3195 + Interview Questions in ELECTROCHEMISTRY IN GENERAL AWARENESS Page 30 InterviewSolution

1451.	At the anode, the species having minimum reduction potential is formed form the oxidation of corresponding oxidizable species.
Answer» Correct Answer - T

Discussion

1452.	In a galvanic cell, the half cell with higher potential provides a reducing agent.
Answer» Correct Answer - T Half cell with higher potential involves reduction and thus provides an oxidizing agent.

Discussion

1453.	Using the data in Table 3.1, predict whether `Cl_(2)` disproportionates in alkaline solution: Strategy: Chlorine can exist in manu oxidation states. Chlorine in the `0` oxidation state, as in `Cl_(2)`, can be reduced. Chorine can also be oxidized to one of several positive oxidation states. so `Cl_(2)` can react with itself in a redox reaction, which means that it is possible for chlorine to disproportionate. Search out two different half reactions involving `Cl_(2)`
Answer» Solution: For the half reaction `overset(0)Cl_(2)(g,1"bar")+2e^(-)hArr overset(-1)(2Cl^(-))(aq.)` Table 3.1 lists `E^(@) = + 1.36`. This large positive reduction potentail shows that `Cl_(2)` can be easily reduced, i.e., it is a realtively powerful oxidizing agent. The table lists the alf-reaction `overset(+1)(2ClO^(-))(aq.1M)+2H_(2)O(1)+2e^(-)hArroverset(0)Cl_(2)("g,1 bar")+4OH^(-)(aq. 1M)` with `E^(@)` of `0.04V`. The reverse of this half-reactions is an oxidation of `Cl_(2)`. When the two half-reaction are combined, the overall reaction is `2Cl_(2)(g,1"bar")+4OH^(-)(aq.,1M)hArrCl^(-)(aq., 1M)+ClO^(-)(aq., 1M)+H_(2)O(1)` or, more simply `Cl_(2)(g)+2OH^(-)(aq.,1M)hArr Cl^(-)(aq.,1M)+ClO^(-)(aq.,1M)+H_(2)O(1)` The standard potential of this reaction is positive : `1.36 V - 0.40 V = 0.96 V`. Therefore, the reaction proceeds spontaneously from left to right. In other words, `Cl_(2)` in unstable in alkaline solution, it is converted to chloride anion and hypochlorite anion.

Discussion

1454.	Tollen reagent is used for the detection of aldehydes. When a solution of `AgNO_(3)` is added to glucose with `NH_(4)OH`, then gluconic acid is formed. `Ag^(o+)+e^(-) rarr Ag," "E^(c-)._(red)=0.8V` `C_(6)H_(12)O_(6)rarr underset(Gluconic aci d)(C_(6)H_(12)O_(7)+)2H^(o+)+2e^(-) , " "E^(c-)._(o x i d ) =-0.05V` `[Ag(NH_(3))_(2)]^(o+)+e^(-) rarr Ag(s)+2NH_(3), " "E^(c-)._(red)=0.337V` `[Use2.303xx(RT)/(F)=0.0592` and `(F)/(RT)=38.92at 298 K ]` When ammonia is added to the solution, `pH` is raised to `11`. Which half cell reaction is affected by `pH` and by how much ?A. `E_("oxd")^(@)` glucose will increase by a factor of `0.463 V`B. `E_("oxd")^(@)` glucose decrease by a factor of `0.463 V`C. `E_(red)^(@)Ag^(+)` will increase by a factor of `0.463 V`.D. `E_(red)^(@)Ag^(+)` will decrease by a factor of `0.463 V`.
Answer» Correct Answer - D `E^(@)` of `Ag^(+)//Ag = 0.8 V` `E^(@)` of `Ag^(+)//Ag` in `NH_(3) = 0.337 V`

Discussion

1455.	An ammeter and copper voltmeter are connected in aseries in an electric circuit through which a constant direct current flows. The ammeter shows `0.525` ampere. If `0.6354 g` of `Cu` is deposited in `1` hr, what is the percentge error of ammeter ? `["At. wt. of "Cu = 63.54]`
Answer» Correct Answer - 2 Current flown `= 0.525` ampere as shown by ammeter Actual current flown, `i = (w xx 96500)/(E xx t)` `=(0.6354 xx 96500)/((63.5)/(2) xx 60 xx 60) = 0.536` ampere Thus, error in current `= 0.536 - 0.525 = 0.011` `:. %` error in ameter `=0.011/0.536xx100` `= 2.05% approx 2.0 %`

Discussion

1456.	Tollen reagent is used for the detection of aldehydes. When a solution of `AgNO_(3)` is added to glucose with `NH_(4)OH`, then gluconic acid is formed. `Ag^(o+)+e^(-) rarr Ag," "E^(c-)._(red)=0.8V` `C_(6)H_(12)O_(6)rarr underset(Gluconic aci d)(C_(6)H_(12)O_(7)+)2H^(o+)+2e^(-) , " "E^(c-)._(o x i d ) =-0.05V` `[Ag(NH_(3))_(2)]^(o+)+e^(-) rarr Ag(s)+2NH_(3), " "E^(c-)._(red)=0.337V` `[Use2.303xx(RT)/(F)=0.0592` and `(F)/(RT)=38.92at 298 K ]` `2Ag^(o+)+C_(6)H^(12)O_(6)+H_(2)O rarr 2Ag^(s)+C_(6)H_(12)O_(7)+2H^(o+)` Find `lnK` of this reaction.A. `12.7`B. `25.33`C. `28.30`D. `46.29`
Answer» Correct Answer - B For reaction `C_(6)H_(12)O_(6) + H_(2)O + 2Ag^(+) rarr C_(6)H_(12)O_(7) + 2H^(+) + 2Ag` `E^(@) = E_(OP_("glucose"))^(@) + E_(RP_(Ag^(+)//Ag))^(@)` `E^(@) = -0.05 + 0.80 = 0.75 V` Now ` nFE^(@) = 2.303 Rtlog_(10)K` `log_(10)K = (nFE^(@))/(2.303 RT) = (nE^(@))/(2.303 RT//F)` `= (2 xx 0.75)/(0.0592) = 25.33`

