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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.
| 2651. |
For the oxidation of ferric oxalate to `CO_(2), 18F` of electricity is required. How mnay moles of ferric oxalate is oxidized ? |
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Answer» Correct Answer - 3 Ferric oxalate `[Fe_(2)(C_(2)O_(4))_(3)rarr2Fe^(3+)+3C_(2)O_(4)^(2-)]` `3C_(2)O_(4)^(2-)rarr 6CO_(2)+6e^(-) [Fe^(3+)` does not change `]` `:. 6F=1 mol of Fe_(2)(C_(2)O_(4))_(3)` `18F=3 mol of Fe_(2)(Fe_(2)(C_(2)O_(4))_(3)` |
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| 2652. |
When lead storage battery discharges , specific gravity of `H_(2)SO_(4)` decreases toA. zeroB. 1.17C. 1.28D. 1.215 |
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Answer» Correct Answer - B During discharging of the cell , specific gravity of `H_(2)SO_(4)` decreases from 1.215 to 1.17 g/cc. |
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| 2653. |
How much charge is required to reduce (a) 1 mole of `Al^(3+)` to Al and (b) 1 mole of `MnO_(4)^(-)` to `Mn^(2+)` ? |
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Answer» (a) The reduction reaction is : `underset("1 mole")(Al^(3+))+ underset("3 mole")(3e^(-)) rarr Al` Thus, 3 mole of electrons are needed to reduce 1 mole of `Al^(3+)` `Q=3xxF` `=3xx96500=289500` coulomb (b) The reduction reaction is : `underset("1 mole")(MnO_(4)^(-))+8H^(+) + underset("5 mole")(5e^(-)) rarr Mn^(2+)+ 4H_(2)O` `Q=5xxF` `=5xx96500=482500` coulomb |
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| 2654. |
Give the mechanism for the decomposition reacton of `H_(2)O_(2)` in alkaline medium catalysed by `I^(-)` ions. |
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Answer» Chemical equation of decomposition of `H_(2) O_(2)` in alkaline medium is `2H_(2) O_(2) overset(I^(-))underset("Alkaline medium")to2H_(2)O+O_(2)` Mechanism : `to` It is a first order reaction w.r.t. both `H_(2)O_(2),I^(-)` `to` This reaction takes place is two steps `i) H_(2)O_(2)+ I^(-) to H_(2)O +IO^(-)` `ii) H_(2)O_(2)+IO^(-) to H_(2)O +I^(-) +O_(2)` |
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| 2655. |
Calculate the time to deposit 1.27 g of copper at cathode when a current of 2A was passed through the solution of `CuSO_(4)`. `("Molar mass of "Cu=63.5g mol^(-1), 1F =96500 C mol^(-1)).` |
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Answer» Correct Answer - 1930 s Reaction at cathode is : `underset(63.5" g")(Cu^(2+))(aq)+2e^(-) to underset(2xx96500" C")(Cu(s))` 63.5 g of copper is deposited by passing charge `=2xx96500" C"` 1.27 g of copper of deposited by passing charge `=((2xx96500C)xx(1.27 g))/((63.5 g))=3860" C"=3860" As"` Time to deposit 1.27 g of copper `=((3860" As"))/((2A))=1930" S "` |
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| 2656. |
Calculate the time to deposit 1.5 g of silver at cathode when a current of 1.5 A is passed through the solution of `AgNO_(3)`. |
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Answer» The reaction at the cathode is : `{:(Ag^(+)(aq)+e^(-)rarr Ag(s)),(108 g " "96500 C):}` 108 g of silver is deposited by passing charge =96500 C 1.5 g of silver is deposited by passing charge `=((96500 C)xx(1.5 g))/((108 g))=1340.27 C =1340.27 `As Time to deposited 1.5 g of silver `=("charge"(Q))/("current"(I))=((1340.27 As))/((1.5 A))` =893.51 s or 14.85 min |
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| 2657. |
The desired amount of charge for obtaining one mole of Al from `Al^(3+)`A. `3xx96500C`B. `96500C`C. `(96500)/(3)C`D. `(96500)/(2)C` |
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Answer» Correct Answer - A `AltoAl^(3+)+3e^(-)` the charged obtained is `3xx96500C`. |
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| 2658. |
