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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 2751. |
The standard potential of the following cell is `0.23V` at `15^(@)C` and `0.21V` at `35^(@)C:` `Pt|H_(2(g)|HCl(aq)|AgCl(s)|Ag(s)` `a.` Write the cell reaction. `b.` Calculate `DeltaH^(c-)` and `DeltaS^(c-)` for the cell reaction by assuming that these quantities remain unchanged in the range `15^(@)C` to `35^(@)C` `c.` Calculate the solubility of `AgCl` in water at `25^(@)C`. Given `:` The standard reduction potential of `Ag^(o+)(aq)|Ag(s)` is `0.80V` at `25^(@)C`. |
| Answer» Correct Answer - `DeltaH^(@) =- 49987 J mol^(-1), DeltaS^(@) =- 96.5 J mol^(-1)K^(-1), s = 1.47 xx 10^(-5)M` | |
| 2752. |
Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox titration. Some half cell reaction and their standard potentials are given below: `MnO_(4)^(-)(aq) +8H^(+)(aq) +5e^(-) rarr Mn^(2+)(aq) +4H_(2)O(l) E^(@) = 1.51V` `Cr_(2)O_(7)^(2-)(aq) +14H^(+) (aq) +6e^(-) rarr 2Cr^(3+)(aq) +7H_(2)O(l), E^(@) = 1.38V` `Fe^(3+) (aq) +e^(-) rarr Fe^(2+) (aq), E^(@) = 0.77V` `CI_(2)(g) +2e^(-) rarr 2CI^(-)(aq), E^(@) = 1.40V` Identify the only correct statement regarding quantitative estimation of aqueous `Fe(NO_(3))_(2)`A. `MnO_(4)^(-)` can be used in aqeuous HClB. `Cr_(2)O_(7)^(2-)` can be used in aqueous HClC. `MnO_(4)^(-)` can be used in aqueous `H_(2)SO_(4)`D. `Cr_(2)O_(7)^(2-)` can be used in aqueous `H_(2)SO_(4)`. |
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Answer» Correct Answer - A When `MnO_(4)^(-)` is used in a queous HCl, `MnO_(2)` reacts wit `Fe^(2+)` as well as with HCl as explained below. `MnO_(4)^(-)+5Fe^(2+)+8H^(+)rarr5Fe^(3+)+ 4H_(2)O+Mn^(2+)` `E_("Cell")^(@)=1.51- 0.77=0.74Vgt0` `MnO_(4)^(-)+16H^(+)+10Cl^(-)rarrMn^(2+)+8H_(2)O+5Cl_(2)` `E_("cell"^(@)=1.51-0.77=0.74Vgt0` As such `MnO_(4)^(-)` cannot be used for the quantitative estimation of aqueous `Fe(NO_(3))_(2)` . Thus option A is not correct. Similarly we can show that `Cr_(2)O_(7)^(2-)` will react with `Fe^(2+)` but not with HCl. `Cr_(2)O_(7)^(2-)+6Fe^(2+)+14H^(+)rarr2Cr^(3+)+7H_(2)O+6 Fe^(3+)` `E_("cell")^(@)=1.38-0.77=0.61Vgt0` `Cr_(2)O_(7)^(2-)+ 6Cl^(-)+14H^(+)rarr2Cr^(3+)+7H_(2)O^(+)3Cl_(2)` `E_("cell")^(@)=1.38-1.40=-0.02Vlt0` Thus, the normal reactions is not possible Thus `Cr_(2)O^(7)^(-)` in aqueous HCl can be used for the quantitative estimation of aqueous `Fe(NO_(3))_(2)`. Option C and D are correct. |
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| 2753. |
Saturated solution of `KNO_(3)` is used to make salt bridge because:A. velocity of `K^(+)` is greater than that of `NO_(3)^(-)`B. velocity of `NO_(3)^(-)` is greater than that of `K^(+)`C. velocity of both `K^(+)` and `NO_(3)^(-)` are nearly the sameD. `KNO_(3)` is higly soluble in water |
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Answer» Correct Answer - C The salt bridge possesses the electrolyte having nearly same ionic mobilities of its cation and anion. |
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| 2754. |
`MnO_(4)^(-) + 8H^(+) + 5e^(-) rightarrow Mn^(2+) + 4H_(2)O`, `E^(@) = 1.51V` `MnO_(2) + 4H^(+) + 2e^(-) righarrow Mn^(2+) + 2H_(2)O` `E^(@) = 1.23V` `E_(MnO_(4)^(-)|MnO_(2)`A. `- 1 V `B. `-1.70` VC. `+ 1` VD. `1.70` V |
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Answer» Correct Answer - D Eq(i) - Eq(ii) `MnO_(4)^(-) + 4H^(+) + 3e^(-) to Mn underset(+4)(O_(2)) + 2 H_(2)O , E_(3)^(@) = ?` `Delta G_(3)^(@) = DeltaG_(1)^(@) - DeltaG_(2)^(@) - n_(3) E_(3)^(@) F = - nE_(1)^(@) F - n_(2) E_(2)^(@) F - 3 E_(3)^(@) = (5 xx 1.51) - (-2 xx 1.23)` `therefore - E_(3)^(@) = ((-5 xx 1.51) + (2 xx 1.23))/(3) = -1.70` V . |
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| 2755. |
For an electrolyte solution of `0.05 mol L^(-1)` , the conductivity has been found to be `0.0110S cm^(-1)`. The molar conductivity isA. `0.05S cm^(2)mol ^(-1)`B. `550 S cm^(2) mol ^(-1)`C. `0.22 S cm^(2) mol ^(-1)`D. `220 S cm^(2) mol ^(-1)` |
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Answer» Correct Answer - d `wedge_(m)=(kxx1000)/(M)=(0.0110xx1000)/(0.05)` `=220S cm^(2) mol ^(-1)` |