Discussion

1457.	The `E^(@)` values for the changes given below are measured againest `NHE` at `27^(@)C`. `Cu^(2+) + e rarr Cu^(+), E^(@) = + 0.15 V`, `Cu^(+) + e rarr Cu, E^(@) = + 0.50 V`, `Zn^(2+) + 2e rarr Zn, E^(@) = - 0.76 V` The temperature coefficient of emf a cell designed as `Zn"\|"underset(1 M)(Zn^(2+))"\|\|"underset(0.1 M)(Cu^(2+))"\|"Cu` is `-1.4 xx 10^(-4)V` per degree. For a cell reaction in equilibrium `DeltaG = 0` and `DeltaG^(@) = -2.303 RTlog_(10)K_(c)`. The heat of reaction and entropy change during the reaction are are related by `DeltaG = DeltaH - TDeltaS`. The heat of reaction for the change `Zn+underset(0.1 M)(Cu^(2+)) hArr underset(1M)(Zn^(2+))+Cu` at `27^(@)C` is :A. `2.037 xx 10^(5)J`B. `2.116 xx 10^(5)J`C. `2.037 xx 10^(6)J`D. `2.116 xx 10^(6)J`
Answer» Correct Answer - A `E_(cell) = E_(cell)^(@) + (0.059)/(2)"log"([Cu^(2+)])/([Zn^(2+)])` `= 1.085 + (0.059)/(2)"log"(0.1)/(1) = 1.0555 V` `:. -DeltaG^(@) = n xx E xx F = 2 xx 1.055 xx 96500` `= 2.037 xx 10^(5)J`

Discussion

1458.	The `E^(@)` values for the changes given below are measured againest `NHE` at `27^(@)C`. `Cu^(2+) + e rarr Cu^(+), E^(@) = + 0.15 V`, `Cu^(+) + e rarr Cu, E^(@) = + 0.50 V`, `Zn^(2+) + 2e rarr Zn, E^(@) = - 0.76 V` The temperature coefficient of emf a cell designed as `Zn"\|"underset(1 M)(Zn^(2+))"\|\|"underset(0.1 M)(Cu^(2+))"\|"Cu` is `-1.4 xx 10^(-4)V` per degree. For a cell reaction in equilibrium `DeltaG = 0` and `DeltaG^(@) = -2.303 RTlog_(10)K_(c)`. The heat of reaction and entropy change during the reaction are are related by `DeltaG = DeltaH - TDeltaS`. If `9.65` ampere current is passed through making `Cu` anode `Zn` cathode for `1000` sec in the cell `Zn"\|"underset(1 M)(Zn^(2+))"\|\|"underset(0.1 M)(Cu^(2+))"\|"Cu`, the e.m.f. of cell after after passage of current would be :A. `1.066`B. `1.076`C. `1.086`D. `1.056`
Answer» Correct Answer - A `(w)/(E) = (i .t)/(96500) = (9.65 xx 1000)/(96500) = 0.1` `2e + Zn^(2+) rarr Zn` (Cathode) `Cu rarr Cu^(2+)+2e` (Anode) `:. [Zn^(2+)] = 1 - 0.10 = 0.9` `[Cu^(2+)] = 0.1 + 0.1 = 0.2` `:. E_(cell) = E_(cell)^(@) + (0.059)/(2)"log"([Cu^(2+)])/([Zn^(2+)])` `= 1.085 + (0.059)/(2)"log"(0.2)/(0.9) = 1.066 V`

Discussion

1459.	The `E^(@)` values for the changes given below are measured againest `NHE` at `27^(@)C`. `Cu^(2+) + e rarr Cu^(+), E^(@) = + 0.15 V`, `Cu^(+) + e rarr Cu, E^(@) = + 0.50 V`, `Zn^(2+) + 2e rarr Zn, E^(@) = - 0.76 V` The temperature coefficient of emf a cell designed as `Zn"\|"underset(1 M)(Zn^(2+))"\|\|"underset(0.1 M)(Cu^(2+))"\|"Cu` is `-1.4 xx 10^(-4)V` per degree. For a cell reaction in equilibrium `DeltaG = 0` and `DeltaG^(@) = -2.303 RTlog_(10)K_(c)`. The heat of reaction and entropy change during the reaction are are related by `DeltaG = DeltaH - TDeltaS`. The equilibrium constant for the disproportionation of `Cu^(+)` is:A. `8.5 xx 10^(-5)`B. `85 xx 10^(5)`C. `8.5 xx 10^(-4)`D. `8.5 xx 10^(6)`
Answer» Correct Answer - B `E^(@) = (0.059)/(1)logK_(c)` or `0.35 = (0.059)/(1)logK_(c)` `:. K_(c) = 8.5 xx 10^(5)`

Discussion

1460.	The `E^(@)` values for the changes given below are measured againest `NHE` at `27^(@)C`. `Cu^(2+) + e rarr Cu^(+), E^(@) = + 0.15 V`, `Cu^(+) + e rarr Cu, E^(@) = + 0.50 V`, `Zn^(2+) + 2e rarr Zn, E^(@) = - 0.76 V` The temperature coefficient of emf a cell designed as `Zn"\|"underset(1 M)(Zn^(2+))"\|\|"underset(0.1 M)(Cu^(2+))"\|"Cu` is `-1.4 xx 10^(-4)V` per degree. For a cell reaction in equilibrium `DeltaG = 0` and `DeltaG^(@) = -2.303 RTlog_(10)K_(c)`. The heat of reaction and entropy change during the reaction are are related by `DeltaG = DeltaH - TDeltaS`. Change in entropy for the reaction : `Zn+underset(0.1 M)(Cu^(2+)) hArrunderset(1M)(Zn^(2+))+Cu` at `27^(@)C` is :A. `14 J`B. `27 J`C. `-14 J`D. `-27 J`
Answer» Correct Answer - D `DeltaG = DeltaH - TDeltaS` `-2.037 xx 10^(5) = -2.118 xx 10^(5) -300 xx DeltaS` `DeltaS = -27J`