Two platinum electrodes were immersed in a solution of `CuSO_(4)` and electric current was passed through the solution. After some time, it was found that colour of `CuSO_(4)` disappeared with evolution of gas at the electrode. The colourless solution contains.A. Platimus sulphateB. Copper hydroxideC. copper sulphateD. sulphuric acid |
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Answer» Correct Answer - D During electrolysis of `CuSO_(4).Cu^(2+)` gets discharged cathode and `OH^(-)` at anode. Tghus solution becomes acidic dur to excess of `H^(+) and SO_(4)^(2-)` or `H_(2)SO_(4)`. |
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| 2659. |
Two platinum electrodes were immersed in a solution of `CuSO_(4)` and electric current was passed through the solution. After some time, it was found that colour of `CuSO_(4)` disappeared with evolution of gas at the electrode. The colourless solution contains.A. Plantinum sulphateB. copper hydroxideC. copper sulphateD. sulphuric acid |
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Answer» Correct Answer - D During electrolysis of `CuSO_(4) , Cu^(2+)` gets discharged at cathode and `OH^(-)` at anode . Thus solution becomes acidic due to excess of `H^(+)` and `SO_(4)^(2-) or H_(2)SO_(4)` . |
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| 2660. |
Consider the following concentration cell`:` `Zn(s)|Zn^(2+)(0.024M)||Zn^(2+)(0.480M)|Zn(s)` which of the following statements is `//` are correct?A. The `EMF` of the cell at `25^(@)C` is nearly `+0.039V.`B. The `EMF` of the cell at `25^(@)C` is nearly `-0.039V`.C. If water is added in `LHE`, so that the `[Zn^(2+)]` is reduced to `0.012 M`, the cell voltage increases.D. If water is added in `LH` , so that the `[Zn^(2+)]` is reduced to `0.012M` , the cell voltage decreases. |
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Answer» Correct Answer - a,c Cell reaction `:Zn^(2+)(0.484M) rarr Zn^(2+)(0.024M)` `E_(cell)=0-(0.059)/(2) log ((0.024)/(0.480))=0.039V` For `:Zn^(2+)(0.484M)rarr Zn^(2+)(0.12M)` `implies E_(cell)=0-(0.059)/(2) log ((0.012)/(0.48))` `=0.03(1.6)=0.048V` |
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| 2661. |
Under what conditions the cell potential is called standard cell potential ? |
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Answer» In the standard cell, the active masses of the substances taking part in the electrochemical reaction have unit value, i.e., 1 M solution or ions and 1 atm gas. |
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| 2662. |
Calculating the cell emf for Nonstandard condition : A galvanic cell is constructed at `298 K` as follows. One half-cell consists of a chlorine gas electrode (i.e. `Cl_(2)//Cl^(-)` couple) with the partial pressure of `Cl_(2) = 0.100` bar and `C_(CI-) = 0.100 M`. The other half-cell involves the `MnO_(4)^(-)//Mn^(2+)` couple in acidic solution with `C_(MnO_(4)^(-)) = 0.100 M, C_(Mn^(2+)) = 0.100 M`, and `C_(H+) = 1.00 M`. Apply the Nernst equation to datermine the cell potenital for this cell. Strategy: First datermine the overall cell reaction and then calculate the standard cell potential `(E_("cell")^(@))` from the standard electrode potenital in Table 3.1. then use the Nernst equation to find the cell potential `(E)` under cited conditions. |
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Answer» The `MnO_(4)^(-)//Mn^(2+)` half-cell reaction has the more positive reduction potential `(+ 1.507 V)`, so we write it first. Then we write the `Cl_(2)//Cl^(-)` half-cell reaction as an oxidation, balance the electron transfer, and add the two half-reactions and their standard potentials to obtain the overall cell reaction and its `E_("cell")^(@)` `{:(2[MnO_(4)^(2-)+8H^(+)+5e^(-)hArrMn^(2)+4H_(2)O]" "+1.507V),(5[2Cl^(-)hArrCl_(2)+2e^(-)]" "-1.360V),(bar(2MnO_(4)^(-)+16H^(+)+10Cl^(-)hArr 8H_(2)O "]"+5Cl_(2)E_("cell")^(@)=+0.147 V)):}` In the overall reaction, `n = 10`, we then apply the Nernst equation to this overall reaction by subsitution suitable concentration and partial pressure values. Bacause `Cl_(2)` is a gaseous substance, its term in the Nernst equation involves its partial pressure, `P_(Cl_(2))`. `E_("cell") = E_("cell")^(@)-(0.592 V)/(n) "log"(C_(Mn^(2+))^(2)P_(Cl_(2))^(5))/(C_(MnO_(4)^(-))^(2)C_(H+)^(16)C_(Cl^(-))^(10))` ` = 0.147 V - (0.0592 V)/(10) "log" ((0.100)^(2)(0.100)^(5))/((0.100)^(2)(0.100)^(16)(0.100)^(10))` ` = 0.147 V - (0.0592 V)/(10) "log" (1.00 xx 10^(21))` ` = 0.147 V - (0.0592 V)/(10) "log" (21.00)` `= 0.017 V` |
