3195 + Interview Questions in ELECTROCHEMISTRY IN GENERAL AWARENESS Page 26 InterviewSolution

1251.	`1F` of electricity is passed through `10 L` of a solution of aqueous solution of `NaCl` . Calculate the `pH` of the solution.
Answer» Aqueous `NaCl` solution on electrolysis gives `Cl_(2)(g)` at anode and `H_(2)(g)` at cathode and `overset(c-)(O)H` ions in the solution. `1F=1` equilvalent of `H_(2)(g)=1` equivalent of `Cl_(2)(g)` and `=1`Equivalent of `overset(c-)(O)H` ions `:. [overset(c-)(O)H]=("Equivalent")/("Volume of solution in L")=(1)/(10L)=10^(-1)N` or `M` `:. pOH=-log (10^(-1))=1` `pH=14-1=13`

Discussion

1252.	The density of copper is `8.95 g mL^(-1)`. Find out the number of coulombs needed to plate an area of `100cm^(2)` to a thickness of `10^(-2)cm` using `CuSO_(4)` solution as electrolyte. `(` Atomic weight of `Cu=63.5g)`
Answer» Correct Answer - `27172C` Mass of `Cu=Vxxd=(10xx10xx10^(-2)xx8.94)g=8.94g` `CuSO_(4) overset(El ectrolysis)rarr Cu^(2+)+SO_(4)^(2-)` `Cu^(2+)+2e^(-) rarr Cu` `2e^(-)=2F=2xx96500C-=1 mol `of `Cu=63.5g` `:. 63.5g `of `Cu` requires `-=2xx96500C` `8.94g` of `Cu` requires `=(2xx96500xx8.94)/(63.5)C=27172.0C`

Discussion

1253.	In an electrolysis of AgNO3 solution, 0.7 g of Ag is deposited after a certain period of time. Calculate the quantity of electricity required in coulomb. (Molar mass of Ag is 107.9 g mol-1.)
Answer» Given : Mass of Ag deposited = 0.7 g Molar mass of Ag = 107.9 g mol^-1 Quantity of electricity = Q = ? Reduction half reaction is, Ag⁺ + e^- → Ag 1 mole of Ag = 107.9 g Ag requires 1 mole of electrons ∴ 0.7 g Ag will require, \(\frac{0.7}{107.9}\) = 6.49 × 10^-3 mole of electrons ∵ 1 mole of electrons carry 96500 C charge ∴ 6.49 × 10^-3 mole of electrons will carry, 96500 × 6.49 × 10^-3 = 626 C ∴ Quantity of electricity required = 626 C,

1253.

In an electrolysis of AgNO3 solution, 0.7 g of Ag is deposited after a certain period of time. Calculate the quantity of electricity required in coulomb. (Molar mass of Ag is 107.9 g mol-1.)

Answer»

Given :

Mass of Ag deposited = 0.7 g

Molar mass of Ag = 107.9 g mol^-1

Quantity of electricity = Q = ?

Reduction half reaction is,

Ag⁺ + e^- → Ag

1 mole of Ag = 107.9 g Ag requires 1 mole of electrons

∴ 0.7 g Ag will require,

\(\frac{0.7}{107.9}\) = 6.49 × 10^-3 mole of electrons

∵ 1 mole of electrons carry 96500 C charge

∴ 6.49 × 10^-3 mole of electrons will carry,

96500 × 6.49 × 10^-3 = 626 C

∴ Quantity of electricity required = 626 C,

Discussion

1254.	The same two separate electrolytic was passed through two separate electrolytic cells containing solutions of nickel nitrate and chromium nitrate respectively . If 0.3 g of nickel was deposited in the first cell , the amount of chromium deposited is (At.mass of Ni= 59 , Cr = 52)A. `0.1` gB. `0.17` gC. `0.3g`D. `0.6g`
Answer» Correct Answer - B `("Wt. of Ni")/("Wt. of Cr") = ("Eq. of mass of Ni")/("Eq. of mass of Cr")` `(0.3)/("Wt. of Cr") = (59//2)/(52//3)` or Wt. of Cr = 0.17 g .

Discussion

1255.	The amount of sodium produced by passage of 2 moles of electrons in the electrolysis of fused NaCl is ______A. 23 gB. 46 gC. `11.5 g`D. `58.5g`
Answer» Correct Answer - B

Discussion

1256.	In a voltaic cell, if iron and silver electrodes are connected with each other then current flowA. from iron to silver outside the cellB. from silver to iron within the cellC. from silver to iron outside the cellD. current does not flow in this cell
Answer» Correct Answer - C

Discussion

1257.	An external current source giving a current of `5.0A` was joined with Daniel cell arrangement opposing the normal current flow and was removed after 10 hrs. Before passing the current the LHE and RHE contained 1L each of `1M Zn^(2+)` and `Cu^(2+)` respectively. Find the EMF supplied by the Daniel cell after removal of teh external curretn source, `E^(@)` of `Zn^(2+)//Zn` and `Cu^(2+)//Cu` at `25^(@)C` is `-0.76` and `+0.34V` respectively.
Answer» Correct Answer - `1.143V` `n_(E) = (5 xx 3600 xx 10)/(96500)` `n_(E) = 1.865` `Zn^(2+) rarr ZN +2e^(-)` mole of `Zn^(2+)` deposited `= 0.932` mole of `Zn^(2+)` left `= 1 - 0.932` `=0.0673` `Cu rarr Cu^(2+) +2e^(-)` mole of `Cu^(2+)` produced `=0.932` so total mole of `Cu^(2+) = 1.932` `E_(cell) = E^(@) -(0.059)/(2)log.(0.0673)/(1.932)` `E_(cell) = 1.143V`

Discussion

1258.	An external voltage opposing the cell potential is applied to a Daniel cell with zinc and copper electrodes. What happen when the external emf is increased from 0.8V to 1.1 V and then 1.54 V ?
Answer» (i) Working principle of Daniel cell. (ii) Calculation of cell emf. (iii) Comparison of the cell emf with the external voltage applied. (iv) Determination of direction of current depending on the magnitude of the external voltage

Discussion

1259.	A galvanic cell has `E_(cell)^(@)=1.5" V"`. If an opposing potential of `1.5" V"` is applied to cell, what will happen to the cell reaction and current flowing through the cell ?
Answer» On applying opposing potential, reaction continues to take palce till the magnitude of opposing e.m.f. reaches the value of 1.1" V". After this , reaction stops and no further reaction will take place.