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| 2756. |
For an electrolyte solution of `0.05 mol L^(-1)` , the conductivity has been found to be `0.0110S cm^(-1)`. The molar conductivity isA. `0.055 cm^(2) mol^(-1)`B. `550 S cm^(2) mol^(-1)`C. `0.22 S cm^(2) mol^(-1)`D. `220 S cm^(2) mol^(-1)` |
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Answer» Correct Answer - D `Lambda_(m) = k xx (1000)/("Molarity") = (0.11 xx 1000)/(0.05)` `220 S cm^(2) mol^(-1)` . |
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| 2757. |
A conductivity cell having cell constant `8.76 cm^(-1)` placed in `0.01` M solution of an electrolyte offered a resistance of 1000 ohms . What is the conductivity of the electrolyte ?A. `8.76 xx 10^(-4) ohm^(-1) cm^(-1)`B. `8.76 xx 10^(-3) ohm^(-1) cm^(-1)`C. `8.76 xx 10^(-2) ohm^(-1) cm^(-1)`D. `8.76 xx 10^(-1) ohm^(-1) cm^(-1)` |
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Answer» Correct Answer - B `k = (1)/(R) xx` cell const = `(1)/(1000) xx 8.76` `= 8.76 xx 10^(-3) Omega^(-1) cm^(-1)` |
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| 2758. |
A standard solution of `KNO_(3)` is used to make salt bridge, becauseA. velocity of `K^(+)` is greaer than that of `NO_(3)^(-)`B. velocity of `NO_(3)^(-)` is greater than that of `K^(+)`C. velocities of both `K^(+)` and `NO_(3)^(-)` are neraly the sameD. `KNO_(3)` is highly soluble in water |
| Answer» Correct Answer - C | |
| 2759. |
A standard solution of `KNO_(3)` is used to make salt bridge, becauseA. Velocity of `K^(+)` is greater than of `NO_(3)^(-)`B. Velocity of `NO_(3)^(-)` is greater than that of `K^(+)`C. Velocities of both `K^(+)` and `NO_(3)^(-)` are nearly the sameD. `KNO_(3)` is highly soluble in water |
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Answer» Correct Answer - C (c ) Velocities (ionic mobilities) of `K^(+)` and `NO_(3)^(-)` ions are nearly the same. |
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| 2760. |
A standard solution of `KNO_(3)` is used to make salt bridge, becauseA. velocity of `K^(+)` is greater than that of `NO_(3)^(-)`B. velocity of `NO_(3)^(-)` is greater than that of `K^(+)`C. Velocities of both `K^(+)` and `NO_(3)^(-)` are neraly the sameD. `KNO_(3)` is highly soluble in water. |
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Answer» Correct Answer - C Velocities of `K^(+)` and `NO_(3)^(-)` ions are newly the same. |
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| 2761. |
Find the volume of `Cl_(2)` at `NTP` produced during electrolysis of `MgCl_(2)` which produces `6.6 g Mg`. `("At.wt. of " Mg = 24.3)` |
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Answer» Correct Answer - 6 At cathode `Mg^(2+) + 2e rarr Mg` At anode `2Cl^(-) rarr Cl_(2) + 2e` `:.` Equivalent of `Mg` at cathode `=` Equivalent of `Cl_(2)` at anode `:. (6.6)/(24.3//2) = w_(Cl_(2))/35.5" "implies w_(Cl_(2))=19.28 g` Now at NTP `PV=w/m RT` `1xxV=(19.28xx0.0821xx273)/(71)` `implies " "V=6.08` litre `~~6` litre |
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| 2762. |
The molar conductivity of `0.05 M` of solution of an electrolyte is `200 Omega^(-1) cm^2`. The resistance offered by a conductivity cell with cell constant `(1//3) cm^(-1)` would be about .A. ` 11. 11 Omega`B. ` 22.22 Omega`C. ` 33.33 Omega`D. ` 44.444 Omega` |
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Answer» Correct Answer - C `k=Lambda_C = (200 s cm2 "mol"^(-1)) (.0.5 xx 10^(-3) "mol" cm^(-1))` ` = 0.01 S cm^(-1)` ` R=1/k (e/A) = 1/((0.01 S cm^(-1)) (1/3 cm^(-1)) = 33. 33 Omega`. |
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| 2763. |
How much electricity in terms of Faraday is required to produce `20 g` of `Ca` from molten `CaCl_(2)` ? |
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Answer» Correct Answer - 1 Eq. of `Ca, (w)/E = (i xx t)/(96500)` `Ca^(2+) + 2e rarr Ca` `:. E_(Ca) = (40)/(2)` or `(20)/(40//2) = (i xx t)/(96500)` `i xx t = 1 xx 96500 = 1 F` |
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| 2764. |