Discussion

1461.	The `E^(@)` values for the changes given below are measured againest `NHE` at `27^(@)C`. `Cu^(2+) + e rarr Cu^(+), E^(@) = + 0.15 V`, `Cu^(+) + e rarr Cu, E^(@) = + 0.50 V`, `Zn^(2+) + 2e rarr Zn, E^(@) = - 0.76 V` The temperature coefficient of emf a cell designed as `Zn"\|"underset(1 M)(Zn^(2+))"\|\|"underset(0.1 M)(Cu^(2+))"\|"Cu` is `-1.4 xx 10^(-4)V` per degree. For a cell reaction in equilibrium `DeltaG = 0` and `DeltaG^(@) = -2.303 RTlog_(10)K_(c)`. The heat of reaction and entropy change during the reaction are are related by `DeltaG = DeltaH - TDeltaS`. The `E^(@)` for `Cu rarr Cu^(2+) + 2e` is :A. `0.325 V`B. `0.35V`C. `-0.325 V`D. `-0.35 V`
Answer» Correct Answer - C `{:(e+Cu^(+)rarrCu,-DeltaG^(@)=1xx0.50xxF),(e+Cu^(2+)rarrCu^(+),-DeltaG^(@)=1xx0.15xxF):}/(Cu^(2+)+2e rarrCu,-DeltaG^(@)=0.65xx1xxF=E^(@)xx2xxF)` `E^(@) = 0.325V`

Discussion

1462.	The `E^(@)` values for the changes given below are measured againest `NHE` at `27^(@)C`. `Cu^(2+) + e rarr Cu^(+), E^(@) = + 0.15 V`, `Cu^(+) + e rarr Cu, E^(@) = + 0.50 V`, `Zn^(2+) + 2e rarr Zn, E^(@) = - 0.76 V` The temperature coefficient of emf a cell designed as `Zn"\|"underset(1 M)(Zn^(2+))"\|\|"underset(0.1 M)(Cu^(2+))"\|"Cu` is `-1.4 xx 10^(-4)V` per degree. For a cell reaction in equilibrium `DeltaG = 0` and `DeltaG^(@) = -2.303 RTlog_(10)K_(c)`. The heat of reaction and entropy change during the reaction are are related by `DeltaG = DeltaH - TDeltaS`. The heat of reaction for the change `Zn+underset(0.1 M)(Cu^(2+)) hArr underset(1M)(Zn^(2+))+Cu` at `27^(@)C` is :A. `-2118 xx 10^(5)J`B. `-2.197 xx 10^(5)J`C. `-2.237 xx 10^(5)J`D. `-1.818 xx 10^(5)J`
Answer» Correct Answer - A `DeltaH = nF[T((deltaE)/(deltaT))_(P)-E]` `= 2 xx 96500[300 xx (-1.4 xx 10^(-4)) -1.0555]` `= -02.118 xx 10^(5)J`

Discussion

1463.	The `E^(@)` values for the changes given below are measured againest `NHE` at `27^(@)C`. `Cu^(2+) + e rarr Cu^(+), E^(@) = + 0.15 V`, `Cu^(+) + e rarr Cu, E^(@) = + 0.50 V`, `Zn^(2+) + 2e rarr Zn, E^(@) = - 0.76 V` The temperature coefficient of emf a cell designed as `Zn"\|"underset(1 M)(Zn^(2+))"\|\|"underset(0.1 M)(Cu^(2+))"\|"Cu` is `-1.4 xx 10^(-4)V` per degree. For a cell reaction in equilibrium `DeltaG = 0` and `DeltaG^(@) = -2.303 RTlog_(10)K_(c)`. The heat of reaction and entropy change during the reaction are are related by `DeltaG = DeltaH - TDeltaS`. The `E^(@)` of `Zn\|Zn^(2+)\|\|Cu^(2+)\|Cu` cell is :A. `1.076 V`B. `1.091 V`C. `1.117 V`D. `0.90 V`
Answer» Correct Answer - A `E_(cell) = 1.085 + (0.059)/(2)"log"(0.1)/(0.2) = (0.1)/(0.2) = 1.076`

Discussion

1464.	The `E^(@)` values for the changes given below are measured againest `NHE` at `27^(@)C`. `Cu^(2+) + e rarr Cu^(+), E^(@) = + 0.15 V`, `Cu^(+) + e rarr Cu, E^(@) = + 0.50 V`, `Zn^(2+) + 2e rarr Zn, E^(@) = - 0.76 V` The temperature coefficient of emf a cell designed as `Zn"\|"underset(1 M)(Zn^(2+))"\|\|"underset(0.1 M)(Cu^(2+))"\|"Cu` is `-1.4 xx 10^(-4)V` per degree. For a cell reaction in equilibrium `DeltaG = 0` and `DeltaG^(@) = -2.303 RTlog_(10)K_(c)`. The heat of reaction and entropy change during the reaction are are related by `DeltaG = DeltaH - TDeltaS`. The `E^(@)` of `Zn\|Zn^(2+)\|\|Cu^(2+)\|Cu` cell is :A. `1.085 V`B. `1.1 V`C. `1.126 V`D. `0.91 V`
Answer» Correct Answer - A `E_(cell)^(@) = E_(OP_(Zn))^(@) + E_(RP_(Cu^(2+)))^(@)` `= 0.76 + 0.325 = 1.085 V`

Discussion

1465.	The `E^(@)` values for the changes given below are measured againest `NHE` at `27^(@)C`. `Cu^(2+) + e rarr Cu^(+), E^(@) = + 0.15 V`, `Cu^(+) + e rarr Cu, E^(@) = + 0.50 V`, `Zn^(2+) + 2e rarr Zn, E^(@) = - 0.76 V` The temperature coefficient of emf a cell designed as `Zn"\|"underset(1 M)(Zn^(2+))"\|\|"underset(0.1 M)(Cu^(2+))"\|"Cu` is `-1.4 xx 10^(-4)V` per degree. For a cell reaction in equilibrium `DeltaG = 0` and `DeltaG^(@) = -2.303 RTlog_(10)K_(c)`. The heat of reaction and entropy change during the reaction are are related by `DeltaG = DeltaH - TDeltaS`. The `E^(@)` for the reaction : `2Cu^(+) rarr Cu^(2+) + Cu` is:A. `+0.35 V `B. `-0.35 V`C. `+0.65 V`D. `-0.65 V`
Answer» Correct Answer - A `{:(Cu^(+)rarrCu^(2+)+e),(Cu^(+)+erarrCu):}/(2Cu^(+)rarrCu^(2+)+Cu)` `:. E^(@) = E_(OP_(Cu^(+)//Cu^(2+)))^(@)+E_(RP_(Cu^(+)//Cu))^(@)` `= -0.15 + 0.50 = +0.35 V`