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| 2663. |
The Nearst Equation: A galvanic cell is constructed at `25^(@)C` as follows. Once half-cell consists of the `Fe^(3+)//Fe^(2+)` couple in which `C_(Fe^(3+)) = 1.00 M` and `C_(Fe^(2+)) = 0.100 M`, the other involves the `MnO_(4)^(-)//Mn^(2+)` couple in which `C_(MnO_(4)^(-)) = 1.00 xx 10^(-2)M, C_(Mn^(2+)) = 1.00 xx 10^(-4) M`, and `C_Mn^(2+) = 1.00 xx 10^(-3) M`. (a) Find the elecrode potential for each half-cell with these concentrations, and (b) calculate the overall cell potential. Strategy: (a) Apply the Nernst equation to find the electrode potential of each half-cell with the given concentrations. Write the half-reaction with the more positive (or less negative) potentikal (after correction) as the cathode reaction along with its potential. Reverse the other half-reaction (to make it anode reaction) and change the sign of its `E` value. Finally balance the electron transfer and then add the half-reactions and their electrode potentials to find the overall cell potentail. |
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Answer» (a) For the `Fe^(3+)//Fe^(2+)` half-cell as a reduction. `Fe^(3+)(aq.)+e^(-)hArrFe^(2+)(aq.)E_(Fe^(3+)//Fe^(2+))^(2+)=+0.771V` `E_(Fe^(3+)//Fe^(2+))^(@) - (0.0592 V)/(n) log (C_Fe^(2+))/(C_Fe^(3+))` ` = +0.77 V - (0.0592 V)/(1) log (0.100)/(1.00)` `= +0.77 V - (0.0592 V) (-1.00)` `= +0.829 V` For the `MnO_(4)^(-) + 8H6(+) + 5e^(-) rarr Mn^(2+) + H_(2)O E_(MnO_(4)//Mn^(2+))^(@) = +1.507` `E_(MnO_(4)//Mn^(2+)) = E_(MnO_(4)//Mn^(2+)) - (0.0592 V)/(n)log (C_(Mn^(2+)))/(C_(MnO_(4)^(-))C_(H+)^(8)` `= +0.77 V - (0.0592 V)/(5)` `"log" (1.00 xx 10^(-4))/(1.00 xx 10^(-2))(1.00 xx 10^(-3))^(8)` `= +1.507 V - (0.0592 V)/(5) "log" (1.00 xx 10^(22))` `= +1.507 V - (0.0592 V)/(5) (22.0) = +1.246 V` (b) Since the corrected electrode potential for the `MnO_(4)^(-)//Mn^(2+)` half-cell is greater than that for the `Fe^(3+)//Fe^(2+)` half-cell, we reaverse the half-cell reaction for the `Fe^(3+)//Fe^(2+)` couple, balance the electron trasfer, and add. `{:(MnO_(4)^(-)(aq.)+8H^(+)(aq.)5e^(-)hArr Mn^(2+)(aq.)4H_(2)P(1)+1.246 V-0.0829V),(5xx[Fe^(2+)(aq)hArr Fe^(3+)(aq)+e^(-)]),(bar(MnO_(4)^(-)(aq,)+8H^(+)(aq.)+5Fe^(2+)(aq.)rarr Mn^(2+)(aq.)+4H_(2)O(1)5Fe^(3+)(aq.)E_("cell")=0.417V)):}` |
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| 2664. |
Two half cells have reduction potentials `-0.76 V` and `-0.13 V` respectively . A galvanic cell is made from these two half cells . Which of the following statements is correct ?A. Electrode of half-cell ptential ` -076 V` serve as cathodeB. Electrode of half-cell potential `-0 76 V` serve as anodeC. Electrode of half-cell potential ` -0 76 V` serve as anodeD. Delctrode of half-cell potential `-076 V` see as positive electrode and ` -`0 13 V` as negative electrode |
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Answer» Correct Answer - B The electrode with more negative reduction potuction potential constitutes the anode. |
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| 2665. |
Metallic anodes smore reactive than platinum tend to pass into the solution instead of `O_(2)` being produced. |
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Answer» Correct Answer - T Metals are more reactive than `H_(2)O` Hence oxidation of metals takes lace to give ions `(` which passes into the solution `)` than the oxidation of `H_(2)O`. |