How many faradays are required to reduce ` 1` mol of `MnO_4^(-)` to `Mn^(2+)`?A. `1`B. `2`C. `3`D. `5` |
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Answer» Correct Answer - D `8H^+ + 5e^- + MnO_4^- rarr Mn^(+2) + 4 H_2O` `(1 "mole")` ` 5` moles of `e^- = 5` Faradays. |
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| 2765. |
The specific conducitvity of a saturated solution of `AgCI` is ` 3.40 xx 10^6 "ohm"^(-1) cm^(-1) ` at ` 25^@C`. If ` lambda_(Ag+) = 62.3 "ohm"^(-1) cm^2 "mol"^(-1)` and `lambda_(CI-) = 67.7 "ohm"^(-1) cm^2 "mol"^(-1)`, the solubility of `AgCI` at `25^@C` is.A. ` 2.6 xx 10^(-5) "mol" L-1`B. ` 3.731 xx 10^(-3) "mol" L-1`C. ` 3.731 xx10^(-5) "mol" L-1`D. ` 2.6 xx 10^(-3) g L-1` |
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Answer» Correct Answer - A `lambda_(sec) = 62.3 + 67.7 = 130 "ohm"^(-1) cm^2 "mol"^(-1)` ` S= ( 3.4 xx 10^(-6))/(130) "mol"//cm^3 = 2.6 xx 10^(-5) "mol"//L`. |
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| 2766. |
One faraday of electricity is passed through molten `Al_(2)O_(3)`, aqueous solution of `CuSO_(4)` and molten NaCl taken in three different electrolytic cells connected in series. The mole ratio of Al, Cu and Na deposited at the respective cathodes isA. `2:3:6`B. `6:2:3`C. `6:3:2`D. `1:2:3` |
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Answer» Correct Answer - A `Ag^(3+)+3e^(-)toAl` `Cu^(2+)+2e^(-)toCu` `Na^(+)+e^(-)toNa` thus 1F will be deposit `(1)/(3)` mole of `Al,(1)/(2)` mole of Cu and 1 mole of Na. Hence, molar ratio `=(1)/(3):(1)/(2):1=2:3:6` |
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| 2767. |
How much electricity in terms of Faraday is required to produce. `a.` `20.0g` of `Ca` from molten `CaCl_(2)` `b.` `40g` of `Al` from molten `Al_(2)O_(3)` |
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Answer» `Ca^(2+)to2e^(-)toCa` thus, 1 mol of Ca, i.e., 40 of Ca require=2F electricity `therefore20`g of Ca requrie=1F of electricity (ii). `Al^(3+)+3e^(-)toAl` Thus, 1 mole of Al, i.e., 27g of Al require= 3F electricity `therefore40g` of Al will require electricity `=(3)/(27)xx40=4.44` F of electricity. |
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| 2768. |
How much electricity in terms of faraday is required to produce (i) 20.0 g of Ca from `CaCl_(2)` (ii) 40.0 g of Al from molten `Al_(2)O_(3)`? |
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Answer» (i) `Cu^(2+)+2e^(-)toCa` Thus, 1 mol of Ca, i.e., 40 g of Ca require electricity=2F`therefore20` g of Ca will require electricity=1F (ii) `Al^(3+)+3e^(-)toAl` . Thus 1 mol of Al, i.e., 27g of Al required electricity=3F `therefore40g` of Al will require electricity=`(3)/(27)xx40=4.44E` |
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| 2769. |
One Faraday of electricity is pa ssed through molten `Al_(2)O_(3)`, aqeusous solution of `CuSO_(4)` and molten NaCl taken in three different electrolytic cells connected in seris. The mole ratio of Al, Cu,Na deposted at the respective cathode isA. `2:3: 6`B. `6:2:3`C. ` 6:3:2`D. `1:2:3` |
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Answer» Correct Answer - A `Al^(3+)+3e^(-)rarrAl` `Cu^(2)+2e^(-)rarrCu` `Na^(+)+e^(-)rarrNa` Hence 1F will deposit 1/3 mole of Al, 1/2 mol of Cu and 1 mole of Na `therefore` molar ratio `=(1)/(3):(1)/(2):1=2:3:6`. |
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| 2770. |
Check the feasibility of the following redox reaction with the help of electrochemical series `Ni(s)+2Ag^(+)(aq) to Ni^(2+)(aq)+2Ag(s)` |
| Answer» The `E^(@)` value of `Ni^(2+)//Ni "is" -0.25" V"` while that of `Ag^(+)//Ag is +0.80" V"`. This means that nickel is placed below silver in the series and can easily reduce `Ag^(+)` ions to silver by releasing electrons. The redox reaction is therefore,feasible. | |
| 2771. |
Resistance of 0.2 M solution of an electrolyte is 50 `Omega`. The specific conductance of the solution is 1.4 S `m^(-1)`. The resistance of 0.5 M solution of the same electrolyte is `280Omega`. The molar conductivity of 0.5 M solution of the electrolyte in S `m^(2)mol^(-1)` isA. `5xx10^(2)`B. `5xx10^(-4)`C. `5xx10^(-3)`D. `5xx10^(3)` |
