discussion
446.

Which of the following statements is true for a cell, `Pt | H_(2 (g)) | H^(+) || Cu^(2+) | Cu` ?A. Reduction occurs at `H_(2)` electrodeB. `H_(2)` is cathode and Cu is anodeC. `H_(2)` is anode and `Cu` is cathodeD. Oxidation occurs at Cu electrode

Answer» Correct Answer - C
`H_(2)` is anode and Cu is cathode .
Discussion
447.

The standard electrode potentials for the reactions, `Ag^(+) (a.q) + e^(-) rarr Ag (s)` `Sn^(2+) (a.q) + 2e^(-) rarr Sn (s)` at `25^(@)C` are 0.80 volt and `-0.14` volt respectively. The emf of the cell `Sn|Sn^(2+) (1M)|| Ag^(+) (1M) | Ag` is:A. 0.66 voltB. 0.80 voltC. 1.08 voltD. 0.94 volt

Answer» Correct Answer - D
Discussion
448.

When lead accumulator is charged, it isA. an electrolytic cellB. a galvanic cellC. a Daniel cellD. None of the above .

Answer» Correct Answer - A
Electrical energy is provided during charging of battery .
Discussion
449.

When lead accumulator is charged, it isA. an electrolytic cellB. a galvanic cellC. a Daniell cellD. none of these

Answer» Correct Answer - D
Discussion
450.

Assertion Decreasing concentration by dilution increases the equivalence conductance of aqueous solution NaCl. Reason With dilution. Degree of ionisation of NaCl increases.A. Both assertion and reason are correct and reason is the correct but reason is the correct explanation of the assertion.B. Both assertion and reason are correct but reason is not the correct explantion of assertion.C. Assertion is correct but reason is wrong.D. Assertion is wrong but reason is correct.

Answer» Correct Answer - c
Discussion