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| 2666. |
In highly alkaline medium, the anodic process during the electrolytic process is `4overset(c-)(O)H rarr O_(2)+2H_(2)O+4^(-)` |
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Answer» Correct Answer - T At anode, oxidation takes place. |
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| 2667. |
Small quantities of solutions of compounds TX, TY and TZ are put into separete test tubes containing X,Y and Z solution. TX does not react with any of these. TY reacts with both X and Z. Does not react with any of these. TY reats with both X and Z. TZ reacts with X. the decreasing order of state of oxidation of the anions `X^(-),Y^(-),Z^(-)`A. `Y^(-),Z^(-),X^(-)`B. `Z^(-),X^(-),Y^(-)`C. `Y^(-),X^(-),Z^(-)`D. `X^(-),Z^(-),Y^(-)` |
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Answer» Correct Answer - A Oxidizing tendency`prop(1)/("Electrode potential")` `Txto`No reaction `TY to X,Z` `TZtoX` `implies`Order of electrode potential is `TY lt TZ lt TX` `implies`Order of oxidation of the anion is `Y^(-) gt Z^(-) gt X^(-)`. |
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| 2668. |
Oxygen and hydrogen gas are produced at the anode and cathode during the electrolysis of fairly concentration aqueous solution of :A. `K_2SO_4`B. `AgNO_3`C. `H_2SO_4`D. NaOH |
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Answer» Correct Answer - A::C::D |
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| 2669. |
During the purification of copper by electrolysis:A. the anode used is made of copper oreB. pure copper is deposited on the cathodeC. the impurities such as Ag, Au present in solution as ionsD. concentration of `CuSO_4` solution remains constant during dissolution of Cu |
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Answer» Correct Answer - A::B::D |
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| 2670. |
A Galvanic cell consits of three compartment as shown in figure. The first compartment contain `ZnSO_4`(1M) and III compartment contain `CuSO_4`(1M). The mid compartment contain `NaNO_3` (1M). Each compartment contain 1L solution: `E_(Zn^(2+)//Zn)^(@)=-0.76`, `E_(Cu^(2+)//Cu)^(@)=+0.34` The concentration of `NO_3^-` in mid compartment after passage of 0.1 F of charge will be:A. 0.95MB. 0.90MC. 0.975MD. 1.05M |
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Answer» Correct Answer - A |
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| 2671. |
A Galvanic cell consits of three compartment as shown in figure. The first compartment contain `ZnSO_4`(1M) and III compartment contain `CuSO_4`(1M). The mid compartment contain `NaNO_3` (1M). Each compartment contain 1L solution: `E_(Zn^(2+)//Zn)^(@)=-0.76`, `E_(Cu^(2+)//Cu)^(@)=+0.34` The concentration of `SO_4^(2-)` ions in III compartment after passage of 0.1 F of charge will be:A. 1.05MB. 1.025MC. 0.95MD. 0.975M |
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Answer» Correct Answer - D |
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| 2672. |
`Zn(s)|Zn(CN)_4^(2-)(0.5M),CN^(-)(0.01)"||"Cu(NH_3)_4^(2+)(0.5M),NH_3(1M)|Cu(s)` Given: `K_f "of"Zn(CN)_4^(-2)=10^(16)` , `K_f"of"Cu(NH_3)_4^(2+)=10^(12)`, `E_(Zn|Zn^(2+))^(@)=0.76V,E_(Cu^(+2)|Cu)^(@)=0.34V , (2.303RT)/F=0.06` The emf of above cell is :A. 1.22VB. 1.10VC. 0.98VD. None of these |
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Answer» Correct Answer - C |
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| 2673. |
A Galvanic cell consits of three compartment as shown in figure. The first compartment contain `ZnSO_4`(1M) and III compartment contain `CuSO_4`(1M). The mid compartment contain `NaNO_3` (1M). Each compartment contain 1L solution: `E_(Zn^(2+)//Zn)^(@)=-0.76`, `E_(Cu^(2+)//Cu)^(@)=+0.34`, The concertation of `Zn^(2+)` in first compartment after passage of 0.1 F charge will be:A. 1MB. 1.05MC. 1.025MD. 0.5M |