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Answer» Correct Answer - B Case I: `C=0.2M,R=50Omega,kappa=1.4" S "cm^(-1)` `kappa`=Conductance`xx`Cell const.`=(1)/(R)xx`cell const. ltBrgt `therefore`Cell constant`=kappaxxR=(1.4" S "m^(-1))(50Omega)=70m^(-1)` Case II: `C=0.5" M ",R=280Omega," cell cont."=70m^(-1)` `kappa=(1)/(R)xx"cell constant"=(1)/(280Omega)xx70m^(-1)` Molar conductivity (in S `m^(2)mol^(-1)`) `=(kappa(S" "m^(-1)))/("Molarity"(mol " "L^(-1))xx1000" L "m^(-3))` `=(0.25" S "m^(-1))/(0.5 " mol "L^(-1)xx1000" L "m^(-3))` `=5xx10^(-4)" S "m^(2)mol^(-1)` |
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| 2772. |
`Delta_rG^@` for the cell with the cell reaction : `Zn_((s))+ Ag_2O_((s)) + H_2O_((l)) to Zn_((aq))^(2+) + 2Ag_((s)) + 2OH_((aq))^(-)` `[E_(Ag_2O // Ag)^(@)=0.344 V, E_(Zn^(2+)//Zn)^@=-0.76 V]`A. `2.13xx10^5 "J mol"^(-1)`B. `-2.13xx10^5 "J mol"^(-1)`C. `1.06xx10^5 "J mol"^(-1)`D. `-1.06xx10^5 "J mol"^(-1)` |
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Answer» Correct Answer - B `E_"cell"^@=E_(Ag_2O //Ag)^@-E_(Ag^(2+)//Zn)^@` =0.344-(-0.76)=1.104 V `DeltaG^@= -FE_"cell"^@= - 2 xx 96500 xx 1.104 = - 2.13 xx 10^5 " J mol"^(-1)` |
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| 2773. |
Resistance of `0.2 M` solution of an electrolyte is `50 Omega`. The specific conductance of the solution is ` 1.3 S m^(-1)`. If resistance of the `0.4 M` solution of the same electrolyte is `260 Omega`, its molar conductivity is .A. ` 6250 S m^2 "mol"^(-1)`B. ` 6.25 xx 10^(-4) Sm^2 "mol"^(-1)`C. `625 xx 10^(-4) S m^2 "mol"^(-1)`D. ` 62.5 S m^2 "mol"^(-1)` |
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Answer» Correct Answer - B For ` 0.2 M` solution. Specific conductance k = Conductance `(1/R)` `x` Cell constant `(l//a)` `rArr k = 1/(50) xx l/a = 1.3 Sm^(-1)` `rArr 1/a = 13. xx 50 m^(-1)` `1/a = 1.3 xx 50 xx 10^(-2) cm^(-1)` For `0.4 M` solutin, `R= 2600 Omega` `R= rho 1/a` `rArr 1/(rho) = k = 1/R xx 1/a` ` k= 1/(260) x 1.3 xx 50 xx 10^(-2) S cm^(-1)` Molar conductivity of the solution is given by `Lambda_m = (kxx 1000)/M` `1/(260) xx (1.3 xx 50 xx 10^(-2) xx 1000)/4` ` =6. 26 S cm^2 "mol"^(-1) 6, 25 xx 10^(-4) m^2 "mol"^(-1)`. |
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| 2774. |
The specific conductance `(kappa)` of an electrolyte of 0.1 N concentration is related to equivalent conductance `(Lambda_(e))` by the following formula :A. `Lambda_(e)= kappa`B. `Lambda_(e)= 10kappa`C. `Lambda_(e)= 100kappa`D. `Lambda_(e)= 10000kappa` |
| Answer» Correct Answer - D | |
| 2775. |
The potential of a hydrogen electrode at pH=10 is :A. `+0.59" V "`B. `0.00" V"`C. `-0.59" V "`D. `-0.059" V "` |
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Answer» Correct Answer - C (c ) `H^(+)+e^(-) to 1//2 H_(2)` `E=E^(@)-(0.059)/(n)"log"(1)/([H^(+)])=-(0.059)/(1)pH` `=(0-0.059)/(1)xx10=0-0.59=-0.59" V"` |
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| 2776. |
The reduction potential of hydrogen half cell will be negative if :A. `p(H_(2)) =2 atm[H^(+)] = 1.0M`B. `p(H_(2)) =2` atm and `[H^(+)] = 2.0M`C. `p(H_(2)) =1` atm and `[H^(+)] = 2.0M`D. `p(H_(2)) =1` atm and `[H^(+)] = 1.0M` |
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Answer» Correct Answer - A `2H^(+) +2e rarr H_(2)(g)` Applying Nernst Equation `E_(H^(+)//H_(2)) = E_(H^(+)//H_(2))^(@) - (0.059)/(2)log.(pH_(2))/([H^(+)]^(2))` `E_(H^(+)//H_(2)) = 0` `:. E_(H^(+)//H_(2))^(@) = 0 - (0.059)/(2) log.(pH_(2))/([H^(+)]^(2))` For the negative nature of `E_(H^(+)//H_(2))` |
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| 2777. |
Electrode potential of the half Pt(s)|Hg(l)|`Hg_(2)Cl_(2)(s)`|`Cl^(-)(aq)` can be incresed by :A. Increasing `[Cl^(-)]`B. decreasing `[Cl^(-)]`C. Increasing `Hg_(2)Cl_(2)(s)`D. decreasing `Hg(l)` |
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Answer» Correct Answer - A |
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| 2778. |