Discussion

1466.	Salt bridge contains:A. calomelB. corrosive sublimateC. `H_2O`D. agar-agar pase
Answer» Correct Answer - D Agar-Agar is a gelatin, if is used in salt bridge aldge along whith KCl electrolyte.

Discussion

1467.	Civen `E^@` for ` Cu^(2+) rarr Cu^+ ` is ` + 0.15 V` and ` Cu^+ rarr Cu ` is ` +0.05 V` Calculate ` E^@` for ` Cu^(2+) rarr Cu`.A. `-0.325 V`B. `+0.325 V`C. `+0.65 V`D. `-0.65 V`
Answer» Correct Answer - B `Cu^(2+) + 2e rarr Cu` `E_(RP_(Cu))^(@) = + 0.325 V`

Discussion

1468.	In wich of the following electrochemical cell the overall cell reaction is :A. ` Zn\|H_2SO_4(aq) \|H_2(g) \|Pt`B. ` Zn\|H_2SO_4(aq) \|H_2(g) \|Pt`C. `Zn\|ZnSO_4(aq)\|H_2SO_4(aq)\|H_2(g)\|Pt`D. `Zn\|ZnSO_4(aq)\|H_2SO_4(aq)\|H_2(g)\|Pt`
Answer» Correct Answer - D For the cell: `Zn\|ZnSO_4 (aq) \|\|H_2SO_4(aq)\|H_2(g)\|Pt` LHE reaction: `Zn rarr Zn^(2+) +2e` RHE reaction : `2H^(+) +2e rarr K_2` Net reaction: `Zn+e2H^+ rarr Zn^+ + H_2` or `Zn(s) + H_2SO_4 (aq)rarr ZnSO_4 (aq) = H_2(g)`

Discussion

1469.	Consider the following ` E^@` values . `E_(Fe^(3+)//Fe^(2+)^@ = + 0.77 V`, `E_(Sn^(2+)//Sn)^@=- 15. V` The `E_(cell)^@` for the reaction , `Sn (s) + 2Fe_(aq)^(3+) rarr 2 Fe_(aq.)^(2+) + Sn_(aq.)^(2+) is :A. ` 0.63 V`B. `1. 40 V`C. ` 0. 91 V`D. ` 1. 668 V`
Answer» Correct Answer - C `E_(cell)^@ =E(OPSn//Sn^(2+)) +E_(Fe^(2+)^@//Fe^(2+)`. ` =0.14 + 0.77 = 0. 91 V`.

Discussion

1470.	Civen `E^@` for ` Cu^(2+) rarr Cu^+ ` is ` + 0.15 V` and ` Cu^+ rarr Cu ` is ` +0.05 V` Calculate ` E^@` for ` Cu^(2+) rarr Cu`.A. ` + 0.325 V`B. ` + 0.125 V`C. ` + 0.250 V`D. ` + 0. 160 V`
Answer» Correct Answer - A ` Cu^2 +le^- rarr Cu^+ E_1 =0.15 `…(i) `Cu +le^- rarr Cu E_2 =0.50` …(ii) `Cu^2 +le^- rarr Cu E_3 =?` …(iii) Clearly (iii) = (i) + (ii) ` -DeltaG_3^0 + (-Delta G_2^0)` ` 2xx Fxx E_3 = 1 xx F E_1 +1 xx Fxx E_2` `E_3 = (0.65)/2 =0.324 V`.

Discussion

1471.	The Nearst Equation: Calcualte the nonstandard electrode potential, `E`, for the `Fe^(3+)//Fe^(2+)` electrode when the concentration of `Fe^(2+)` is exactly firve time that of `Fe^(3+)` Strategy: The Nearst equation helps us calculate electrode potentials for concentrations other than one molar. The Tabulation of standard electrode potentials gives us the value of `E^(@)` for the reduction half-reaction. Use the balanced half-reaction and the given concentration ratio to calculate the value of `Q`. Finally substitude this into the Nearst equation with `n` equal to the number of moles of elecrons involved in the half-reaction.
Answer» The reduction half-reaction is `Fe^(3+)(aq.) + e^(-) rarr Fe^(2+)(aq.)E_(Fe^(3+)//Fe^(2+)) = + 0.77V` Calculating the value of `Q` by using the fact that the concentration of `Fe^(2+)` is five times that of `Fe^(3+)`: `Q = ["Red"]^(b)/["Ox"]^(a) = C_(Fe^(2+))/C_(Fe^(3+)) = (5C_(Fe^(3+)))/(C_(Fe^(3+))) = 5` The balanced half-cell reaction shows one mole of electrons, or `n = 1`. Putting values into the Nearst equation, `E_(Fe^(3+)//Fe^(2+))^(@) = E_(Fe^(3+)//Fe^(2+))^(@) - (0.0592 V)/(n) log Q` `= +0.77 V - (0.0592 V)/(1) log 5` `= 0.77 V - 0.04 V` `= + 0.73`

Discussion

1472.	In what ways are fuel cells and galvanic cells similar and in what ways are they different ?
Answer» Similarity between fuel cells and galvanic cells : In both the cells, there is oxidation at anode and j reduction at cathode. The cell potential is developed due to net redox reactions. Both are galvanic cells. Difference in fuel cells and galvanic cells : Fuel cells involve electrodes with large surface area while galvanic cells involve electrodes with j compact surface area. Fuel cells involve gaseous materials on a large scale while galvanic cells involve gaseous materials at a definite pressures along with electrolytes or there may not be gases. In fuel cells, the cell potential is developed due to exothermic combustion reactions while in galvanic cell, cell potential is developed due to normal redox reactions. In fuel cells gaseous electrode materials are continuously supplied from outside while in galvanic cells electrode materials have constant concentration or may change due to reactions.