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Answer» Correct Answer - C |
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| 2674. |
One gm metal `M^(3+)` was discharged by the passage of `1.81xx10^(23)` electrons.What is the atomic mass of metal?A. 33.35B. 133.4C. 66.7D. None of these |
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Answer» Correct Answer - D |
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| 2675. |
How many Faradays are required to reduce 0.25g of Nb (V) to the metal?A. `2.7xx10^(-3)`B. `1.3xx10^(-2)`C. `2.7xx10^(-2)`D. `7.8xx10^(-3)` |
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Answer» Correct Answer - B |
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| 2676. |
Calculate the current (in mA) required to deposite 0.195g of platinum metal in 5.0 hours from a solution of `[PtCl_6^(2-)`:(Atomic mass:Pt=195)A. 310B. 31C. 21.44D. 5.36 |
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Answer» Correct Answer - C |
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| 2677. |
Amount of electricty that can depostit `108 g` of silver from `AgNO_3` aikyruib uaA. `1` ampereB. ` 1` freadayC. `1` coulombD. None of the above |
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Answer» Correct Answer - B ` Ag^(+) + e^(-) rarr Ag, E_(Ag) = ("Atomic Mass")/1 = 108` Number of faraday `=(W_(Ag))/(E_(Ag)) = ( 108)/(108) = 1` `Al^(3+) + 3 e^(-) + rarr Al` `E_(AI) = (27)/3 =9` `W_(AI) = E_(AI) xx` No.of faradays `=9 = 4 45 g`. |
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| 2678. |
Amount of electricty that can depostit `108 g` of silver from `AgNO_3` aikyruib uaA. 1 ampereB. 1 coulombC. 1 faradayD. None of the above |
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Answer» Correct Answer - C `Ag^(+)+e^(-)Ag,E_(Ag)=("Atomic mass")/(1)=108` Number of faraday`=(W_(Ag))/(E_(Ag))=(108)/(108)=1`. |
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| 2679. |
When aluminium oxide `(Al_(2)O_(3))` is electrolysed for the production of aluminium metal. For a given quantity of electricty, the number of moles of aluminium obtained if the volume of `O_(2)` gas obtained is 201.6 litre measured at NTP, isA. 9B. 6C. 12D. 4.5 |
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Answer» Correct Answer - C Number of equivalents of oxygen `= (201.6)/(5.6) = 36` (Equivalent volume of oxygen `= 5.6` litre at NTP) `:.` Number of equivalents of `Al = 36` Mass of aluminium `= 36 xx 9 g` Number of moles of aluminium `= (36 xx 9)/(27) = 12` |
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| 2680. |
Calculate the number of moless of `Ag` formed above in Question.A. `10^(-4)`B. `10^(-5)`C. `10^(-6)`D. `10^(-7)` |
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Answer» Correct Answer - c mmoles of `Ag` formed `=` mmoles of `Ag^(o+)` used `=Mxx` Volume in `mL` `=10^(-5)M xx 100 mL` `=10^(-3)` Moles of `Ag` formed `=(mmol es )/(1000)=(10^(-3))/(10^(3))=10^(-6) mol es` |
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| 2681. |
Total charge in faraday present in 54 gm of `AI^(3+)` ion. |
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Answer» Correct Answer - 6 6 faraday |
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| 2682. |
How much charge is present on 1 mole of `Cu^(+2)` ion in faraday. (1 Faraday `= 96500` coulomb) |
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Answer» Correct Answer - 2 2 faraday |
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| 2683. |