One Faraday of electricity is passed through molten `Al_(2)O_(3)`, aqueous solution of `CuSO_(4)` and molten NaCl taken in three different electrolytic cells connected in series . The mole ratio of Al , Cu and Na deposited at the respective cathode is :A. `2: 3: 6`B. `6 : 2 : 3`C. `6 : 3 : 2 `D. `1 : 2 : 3` |
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Answer» Correct Answer - A (a) `Al^(3+)+underset(1F)underset(3F)3e^(-) to nderset(1//3"mol")underset(1"mol")(Al)` `Cu^(2+)+underset(1F)underset(2F)2e^(-) to underset(1//2mol)underset(1mol)(Cu)` `Na^(+)+underset(1F)e^(-) to underset(1mol)(Na)` Mole ratio of Al, Cu and Na deposited at the respective cathodes is : `1//3 : 1//2 : 1//1 "or" 2:3:6` |
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| 2779. |
`E^@` values for the half cell reactions are given below : `Cu^(2+) + e^(-) to Cu^(+) , E^@`=0.15 V `Cu^(2+) + 2e^(-) to Cu, E^@`=0.34 V What will be the `E^@` of the half-cell : `Cu^(+) + e^(-) to Cu` ?A. `+0.49 V`B. `+0.19 V`C. `+0.53V`D. `+0.30V` |
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Answer» Correct Answer - C `{:(Cu^(2+)+e^(-) to Cu^(+) ,,E_(1)^(@)=0.15 V","DeltaG_1^@","n_1=1),(Cu^(2+)+2e^(-) to Cu , ,E_2^(@)=0.34 V","DeltaG_2^(@)","n_2=2),(Cu^(+)+e^(-) to Cu,,E_3^(@)=?"," DeltaG_3^(@)","n_3=1),(DeltaG_3^@ = DeltaG_2^@-DeltaG_1^@ rArr ,-n_3FE_3^@=-n_2FE_2^@+n_1FE_1^@):}` `-E_3^@` = -2 x 0.34 + 1 x 0.15 `E_3^@`= 0.68 - 0.15 = +0.53 V |
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| 2780. |
`E^@` value of `Ni^(2+)//Ni` is -0.25 V and `Ag^+ //Ag` is +0.80 V . If a cell is made by taking the two electrodes what is the feasibility of the reaction?A. Since `E^@` value for the cell will be positive, redox reaction is feasible.B. Since `E^@` value for the cell will be negative, redox reaction is not feasible.C. Ni cannot reduce `Ag^+` to Ag hence reaction is not feasible.D. Ag can reduce `Ni^(2+)` to Ni hence reaction is feasible. |
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Answer» Correct Answer - A The cell reaction will be `Ni_((s))+2Ag_((aq))^+ to Ni_((aq))^(2+) + 2Ag_((s))` `E_"cell"^@=E_"cathode"-E_"anode"=0.80-(-0.25)= pm 1.05` V `DeltaG^@= -nFE_"cell"^@ As E_"cell"^@ = +ve`, `DeltaG^@=-ve` , hence reaction is feasible . |
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| 2781. |
Given,`{:(E_(Fe^(3+) //Fe)^o + 3e Cr E^o =- 0.036 V),(E_(Fe^(3+) //Fe)^o =- 0.439 V):}` The value of standard electrode ptoential for the charge,A. `0.770V`B. `-0.27V`C. `-0.072V`D. `0.385V` |
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Answer» Correct Answer - A `Fe^(3+) +3e rarr Fe` ...(1) `E^(@) = 0.036V` `:. DeltaG_(1)^(@) = nFE^(@)` `=- 3 xx F(-0.036V)` `= 0.108F` Also, `Fe^(2+) +2e rarr Fe` ..(2) `E^(@) =- 0.439V` `:. DeltaG_(2)^(@) = nFE^(@)` `=- 2 xx F(0.439)` `=08.78 F` For `Fe^(3+) (aq) +e^(-) rarr Fe^(2+) (aq)` ..(3) `:.` Assume `E^(@) = xV` eq.(1)-eq.(2) produces eq.(3) `:. DeltaG^(@) = DeltaG_(1)^(@) - DeltaG_(2)^(@)` `-1 xx F xx x = 0.108 F - 0.878F` `- x =- 0.77` `x = 0.77V` |
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| 2782. |
The Gibbs energy for the decomposition of `Al_(2)O_(3)` at `500^(@)C` is as follows: `(2)/(3)Al_(2)O_(3) rarr (4)/(3)Al + O_(2), Delta_(r)G = +966kJ mol^(-1)` The potential difference needed for electrolytic reeduction of `Al_(2)O_(3)` at `500^(@)C` is at least:A. 5.0 VB. 4.5 VC. 3.0 VD. 2.5 V |
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Answer» Correct Answer - D The ionic reactions are : 2/3 x `(2Al^(3+)) + 4e^(-) to ` 4/3 Al 2/3 x `(3O^(2-)) to O_2 + 4e^(-)` Thus, no of electrons transferred, n=4, `DeltaG`= -nFE = - 4 x 96500 x E or 966 x `10^3`= -4 x 96500 x E `rArr E= - (966xx10^3)/(4xx96500)`=-2.5 V |
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| 2783. |
Resistance of `0.2 M` solution of an electrolyte is `50 Omega`. The specific conductance of the solution is ` 1.3 S m^(-1)`. If resistance of the `0.4 M` solution of the same electrolyte is `260 Omega`, its molar conductivity is .A. `6250 S m^(2) mol^(-1)`B. `6.25 xx 10^(-4) S m^(2) mol^(-1)`C. `625 xx 10^(-4) S m^(2) mol^(-1)`D. `62.5 S m^(2) mol^(-1)` |