1472.

In what ways are fuel cells and galvanic cells similar and in what ways are they different ?

Answer»

Similarity between fuel cells and galvanic cells :

In both the cells, there is oxidation at anode and j reduction at cathode.
The cell potential is developed due to net redox reactions.
Both are galvanic cells.

Difference in fuel cells and galvanic cells :

Fuel cells involve electrodes with large surface area while galvanic cells involve electrodes with j compact surface area.
Fuel cells involve gaseous materials on a large scale while galvanic cells involve gaseous materials at a definite pressures along with electrolytes or there may not be gases.
In fuel cells, the cell potential is developed due to exothermic combustion reactions while in galvanic cell, cell potential is developed due to normal redox reactions.
In fuel cells gaseous electrode materials are continuously supplied from outside while in galvanic cells electrode materials have constant concentration or may change due to reactions.

Discussion

1473.	In an electrolchemical cellA. Potential energy changes into kinetic energyB. Kinetic energy changes into potential energyC. Chemical energy changes into electrical energyD. Electrical energy changes into chemical energy
Answer» Correct Answer - C In the electrochemical cell chemical energy changes into electrical energy.

Discussion

1474.	Given `Delta G^(@) = -n F E_("cell")^(@)` and `DeltaG^(@) = - RT` ln K The value of n = 2 will be given by the slope of which line in the fig. A. OAB. OBC. OCD. OD
Answer» Correct Answer - B Plot of ln K vs. `E_("cell")^(@)` will have slope = `(1)/(2) (RT)/(F)` . `Delta G^(@) = - n FE _(cell)^(@) and Delta G^(@) = - RT ln K ` `therefore - n FE_(cell)^(@) = - RT ln K ` `therefore E_("cell")^(@) = (RT)/(nF) ln K = (RT)/(2F) ln K ` This is an equation of straight line Slope = `(RT)/(2F) = (8.314 xx 298)/(2 xx 96500) = 0.0128`

Discussion

1475.	Write the cell reaction for which `E_(cell) =E_(cell)^(0) -(RT)/(2F)ln ""([Mg^(2+)])/([Ag^(+)]^(2))`
Answer» Given `E_(cell) =E_(cell)^(0) -(RT)/(2F) ln ""([Mg ^(2+)])/([Ag^(+)]^(2))` The cell reaction is `Mg_((s))//Mg _((aq))^(2+)\|\|Ag_((aq))^(+)\|Ag_((s))` `Mg yo Mg^(2+)+2e^(-)` (Oxidation) `2 Ag^(+) +2e^(-) to Ag` (Reduction) `overline underline(Mg+ 2 Ag^(+)to Mg^(+2)+Ag)`

Discussion

1476.	How is `E^(0)` cell related mathematically to the equilibrium constant `K_(c)` of the cell reaction ?
Answer» Relation between `E_(0)` cell and equilibrium constant `K_(C)` of the cell reaction `E_(cell)^(0)=(2.303RT)/(nF)log K_(C)` N = number of electrons involved F = Faraday = `96500 C mol ^(-1)` T = Temperature R = gas constant

Discussion

1477.	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 : Electrolyss of an aqueous solution of KI gives `I_2` at the anode but that of ` KF` gives ` O_2` at the anode and not `f_2` `F_2` is more reactive than `I_2`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 - B Correct explanation. `I^-` ions have much higher oxidation potential than of qater while `F^-` ions have much lower oxidation potential than that of water.

Discussion

1478.	Assertion `(A):` Equivalent conductance increase with dilution for an electrolyte solution. Reason `(R):` The number of ions per litre of electrolyte increases with dilution.A. If both the assertion and reason are true but the reason is not 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 true
Answer» Correct Answer - B Correct explanation. Molar conductance increases because with dilution in a weak electrolyte , dissociation increases while in strong electrolyte, interionic attractions derccrease .

Discussion

1479.	If the specific resistance of a solution of concentration C g equivalent `"litre"^(-1)` is R, then its equivalent conductance is:A. `(100R)/(C)`B. `(RC)/(1000)`C. `(1000)/(RC)`D. `(C)/(1000 R)`
Answer» Correct Answer - C

Discussion

1480.	Given below are a set of half-cell reactions (in acidic medium) alongwith their E° (in volt) values.l2+2e- → 2l-E°= 0.54Cl2+2e- → 2Cl-E°= 1.3Mn+3 +e- →Mn+2E°= 1.50Fe+3+e- → Fe+2E°= 0.7O2+4H+4e →2H2OE°= 1.231.Among the following, identify the correct statement (1) Cl– is oxidised by O2 (2) Fe+2 is oxidised by iodine (3) I– is oxidised by chlorine (4) Mn+2 is oxidised by chlorine2. While Fe+3 is stable, Mn+3 is not stable in acid solution because (1) O2 oxidises Mn+2 to Mn+3 (2) O2 oxidises both Mn+2 to Mn+3 and Fe+2 to Fe+3 (3) Fe+3 oxidises H2O to O2 (4) Mn+3 oxidises H2O to O23.The strongest reducing agent in aqueous solution is (1) I– (2) Cl– (3) Mn+2 (4) Fe+2
Answer» The Correct option is 1.(3) I^– is oxidised by chlorine 2.(4) Mn⁺³ oxidises H₂O to O₂ 3. (1) I^–

1480.