The magnitude `(` but not the sign `)` of the standard reduction potentials of two metals `X` and `Y` are `:` `Y^(2)+2e^(-) rarr Y |E_(1)^(c-)|=0.34V` `X^(2)+2e^(-) rarr X |E_(2)^(c-)|=0.25V` When the two half cells of `X` and `Y `are connected to construct a cell, eletrons flow from `X` to `Y`. When `X` is connected to a standard hydrogen electrode `(SHE)`,electrons flow from `X` to `SHE`. If standard emf `(E^(c-))` of a half cell `Y^(2)|Y^(o+)` is `0.15V`, the standard emf of the half cell `Y^(o+)|Y` will beA. `0.19V`B. `0.53V`C. `0.49V`D. `0.64V` |
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Answer» Correct Answer - b Also, `{:(Y^(o+)+e^(-)rarr Y(s),,,=,E^(c-)._(req)=?),(,,,implies,DeltaG^(c-)._(req)=-1xxFxxE^(c-)._(req)):}` ` {:(Y^(2+)+e^(-)rarrY^(o+),,,=,E^(c-)._(1)=0.15V),(,,,implies,DeltaG^(c-)._(1)=-1xxFxxE^(c-)._(1)):} ` ` {:(Y^(2+)+e^(-)rarrY(s),,,=,E^(c-)._(2)=0.34V),(,,,implies,DeltaG^(c-)._(2)=-2xxFxxE^(c-)._(2)):} ` ` {:(ulbar(Y^(2+)+e^(-)rarrY(s)),,,=,-FE^(c-)._(req)=-2FE^(c-)._(2)-(-FE^(c-)._(1))),(,,,implies,E^(c-)._(req)=-2E^(c-)._(2)-E^(c-)._(1)):} ` `=2(0.34)-0.15` `=0.53V` |
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| 2684. |
The magnitude `(` but not the sign `)` of the standard reduction potentials of two metals `X` and `Y` are `:` `Y^(2)+2e^(-) rarr Y |E_(1)^(c-)|=0.34V` `X^(2)+2e^(-) rarr X |E_(2)^(c-)|=0.25V` When the two half cells of `X` and `Y `are connected to construct a cell, eletrons flow from `X` to `Y`. When `X` is connected to a standard hydrogen electrode `(SHE)`,electrons flow from `X` to `SHE`. If a half call `X|X^(2)(0.1M)` is connected to another half cell `Y|Y^(2+)(1.0M)` by means of a salt bridge and an external circuit at `25^(@)C`, the cell voltage would beA. `0.06V`B. `0.12V`C. `0.62V`D. `0.72V` |
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Answer» Correct Answer - c When `X` is connected to `SHE`, electons flow from `X` to `SHE`. This means that `X` is acting as anode and `SHE` as cathode and its oxidation potential is positive. Also, the reduction potential of `Y` is greater than the reduction potential of `X(` as electrons flow from `X` to `Y).` `implies Y^(2)+2e^(-) rarr Y(s)` `E^(c-)._(Y^(2+)|Y)=0.34V,` `X^(2+)+2e^(-) rarr X(s)` `E^(c-)._(X^(2+)|X)=-0.25V` Consider`: X(s)|X^(2+)(0.1M)||Y^(2+)(1.0M)|Y(s)` `E^(c-)._(cell)=E^(c-)._(cell)-(0.059)/(2) log .([X^(2+)])/([Y^(2+)])` `=[0.34-(-0.25)]-(0.059)/(2)log_(10).(0.1)/(1)=0.62V` |
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| 2685. |
Estimate the `E^(c-)` reduction for `Cu|CuS` electrode. Given `: K_(sp) ` of `CuS=8.0 xx 10^(-36),E^(c-)._((Cu|Cu^(2+)))=-0.34V`A. 1.034 VB. 1.0 VC. `-0.694V`D. 0.694 V |
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Answer» Correct Answer - C |
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| 2686. |
The magnitude `(` but not the sign `)` of the standard reduction potentials of two metals `X` and `Y` are `:` `Y^(2)+2e^(-) rarr Y |E_(1)^(c-)|=0.34V` `X^(2)+2e^(-) rarr X |E_(2)^(c-)|=0.25V` When the two half cells of `X` and `Y `are connected to construct a cell, eletrons flow from `X` to `Y`. When `X` is connected to a standard hydrogen electrode `(SHE)`,electrons flow from `X` to `SHE`. Given the following half cell `: YI+e^(-) rarr Y-I^(c-): " "E^(c-)=-0.27 V` Solubility product of the iodide salt `YI` isA. `2 xx 10^(-3)`B. `2 xx 10^(-12)`C. `10^(-15)`D. `6.8 xx 10^(-16)` |
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Answer» Correct Answer - C When connected, electron flow from `X` to `Y` means, `E_(Y^(2+)//Y)^(@) gt E_(X^(2+)//X)^(@)` and when `X` is connected to `SHF`, e flow from `X` to `SHE` means. `E_(X^(2+)//X)^(@)` `{:(YI +e^(-) rarr Y +I^(-)E^(@) =- 0.37 V),(Y rarr Y^(+) +e^(-) " "E^(@) = - 0.53 V):}/(Yi hArr Y^(+) +I^(-) " "E^(@) =- 0.90 V)` So, `E^(@) = (0.06)/(n) log K_(p)` `- 0.90 = (0.06)/(1) log K_(sp)` `log K_(sp) =- 15` `K_(sp) = 10^(-15)` |
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| 2687. |
`Cu^(2+)+2e^(-) rarr Cu`. For this, graph between `E_(red)` versus `ln[Cu^(2+)]` is a straight line of intercept `0.34V`, then the electrode oxidation potential of the half cell `Cu|Cu^(2+)(0.1M)` will beA. `-0.34 + (0.0591)/(2) V`B. `0.34 + 0.0591` VC. 0.34 VD. None |