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Answer» Correct Answer - B `k = (1)/(R) xx(1)/(a)` `1.3 = (1)/(50) xx (1)/(a)` `:. (l)/(a) = 65 m^(-1)` k of `0.4m` solution is `k = (1)/(R) xx(1)/(a)` `= (1)/(260) xx 65 = (1)/(4) ohm^(-1) m^(-1) = (1)/(4) Sm^(-1)` `^^_(m) = (k)/("molarity") = (1)/(4 xx 0.4) S m^(-1) mol^(-1) sm^(3)` `= (1)/(4 xx 0.4 xx 1000) Sm^(-) mol^(-1) m^(3)` `=(1)/(1.6) xx 10^(-3)` `=6.25 xx 10^(-4) S m^(2) mol^(-1)` |
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| 2784. |
The Gibbs energy for the decomposition of `Al_(2)O_(3)` at `500^(@)C` is as follows: `(2)/(3)Al_(2)O_(3) rarr (4)/(3)Al + O_(2), Delta_(r)G = +966kJ mol^(-1)` The potential difference needed for electrolytic reeduction of `Al_(2)O_(3)` at `500^(@)C` is at least:A. `5.0 "V "`B. `4.5 "V"`C. `3.0" V "`D. `2.5" V "` |
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Answer» Correct Answer - D (d) The ionic reactions are : `2//3 Al_(2)^(3+)+4e^(-) to 4//3 Al` `2//3 O_(3)^(2-) to O_(2)+4e^(-)` No. of electrons =4 `DeltaG=-nFE_(cell)` `E_(cell)=-(DeltaG)/(nF)=(966xx10^(3)(CV))/(4xx96500(V))=2.5" V"`. |
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| 2785. |
STATEMENT-1: Conductivity decreases with the decreases is concentration both the weak and strong electolytes. STATEMENT-2: No. of ions per unit volume linearly decreases in both electrolytes.A. If both the statements are TRUE and STATEMENTS-2 is the correct explantion of STATEMENTS-13B. If both the statements are TRUE but STATEMENTS-2 is NOT the correct explanation of STATEMENTS-15C. If STATEMENTS-1 is TRUE and STATEMENTS-2 is FALSED. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE |
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Answer» Correct Answer - C |
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| 2786. |
The equivalent conductance of `M//32` solution of a weak monobasic acid is `8.0 ` and at infinite dilution is `400 `. The dissociation constant of this acid is :A. `1.25xx10^(-6)`B. `6.25xx10^(-4)`C. `1.25xx10^(-4)`D. `1.25xx10^(-5)` |
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Answer» Correct Answer - D (d) `alpha=(Lambda_(E))/(Lambdaoverset(@)E)=((8.0" ohm"^(-1)cm^(2)"equiv"^(-1)))/((400.0" ohm"^(-1)cm^(2)" equiv"^(-1)))` `=2xx10^(-2)` `K=Calpha^(2)=(1)/(32)xx(2xx10^(-2))=1.25xx10^(-5)` |
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| 2787. |
The equivalent conductance of `M//32` solution of a weak monobasic acid is `8.0 ` and at infinite dilution is `400 `. The dissociation constant of this acid is :A. `1. 25xx10^(-4)`B. `1.25xx10^(-5)`C. `1.25xx10^(-6)`D. `6.25xx10^(-4)` |
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Answer» Correct Answer - B ` alpha=(80)/(400)=0.02` `K=Calpha^(2)=(1)/(32)xx(0.02)^(2)` `K=0.03125xx(0.02)^(2)=1.25xx 10^(-5)` |
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| 2788. |
The equivalent conductance of `M//32` solution of a weak monobasic acid is `8.0 ` and at infinite dilution is `400 `. The dissociation constant of this acid is :A. `1.25 xx 10^(-6)`B. `6.25 xx 10^(-4)`C. `1.25 xx 10^(-4)`D. `1.25 xx 10^(-5)` |
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Answer» Correct Answer - D Given , `Lambda = 8 ` mho `cm^(2)` `Lambda_(0) = 400 mho cm^(2)` Degree of dissociation , `alpha = (Lambda)/(Lambda_(0)) implies alpha = (8)/(400) = 2 xx 10^(-2)` Dissociation constant , `K = C alpha^(2)` Given , C = M/32 `therefore K = (1)/(32) xx 2 xx 10^(-2) xx 2 xx 10^(-2)` `K = 1.25 xx 10^(-5)` |
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| 2789. |
The equivalent conductance of `M//32` solution of a weak monobasic acid is `8.0 ` and at infinite dilution is `400 `. The dissociation constant of this acid is :A. `1.25xx10^(-5)`B. `1.25xx10^(-6)`C. `6.25xx10^(-4)`D. `1.25xx10^(-4)` |
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Answer» Correct Answer - A `alpha=(^^)/(wedge_(D))=(8.0)/(400)=2xx10^(-2)` `K_(a)=(Calpha^(2))/((1-alpha))~~Calpha^(2)=(1)/(32)xx(2xx10^(-2))=1.25xx10^(-5)`. |
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| 2790. |