Given below are a set of half-cell reactions (in acidic medium) alongwith their E° (in volt) values.l2+2e- → 2l-E°= 0.54Cl2+2e- → 2Cl-E°= 1.3Mn+3 +e- →Mn+2E°= 1.50Fe+3+e- → Fe+2E°= 0.7O2+4H+4e →2H2OE°= 1.231.Among the following, identify the correct statement (1) Cl– is oxidised by O2 (2) Fe+2 is oxidised by iodine (3) I– is oxidised by chlorine (4) Mn+2 is oxidised by chlorine2. While Fe+3 is stable, Mn+3 is not stable in acid solution because (1) O2 oxidises Mn+2 to Mn+3 (2) O2 oxidises both Mn+2 to Mn+3 and Fe+2 to Fe+3 (3) Fe+3 oxidises H2O to O2 (4) Mn+3 oxidises H2O to O23.The strongest reducing agent in aqueous solution is (1) I– (2) Cl– (3) Mn+2 (4) Fe+2

Answer»

The Correct option is

1.(3) I^– is oxidised by chlorine

2.(4) Mn⁺³ oxidises H₂O to O₂

3. (1) I^–

Discussion

1481.	Redox reactions play a pivotal role in chemistry and biology. The value of standard reduction potetials `(E^(@))` of the two half cells reactions decide which way the reaction is expected to proceed. A simple example is of Daniell cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with with `E^(@)` values. Using the data obtain the correct explanation to the questions that are mentioned. `I_(2)+e^(-) to 2I^(-) , E^(@)=0.54" V "` `Cl_(2)+2e^(-) to 2Cl^(-) , E^(@)=1.36" V"` `Mn^(2+)+2e^(-) to Mn , E^(@)=1.50" V"` `Fe^(3+)+e^(-) to Fe^(2+) , E^(@)=0.77" V"` `O_(2)+4H^(+)+4e^(-) to 2H_(2)O , E^(@)=1.23" V"`. Among the following, identify the correct statement :A. `Cl^(-)` ion is oxidised by `O_(2)`B. `Fe^(2+)`ion is oxidised by iodine.C. `I^(-)` ion is oxidised by chlorine.D. `Mn^(2+)` ion is oxidised by chlorine.
Answer» Correct Answer - C (c ) `2I^(-)+Cl_(2) to 2Cl^(-)+l_(2)` `I^(-)` ions is oxidised by `Cl_(2)` because e.m.f. of the cell comes out to be positive. `E_(cell)^(@)=E_(Cl_(2)//2Cl^(-))^(@)-E_(2I^(-)//I_(2))^(@)` `=1.36-0.54=0.82" V"`

Discussion

1482.	Redox reactions play a pivotal role in chemistry and biology. The value of standard reduction potetials `(E^(@))` of the two half cells reactions decide which way the reaction is expected to proceed. A simple example is of Daniell cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with with `E^(@)` values. Using the data obtain the correct explanation to the questions that are mentioned. `I_(2)+e^(-) to 2I^(-) , E^(@)=0.54" V "` `Cl_(2)+2e^(-) to 2Cl^(-) , E^(@)=1.36" V"` `Mn^(2+)+2e^(-) to Mn , E^(@)=1.50" V"` `Fe^(3+)+e^(-) to Fe^(2+) , E^(@)=0.77" V"` `O_(2)+4H^(+)+4e^(-) to 2H_(2)O , E^(@)=1.23" V"`. While `Fe^(2+)` ion is stable, `Mn^(2+)` ion is not stable in acid solution because :A. `O_(2)` oxidises `Mn^(2+)` to `Mn^(3+)`B. `O_(2)` oxidises both `Mn^(2+)` to `Mn^(3+)` to `Fe^(3+)`C. `Fe^(3+)` oxidises `H_(2)O` to `O_(2)`D. `Mn^(3+)` oxidises `H_(2)O` to `O_(2)`
Answer» Correct Answer - D (d) The following reaction is feasible because `E_(cell)` is positive `Mn^(3+)+e^(-) to Mn^(2+)]xx4 " " cathode` `2H_(2)O to 4H^(+)+O_(2)+4e^(-) " " anode` `4Mn^(3+)+2H_(2)O to 4Mn^(2+)+O_(2)+4H^(+)` `E_(cell)^(@)=E_(Mn^(3+)//Mn^(2+))^(@)-E_(H_(2)"O"//O_(2))^(@)` `=1.50+(-1.23)=0.27" V"`

Discussion

1483.	Redox reactions play a pivotal role in chemistry and biology. The value of standard reduction potetials `(E^(@))` of the two half cells reactions decide which way the reaction is expected to proceed. A simple example is of Daniell cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with with `E^(@)` values. Using the data obtain the correct explanation to the questions that are mentioned. `I_(2)+e^(-) to 2I^(-) , E^(@)=0.54" V "` `Cl_(2)+2e^(-) to 2Cl^(-) , E^(@)=1.36" V"` `Mn^(2+)+2e^(-) to Mn , E^(@)=1.50" V"` `Fe^(3+)+e^(-) to Fe^(2+) , E^(@)=0.77" V"` `O_(2)+4H^(+)+4e^(-) to 2H_(2)O , E^(@)=1.23" V"`. Sodium fusion extract obtained from aniline, on treatment with iron (II) sulphate and `H_(2)SO_(4)` in the presence of air, gives a prussian blue precipitate. The blue colour is due to the formation of :A. `Fe_(4)[Fe(CN)_(6)]_(3)`B. `Fe_(3)[Fe[(CN)_(6)]_(2)`C. `Fe_(4)[Fe(CN)_(6)]_(2)`D. `Fe_(3)[Fe(CN)_(6)]_(3)`.
Answer» Correct Answer - A (a) `Na +C+N to NaCN` `Fe^(2+)+6CN^(-) to [Fe(CN)_(6)]^(4-)` `4Fe^(3+)+3[Fe(CN)_(6)]^(4-) to underset("Prussian blue")(Fe_(4)[Fe(CN)_(6)]_(3))`

Discussion

1484.	Calculate the equilibrium constant for the reaction, `Zn+Cu^(2+)hArrCu+Zn^(2+)` Given: `E^(@)` for `Zn^(2+)//Zn=-0.763V` and for `Cu^(2+)//Cu=+0.34V,R=8.314JK^(-1)mol^(-1),F=96500" C "mol^(-1)`
Answer» Correct Answer - `2.121xx10^(37)` `E_(cell)^(@)=E_(Cu^(2+)//Cu)^(@)-E_(Zn^(2+)//Zn)^(@)=0.34-(-0.0763)V=1.103V`.