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Answer» Correct Answer - A `E_(Cu | Cu^(2+)) = E_(Cu | Cu^(2+))^(@) - (0.0591)/(2) "log" ([Cu^(2+)])/([Cu(s)])` if `[Cu (s) ] = 1 and [Cu^(2+) ] = 0.1` M or `OA = E_(Cu^(2+) | Cu)^(@) therefore E_(Cu| Cu^(2+))^(@) = -0.34 V` Now `E_(Cu| Cu^(2+)) = -0.34 - (0.0591)/(2)"log" 0.1` `= -0.34 + (0.0591)/(2)` |
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| 2688. |
`Cu^(2+)+2e^(-) rarr Cu`. For this, graph between `E_(red)` versus `ln[Cu^(2+)]` is a straight line of intercept `0.34V`, then the electrode oxidation potential of the half cell `Cu|Cu^(2+)(0.1M)` will beA. `-0.34 + (0.0591)/(2)V`B. `0.34 + 0.0591 V`C. `0.34 V`D. None of these |
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Answer» Correct Answer - A `E_(Cu//Cu^(2+)) = E_(Cu//Cu^(2+))^(@) -(0.059)/(1)log[Cu^(2+)]` If `log [Cu^(2+)] = 0`, i.e., `[Cu^(2+)] = 1`, Then `E_(Cu//Cu^(2+)) = E_(Cu//Cu^(2+))^(@)` or Intercept `= E_(Cu^(2+)//Cu) = 0.34` Now `E_(Cu//Cu^(2+))^(@) = -0.34 - (0.059)/(2) log 0.1` `= -0.34 + (0.059)/(2) V` |
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| 2689. |
A voltaice cell constructed form two half-cells composed of the same material but differing in ion concentration is calledA. a chemical cellB. a concentration cellC. electrolytic cellD. an electrochemical cell. |
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Answer» Correct Answer - B Any galvanic cell in which two half-cell reactions use different reactants is called a chemical cell. Deniell cell is one such example. If the chemical reactions in two half-cells are the same, the concentrations of the ionic species are different, then the net result is the transfer of species form higher concentration to lower concentration. There is no net chemical reaction in the cell. Such cells are commmonly known as concentration cells For example, `{:(" "Zn(s)|ZnSO_(4)(C_(1))||ZnSO_(4)(C_(2))|Zn(s)),("Anode":" "Zn(s) hArr Zn^(2+)(C_(1))+2e^(-)),("Cathode" :Zn^(2+)(C_(2))+2e^(-)hArrZn(s)),(bar("Cell reaction :"Zn^(2+)(C_(2))hArr Zn^(2+)(C_(1)))):}` At `298 K` `E_("cell") = E_("cell")^(@) - (0.0592 V)/(2) log (C_(1))/(C_(2))` Since `E_("cell")^(@)` for a concentration cell is zero because the same electrodes and the same type of ions are involved, we have in general `E_("cell") = (0.0592 V)/(n) "log" (C_(2))/(C_(1))` Note that for the process to be feasible, `E_("cell")` should the posi-tive i.e. `C_(2) gt C_(1)`. |
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| 2690. |
Assertion: Galvanished iron does not rust. Reason: Zinc ha a more negative electrode potential than iron.A. if both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is falseD. If the assertion and reason both are false. |
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Answer» Correct Answer - A Zinc metal which has a more negative electrode potential than iron will provide electrons in preference of the iron, and therefore corrode first. Only when all the zinc has been oxidised does the iron start to rust. |
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| 2691. |
A certain amount of charge is passed through acidulated water. A total of `504mL` of hydrogen and oxygen were collected at STP. Find the magnitude of charge that is passed during electrolysis in coulombs. |
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Answer» `1F=1 Eq of O_(2)+1 Eq of h_(2)` `=5.6L O_(2)+11.2L of H_(2)=16.8L` `16.8 xx 10^(3)mLimplies1F` `504 mLimplies(1xx504)/(16.8xx10^(3))` `=0.03F=0.03xx96500=2895C` |
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| 2692. |