Which of the followig expressions correctly represents the equivalent conductance at infinite dilution of `Al_(2)(SO_(4))_(3)`. Given that `wedge_(Al3+)^(0) and wedge_(SO_(4)^(2-))^(0)` ar the equivalent conductances at infinite dilution of the respective ionsA. `2wedge_(Al^(3+))^(0)+3wedge_(SO_(4)^(2-))^(0)`B. `wedge_(Al^(3+))^(0)+wedge_(SO_(4)^(2-))^(0)`C. `(wedge_(Al^(3+))^(0)+wedge_(SO_(4)^(2-))^(0))xx6`D. `(1^(wedge^(o)))/(3)Al^(3+)+(1wedge^(o)))/(2)SO_(4)^(2-)` |
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Answer» Correct Answer - B As equivalent conductance are given for ions. |
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| 2791. |
The resistance of a conductivity cell contaning `0.001M KCl` solution at `298K` is `1500Omega`. What is the cell constant if conductivity of `0.001M KCl` solution at `298K` is `0.146 xx 10^(-3) S cm^(-1)`. |
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Answer» Cell constant `=("Conductivity")/("Conductance")` `="Conductivity"xx"Resistance"` `=0.146xx10^(-3)" S "cm^(-1)xx1500Omega` `=0.219cm^(-1)`. |
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| 2792. |
The resistance of a conductivity cell containing 0.001 M KCl solution at 298K is 1500 `Omega`. What is the cell constant if conductivity of 0.001 M KCl solution at 298K is 0.146`xx10^(-3)"S "cm^(-1)`? |
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Answer» Cell constant`=("Conductivity")/("Conductance")=`Conductivity`xx`Resistance=(0.146`xx10^(-3)`) S `cm^(-1)xx1500Omega` `=0.219cm^(-1)` |
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| 2793. |
The resistance of a conductivity cell containing 0.001M KCl solution at 298K is 1500`Omega`. What is the cell constant ( in `mm^(-1)`) if the conductivity of 0.001M KCl solution is `2xx10^(-3)Smm^(-1)` |
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Answer» Correct Answer - C |
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| 2794. |
A solution containing `1MXSO_4(aq)` and `1MYSO_4(aq)` is electrolysed. If conc. Of `X^(2+)` is `10^(-z)M` when deposition `Y^(2+)` and `X^(2+)` starts simultaneously, calculate the value of Z. Given: `(2.303RT)/F=0.06` `E_(x^(2+)"|"X)^(@)=-0.12V,E_(Y^(2+)"|"Y)^(@)=-0.24V` |
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Answer» Correct Answer - D |
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| 2795. |
Determine of `pH`: The following cell has a potential of `0.55 V` at `25^(@)C`: `Pt(s)|H_(2)(1 bar)|H^(+)(aq. ? M)||Cl^(-)(1 M)|Hg_(2)Cl_(2)(s)|Hg(l)` What is the `pH` of the solution in the anode compartment? Strategy: First, read the shorthand notation to obtain the cell reaction. Then, calculate the half cell potential for the hydrogen electrode from the observed cell potential and the half cell poten-tial for the calomel reference electrode. Finally, apply the Nernst equation to find the `pH`. |
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Answer» The cell reaction is `H_(2)(g)+Hg_(2)Cl_(2)(s)hArr 2H^(+)(aq.)+2Hg(l)+2Cl^(-)(aq.)` and the potential is `E_("cell") = E_("anode") + E_("cathode")` `E_("cell") = E_(H_(2)//H^(+)) + E_(Hg_(2)Cl_(2)//Hg = 0.55 V` Because the reference electrode is the standard calomel electode, whic has `E = E^(@) = 0.28 V`, the half-cell potential for the hydrogn electode is `0.27 V`: `E_(H_(2)//H^(+)) = E_("cell") - E_(Hg_(2)Cl_(2)//Hg` `= (0.55 V) - (0.28 V)` `= 0.27 V` Applying the Nernst equation to the half-reaction `H_(2)(g) rarr 2H^(+)(aq.) + 2e^(-)`: `E_(H_(2)//H^(+)) = E_(H_(2)//H^(+))^(@) - ((0.0592 V)/(n)) "log" ((C_(H^(+))^(2))/(P_(H_(2))))` Substituting in the values of `E, E^(@)`, n and `P_(H_(2))` gives `0.27 V = (0V) - ((0.0592 V)/(2)) "log"((C_(H+)^(2))/(1))` `= (0.0592 V) (pH)` Therefore, the `pH` is `pH = (0.27 V)/(0.0592 V) = 4.6` |
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| 2796. |
The Gibbs energy for the decomposition of `Al_(2)O_(3)` at `500^(@)C` is as follows: `(2)/(3)Al_(2)O_(3) rarr (4)/(3)Al + O_(2), Delta_(r)G = +966kJ mol^(-1)` The potential difference needed for electrolytic reeduction of `Al_(2)O_(3)` at `500^(@)C` is at least:A. `5.0V`B. `4.5V`C. `3.0V`D. `2.5V` |