Discussion

1485.	An apparatus used for the measurement of quantity of electricity is known as aA. CalorimeterB. CathetometerC. CoulometerD. Colorimeter
Answer» Correct Answer - C Cu voltmeter or Cu or Ag coulometer are used to detect the amount deposited on an electrode during passage of known charge through solution .

Discussion

1486.	In which of the following cases `(E_(cell)-E_(cell)^(@))` is zero :A. `Cu\|Cu^(2+)(0.01 M)\|\|Ag^(+)(0.1 M)\|Ag`B. `Pt(H_(2))\|pH=1\|\|Zn^(2+)(0.01 M)\|Zn`C. `Pt(H_(2))\|pH=1\|\|Zn^(2+)(1 M)\|Zn`D. `Pt(H_(2))\|H^(+)=0.01 M)\|\|Zn^(2+)(0.01 M)\|Zn`
Answer» Correct Answer - A::B (a,b) are correct. (a) `E_(cell)=E_(cell)^(@)-(2.303" RT")/(2F)"log"([Cu^(2+)])/([Ag^(+)]^(2))` `E_(cell)-E_(cell)^(@)=(2.303" RT")/(2F)"log"(0.01)/((0.1)^(2))` `E_(cell)-E_(cell)^(@)=0` (b) `E_(cell)-E_(cell)^(@)=-(2.303" RT")/(2F)"log"([H^(+)]^(2))/([Zn^(2+)])` `E_(cell)-E_(cell)^(@)=-(2.303" RT")/(2F)"log"((1xx10^(-1))^(2))/((0.01))=0` (c ) `E_(cell)-E_(cell)^(@)=-(2.303" RT")/(2F)"log"([H^(+)]^(2))/([Zn^(2+)])` `=-(2.3030" RT")/(2F)"log"((1xx10^(-1))^(2))/(1)ne 0` (d) `E_(cell)-E_(cell)^(@)=-(2.303" RT")/(2F)"log"([H^(+)]^(2))/([Zn^(2+)])` `=-(2.303" RT")/(2F)"log"((0.1)^(2))/((0.01))ne0`

Discussion

1487.	The unit of charge is :A. voltB. ampereC. coulombD. none
Answer» Correct Answer - C 1 C = 1 ampere `xx` 1 sec

Discussion

1488.	One gram equivalent of a substance is liberated at an electrode byA. `6.22xx10^(23)` electronsB. 96500 CC. 1 amp of current for one secondD. 1 amp current for 96500 C
Answer» Correct Answer - A::B::D (a,b,d) are correct.

Discussion

1489.	The product (ampere `xx` seconds) is equal to the number ofA. coulomb transferredB. electrons transferredC. faradays transferredD. volt
Answer» Correct Answer - A Coulomb = ampere `xx` second .

Discussion

1490.	Electrochemical equivalent of a substance is equal to its quantity liberated at electrode on passing electricity equal toA. 1 coulombB. 1 ampereC. 1 voltD. 96,500 coulomb
Answer» Correct Answer - A `w = Z xx i.t` if `ixx t = 1 ` coulomb , when w = Z .

Discussion

1491.	The quantity of electricity required to librate 0.1 g equivalent of an element at the electrode isA. 9650 CB. 96500 CC. 965 CD. 96.5 C
Answer» Correct Answer - A Electricity required to liberate 0.1 g equivalent of an element `=96500xx0.1=9650 C`

Discussion

1492.	Faraday has the dimension ofA. CoulombsB. Coulomb equivalentC. Coulomb per equivalentD. Coulomb per degree kelvin
Answer» Correct Answer - C

Discussion

1493.	On passing 3 A of electricity for 50 min, 1.8 g of metal deposits. The equivalent mass of metal isA. 20.5B. 25.8C. 19.3D. 30.7
Answer» Correct Answer - C We know that, `E=(F.w)/(it)=(96500xx1.8)/(3xx50xx60)=19.3`

Discussion

1494.	When one of ampere current flows for 1 sec through a conductor, this quantity of electricity is known as:-A. FaradayB. CoulombC. E.M.F.D. Ohm
Answer» Correct Answer - B

Discussion

1495.	How much charge is required for the reduction of 1 mole of Zn2+ to Zn?
Answer» Zn²⁺ 2e^- → Zn(s). 2 Faradays or 2 x 96500 coloumb charge is required.

Discussion

1496.	In the equation `overset(@)(Lambda) = Lambda + BsqrtC`, the constant `B` depends upon :A. `C^(1)//(2)`B. Stoichiometry of electrolyteC. ResistanceD. Conductivity
Answer» Correct Answer - B The value of `B` for uni-univalent `(NaCl)`, bi-bivalent `(MgO)`, uni-bivalent `(MgCl_(2))` is altogether different.

Discussion

1497.	Which represents disproportionation?A. `2Cu^(+) rarr Cu^(2+) + Cu`B. `3I_(2) rarr 5I^(-) + I^(5+)`C. `H_(2)O + Cl_(2) rarr Cl^(-) + ClO^(-) + 2H^(+)`D. All of these
Answer» Disproportionation is the process in which one molecule is same element is oxidised on the cost of same element of other molecule.

Discussion

1498.	Which represents disproportionation?A. `2 Cu^(+) to Cu^(2+) + Cu`B. `3 I_(2) to 5 I^(-) + I^(5+)`C. `H_(2)O + Cl_(2) to Cl^(-) + ClO^(-) + 2H^(+)`D. All
Answer» Correct Answer - D Disproportionation is the process in which same element is oxidised and reduced .

Discussion

1499.	Which of the following solution will turn blue when placed in copper vessel?A. `AgNO_(3)`B. `NaCl`C. `ZnSO_(4)`D. `KNO_(3)`
Answer» Correct Answer - A `2AgNO_(3)+CurarrCu(NO_(3))_(2)+2Ag` Due to the presence of `Cu^(2+)(aq)` ions in solution it will turn blue.