Peroxodisulphate salts, `(e.g., Na_(2)S_(2)O_(8))` are strong oxidizing agents used be bleaching agents for fats, oils, ets. Given `:` `O_(2)(g)+4H^(o+)(aq)+4e^(-)rarr 2H_(2)O(l)" "E^(c-)=1.23V` `S_(2)O_(8)^(2-)(aq)+2e^(-)rarr 2SO_(4)^(2-)(aq)" "E^(C-)=2.01V` Which of the following statement is `(` are `)` correct ?A. Oxygen gas can oxidize sulphate ion to per`-` oxo disulphate ion `(S_(2)O_(8)^(2-))` in acidic solution.B. `O_(2)(g)` is reduced to water.C. Water is oxidized to `O_(2)`D. `S_(2)O_(8)^(2-)` ions are reduced to `SO_(4)^(2-)` ions. |
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Answer» Correct Answer - c,d Create a cell with required cell reaction `:` `O_(2)+SO_(4)^(2-)rarr S_(2)O_(8)^(2-)+H_(2)O, E^(c-)._(cell)=1.23-2.01lt0` `implies ` Spontaneous cell reaction `: (` reverse reaction `),i.e.,` `H_(2)O+S_(2)O_(8)^(2-) rarr O_(2)+SO_(4)^(2-)` |
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| 2693. |
A current 0.5 ampere when passed through `AgNO_(3)` solution for 193 sec. Deposited 0.108 g of Ag. Find the equivalent weight of Ag.A. 108B. 54C. 10.8D. 5.4 |
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Answer» Correct Answer - A `W = ( I xx t(s) xx E)/(F)` `therefore E = (W xx F)/(I xx t) = (0.108 xx 96500)/(0.5 xx 193) = 108` |
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| 2694. |
Arrange the following metals in the order in which they displace each other from the solution of their salts. Ag, Al, Cu, Fe, Mg and Zn. |
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Answer» Mg, Al, Zn, Fe, Cu, Ag. |
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| 2695. |
Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn. |
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Answer» Mg, Al, Zn, Fe, Cu. |
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| 2696. |
Explain how rusting of iron is envisaged as setting up of an electroCHMemical cell. |
| Answer» Iron (Fe) is involved in the redox-reaction that is carried in the electrochemical cell which is set up. As a result, it slowly dissolves and the metal surface gets rusted or corroded. For more details, consult corrpsion (Section 23). | |
| 2697. |
Arrange the following metals in the order in which they displace each other from the solution of their salts. `Al, Cu, Fe, Mg, ` and `Zn`. |
| Answer» Correct Answer - Mg,Al,Zn,Fe,Cu,Ag. | |
| 2698. |
When an aqueous solution of `LiCl` is electrolyzed using graphite electrodes, as the current flows, the `pH` of the solution around cathode `……………………`A. `CI_(2)` is liberated at the anodeB. `Li` is deposited at the cathodeC. as the current flow, `pH` of the solution around the cathode remains constantD. as the current flows, `pH` of the solution around the cathode decreases |
| Answer» Correct Answer - A | |
| 2699. |
A cell is prepared by dipping a copper rod in `1 MCuSO_(4)` solution and a nickel rod in `1M NiSO_(4)`. The standard reduction potentials of copper and nickel electrodes are `+0.34 V` and `-0.25 V` respectively. (i) Which electrode will work as anode and which as cathode? (ii) What will be the cell reaction? (iii) How is the cell represented ? (iv) Calculate the `EMF` of the cell. |
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Answer» (i) The nickel electrode with smaller `E^(@)` value `(-0.25V)` will work as anode while copper electrode with more `E^(@)` value `(+0.34V)` will work as cathode. (ii) The cell reaction may be written as: `{:("A anode", :, Ni(s) rarr Ni^(2+)(aq).+2e^(-)),("At cathode",:, Cu^(2+) (aq) +2e^(-) rarr Cu(s)),("Cell reaction",:, Ni(s) +Cu^(2+) (aq) rarr Ni^(2+)(aq) +Cu(s)):}` The cell may be represented as: `Ni(s)//Ni^(2+) (aq) || Cu^(2+)(aq)//Cu(s)` (iv) `CMF` of cell `=E_("cathode") -E_("anode")^(@) =(+0.34) -(0.25) = 0.59V` |
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| 2700. |
Arrange the following metals in the order in whiCHM they displace eaCHM other from the solution of their salts. `Al, Cu, Fe, Mg, ` and `Zn`. |
| Answer» Correct Answer - Mg, Al, Zn, Fe, Cu. | |