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Answer» Correct Answer - D `DeltaG =- nFE :. E = (-DeltaG)/(nF)` `E = (-966 xx 103)/(4 xx 96500) =- 0.25` `:.` The potential difference needed for the reduction `= 2.5V` |
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| 2797. |
The following electrochemical cell is represented as `:` `Pt(1)|Hg^(2+),Hg_(2)^(2+)(a=1)||Ce^(4+),Ce^(3+)(a=1)|Pt(2)` `a.` If an ammeter is connected between two platinum electrodes, predict the direction of flow of current. `b.` Will the current decrease or increase with time? Given `:E^(c-)._(2Hg^(2+)|Hg_(2)^(2+))=0.92V,E^(c-)._(Ce^(4+)//Ce^(3+))=1.61V` |
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Answer» The elctrode reactions are as follows `:` Anode reaction `:Hg_(2)^(2+) rarr 2Hg^(2+)+cancel(2e^(-) (` oxidation `)` Cathode reaction `:` `2Ce^(4+)+cancel(2e^(-)) rarr 2Ce^(3+) (` Reduction `)` `ulbar(Cell reaction :Hg_(2)^(2+),2Ce^(4+) rarr2Hg^(2+)+2Ce^(3+))` `:. E^(c-)._(cell)=(E^(c-)._(reduction))_(c)-(E^(c-)._(reduction))_(a)` `=1.61V-0.92V=0.69V` Since `E^(c-)` cell is positive, the cell reaction is feasible or spontaneous. ltbr. Current in electrochemical cell flows from cathode to anode whereas the electrons flow from anode to cathode. So the current in the circuit will flow from `Pt(2)(` cathode`)` to `Pt(1)(` anode `).` `b.` With the passage of time, `EMF` of the cell will drecrease and so the current will also decrease in the circuit. |
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| 2798. |
The standard reduction potential for the half-cell, `NO_(3(aq.))^(-)+2H_((aq.))^(+)+e^(-)rarrNO_(2(g))+2H_(2)O` is `0.78V`. (i) Calculate the reduction potential in 8M `H^(+)`. (ii) What will be the reduction potential of the half-cell in a neutral solution. Assume all the other species to be at unit concentration |
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Answer» (i) The redox reaction is : `NO_(3)^(-)(aq)+2H^(+)(aq)+e^(-) to NO_(2)(g)+H_(2)O(l)` According to Nernst equation : `E=E^(@)-(0.0591)/(n)"log"([NO_(2)(g)][H_(2)O(l)])/([NO_(3)^(-)(aq)][H^(+)(aq)]^(2))` According to available data, `[NO_(2)(g)]=1,[H_(2)O(l)]=1, [NO_(3)^(-)(aq)]=1,n=1,E^(@)=0.78" V", H^(+)=8M` `:. " " E=0.78-(0.0591)/(1)"log"(1)/((8)^(2))=0.78+0.0591 log 8^(2)=0.78+0.0591(2xx"log "8)` `=0.78+0.0591xx2xx0.9031=0.78+0.1067=0.8867" V"` (ii) For neutral solution `(H^(+)]=10^(-7)M` `E=0.78-(0.0591)/(1)"log"(1)/((10^(-7))^(2))=0.78-0.0591(log 10^(14))` `=0.78-0.0591(14" log "10)=0.78-0.0591xx14" "(because "log" 10=1)` `=0.78-0.8274=-0.0474" V"`. |
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| 2799. |
For the reduction of silver ions with copper metal, the standard cell potential was foound to be `+0.46 V` at `25^(@) C`. The value of standard Gibbs energy, `DeltaG^(@)` will be `(F = 96,500C mol^(-1))`:A. `-98.00 kJ`B. `-89.0 kJ`C. `-89.0 J`D. `-44.5 kJ` |
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Answer» Correct Answer - B The relationship between `E_("cell")^(@)` for a galvanic cell and `Delta_(r)G^(@)`, the standard Gibbs energy change, for the chemical reaction of the cell is `Delta_(r)G^(@) = -nFE_("cell")^(@)` For the cell reaction `2Ag^(+)(aq.)+Cu(s) rarr Cu^(2+)(aq.)+2Ag(s)` We have `n = 2 mol` `E_("cell")^(@) = +0.46 V` Therefore `Delta_(r)G^(@) = (-2 mol)(96,500 C mol^(-1))(+0.46 V)` `= -88780 J = -88.780 kJ` `= -89.0 kJ` |
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| 2800. |
The Gibbs energy for the decomposition of `Al_(2)O_(3)` at `500^(@)C` is as follows: `(2)/(3)Al_(2)O_(3) rarr (4)/(3)Al + O_(2), Delta_(r)G = +966kJ mol^(-1)` The potential difference needed for electrolytic reeduction of `Al_(2)O_(3)` at `500^(@)C` is at least:A. 2.5VB. 5.0VC. 4.5VD. 3.0V |
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Answer» Correct Answer - A `2//3Al_(2)O_(3)rarr4//3Al+O_(2)`. `DeltaG= +966kJ "mol"^(-1)` `4//3Al^(3+)+4e^(-)rarr4//3Al` `DeltaG=-nFE` ` therefore 966 xx10^(3)=-4xx96500xxE`. `therefore E=2.50V` `therefore` the minimum potential difference required = 2.50V. |
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