Discussion

1500.	Which of the following process represents disproportionation?A. `2C u^(+)rarrCu^(2+)+Cu`B. ` Fe^(2+)rarr(1)/(2)Fe^(3+)+e^(-)`C. `Cu^(+2)+ZnrarrZn^(2+)+Cu`D. All the above.
Answer» Correct Answer - A Disproportionation refers to a process in which same species increases as well decreases its electrons. `Cl^(-)rarrCl+e^(-)` `Cl+ClrarrCl_(2)uparrow`.

Discussion

Explore topic-wise InterviewSolutions in .

At the anode, the species having minimum reduction potential is formed form the oxidation of corresponding oxidizable species.

In a galvanic cell, the half cell with higher potential provides a reducing agent.

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In wich of the following electrochemical cell the overall cell reaction is :A. ` Zn|H_2SO_4(aq) |H_2(g) |Pt`B. ` Zn|H_2SO_4(aq) |H_2(g) |Pt`C. `Zn|ZnSO_4(aq)|H_2SO_4(aq)|H_2(g)|Pt`D. `Zn|ZnSO_4(aq)|H_2SO_4(aq)|H_2(g)|Pt`

Consider the following ` E^@` values . `E_(Fe^(3+)//Fe^(2+)^@ = + 0.77 V`, `E_(Sn^(2+)//Sn)^@=- 15. V` The `E_(cell)^@` for the reaction , `Sn (s) + 2Fe_(aq)^(3+) rarr 2 Fe_(aq.)^(2+) + Sn_(aq.)^(2+) is :A. ` 0.63 V`B. `1. 40 V`C. ` 0. 91 V`D. ` 1. 668 V`

Civen `E^@` for ` Cu^(2+) rarr Cu^+ ` is ` + 0.15 V` and ` Cu^+ rarr Cu ` is ` +0.05 V` Calculate ` E^@` for ` Cu^(2+) rarr Cu`.A. ` + 0.325 V`B. ` + 0.125 V`C. ` + 0.250 V`D. ` + 0. 160 V`

In what ways are fuel cells and galvanic cells similar and in what ways are they different ?

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Given `Delta G^(@) = -n F E_("cell")^(@)` and `DeltaG^(@) = - RT` ln K The value of n = 2 will be given by the slope of which line in the fig. A. OAB. OBC. OCD. OD

Write the cell reaction for which `E_(cell) =E_(cell)^(0) -(RT)/(2F)ln ""([Mg^(2+)])/([Ag^(+)]^(2))`

How is `E^(0)` cell related mathematically to the equilibrium constant `K_(c)` of the cell reaction ?

If the specific resistance of a solution of concentration C g equivalent `"litre"^(-1)` is R, then its equivalent conductance is:A. `(100R)/(C)`B. `(RC)/(1000)`C. `(1000)/(RC)`D. `(C)/(1000 R)`

Calculate the equilibrium constant for the reaction, `Zn+Cu^(2+)hArrCu+Zn^(2+)` Given: `E^(@)` for `Zn^(2+)//Zn=-0.763V` and for `Cu^(2+)//Cu=+0.34V,R=8.314JK^(-1)mol^(-1),F=96500" C "mol^(-1)`

An apparatus used for the measurement of quantity of electricity is known as aA. CalorimeterB. CathetometerC. CoulometerD. Colorimeter

In which of the following cases `(E_(cell)-E_(cell)^(@))` is zero :A. `Cu|Cu^(2+)(0.01 M)||Ag^(+)(0.1 M)|Ag`B. `Pt(H_(2))|pH=1||Zn^(2+)(0.01 M)|Zn`C. `Pt(H_(2))|pH=1||Zn^(2+)(1 M)|Zn`D. `Pt(H_(2))|H^(+)=0.01 M)||Zn^(2+)(0.01 M)|Zn`

The unit of charge is :A. voltB. ampereC. coulombD. none

One gram equivalent of a substance is liberated at an electrode byA. `6.22xx10^(23)` electronsB. 96500 CC. 1 amp of current for one secondD. 1 amp current for 96500 C

The product (ampere `xx` seconds) is equal to the number ofA. coulomb transferredB. electrons transferredC. faradays transferredD. volt

Electrochemical equivalent of a substance is equal to its quantity liberated at electrode on passing electricity equal toA. 1 coulombB. 1 ampereC. 1 voltD. 96,500 coulomb

The quantity of electricity required to librate 0.1 g equivalent of an element at the electrode isA. 9650 CB. 96500 CC. 965 CD. 96.5 C

Faraday has the dimension ofA. CoulombsB. Coulomb equivalentC. Coulomb per equivalentD. Coulomb per degree kelvin

On passing 3 A of electricity for 50 min, 1.8 g of metal deposits. The equivalent mass of metal isA. 20.5B. 25.8C. 19.3D. 30.7

When one of ampere current flows for 1 sec through a conductor, this quantity of electricity is known as:-A. FaradayB. CoulombC. E.M.F.D. Ohm

How much charge is required for the reduction of 1 mole of Zn2+ to Zn?

In the equation `overset(@)(Lambda) = Lambda + BsqrtC`, the constant `B` depends upon :A. `C^(1)//(2)`B. Stoichiometry of electrolyteC. ResistanceD. Conductivity

Which represents disproportionation?A. `2Cu^(+) rarr Cu^(2+) + Cu`B. `3I_(2) rarr 5I^(-) + I^(5+)`C. `H_(2)O + Cl_(2) rarr Cl^(-) + ClO^(-) + 2H^(+)`D. All of these

Which represents disproportionation?A. `2 Cu^(+) to Cu^(2+) + Cu`B. `3 I_(2) to 5 I^(-) + I^(5+)`C. `H_(2)O + Cl_(2) to Cl^(-) + ClO^(-) + 2H^(+)`D. All

Which of the following solution will turn blue when placed in copper vessel?A. `AgNO_(3)`B. `NaCl`C. `ZnSO_(4)`D. `KNO_(3)`

Which of the following process represents disproportionation?A. `2C u^(+)rarrCu^(2+)+Cu`B. ` Fe^(2+)rarr(1)/(2)Fe^(3+)+e^(-)`C. `Cu^(+2)+ZnrarrZn^(2+)+Cu`D. All the above.