Discussion

1260.	Given that, `Co^(3+) +e^(-)rarr Co^(2+) E^(@) = +1.82 V` `2H_(2)O rarr O_(2) +4H^(+) +4e^(-), E^(@) =- 1.23 V`. Explain why `Co^(3+)` is not stable in aqueous solutions.
Answer» The `E_("cell")` can be calculated as follows: `4 xx [ Co^(3+) +e^(-) rarr Co^(2+)], E^(@) = +1.82 V` `1 xx [2H_(2)O rarr O_(2) +4H^(+) +4e^(-)] , E^(@) =- 1.23 V`. Add: `4Co^(3+) +2H_(2)O rarr 4Co^(2+) +4H^(+) +O_(2), E^(@) = 1.82 - 1.23 = +0.59 V`. Since, `E_("cell")` is positive the cell reaction is spontaneous. This means that `Co^(3+)` ions will take part in the reaction. Therefore, `Co^(3+)` is not stable.

Discussion

1261.	Assume that during the electrolysis of `AgNO+(3)`, only `H_(2)O` is electrolyzed and `O_(2)` is formed as `2H_(2)O rarr 4H^(o+)+O_(2)+4e(-)` `O_(2)` formed at `NTP` due to passage of 2 amperes of current for 96 second isA. `0.112 L`B. `0.224L`C. `11.2L`D. `22.4L`
Answer» Correct Answer - a `ZIt or El t. [{:(Ew,of ,O_(2),=,(32)/(4)),(n,fact o r,,=,4):}]` `W_((O_(2)))=(32xx2xx965)/(4xx96500)` `=0.16g=0.005mol=0.112L at STP`.

Discussion

1262.	Given:`Pt(s)"\|"underset(P_1atm)Cl_2(g)"\|"Cl^(-)(C_1)"\|\|"Cl^(-)(Cl_2)"\|"underset(P_2atm)(Cl)_(2)(g)"\|"Pt(s)` identify in which of following condition working of cell takes place:A. `C_1gtC_2"and"P_1=P_2`B. `P_1gtP_2"and"C_1=C_2`C.D. `P_1ltP_2"and"C_1=C_2`
Answer» Correct Answer - A::B

Discussion

1263.	How will the pH of brine (aq. NaCl solution) be affected when it is electrolysed?
Answer» The pH of the solution will rise as NaOH is formed in the electrolytic cell.

Discussion

1264.	A galvanic cell has electrical potential of 1.1V. If an opposing potential of 1.1V is applied to this cell, what will happen to the cell reaction and current flowing through the cell?
Answer» When the opposing potential becomes equal to electrical potential, the cell reaction stops and no current flows through the cell. Thus, there is no chemical reaction. See NCERT textbook, page no. 64

Discussion

1265.	Which one of the following will increase the voltage of the cell ? `(T = 298 K)` ` Sn+ 2Ag^+ rarr Sn^(2+) +2 Ag`.A. increase in concentration of ` Sn^(2+)`B. Increase in size of silver rodC. Increase in the concentration of `Ag^+`D. none of these
Answer» Correct Answer - C ` C_(cell) = F_(cell)^@ -(2.303RP)/(nF) log. ([Sn^(2+)])/([Ag^(2+)]^@) `. Increase in the concentration of `Ag^+` decreases the value of log `([Sn^(2+)])/([Ag^(2+)])` This results in the increase value of ` E_("cell")`.

Discussion

1266.	`Pt \| Cl_2 (P_2) \|HCl (0.M) \|Pt\|Cl_2 (P_2)` , cell reaction will be spontaneous if :A. ` P_1 =P_2`B. ` P_1 gt P_2`C. ` P_2 gt P_1`D. ` P_1 = P_2 =1 ` atm
Answer» Correct Answer - C `Delta G=- 2.303 RT log K` for spontaneous `Delta G =- ve` ` - (Delta G)/( 2.303) RT = log P_1/P_2` `P_2/(((Delta G)/(2. 303 RT))) gt P_1` ` P_2 gt P_1`.

Discussion

1267.	The half-cell reduction potential of a hudrogen electrode at `pH= 10` will be.A. ` 0. 95 V`B. ` -0. 59 V`C. ` 0.059 V`D. ` -0 .059 `
Answer» Correct Answer - B `E_(red = E_(cell) ^@ + (0. 0591)/n log [Mn^(=)` `= 0 + (0.0591)/1 log [10^(-10)]` ` =+ 0.0591 xx - 10 - 0. 59 V`.

Discussion

1268.	In an electrolytic cell, one litre of a 1 M aqueous solution of `MnO_(4)^(-)` is reduced at the cathode the quantity of electricity required so that the final solution is `0.1M MnO_(4)^(2-)` will beA. 0.1FB. 1FC. 10FD. 0.01F
Answer» Correct Answer - A `MnO_(4)^(-)+e^(-)rarrMnO_(4)^(2-)` To reduce 1 mol of `MnO_(4)^(-)` to `MnO_(4)^(2-)` 1F is required As final solution contains 0.1 M `MnO_(4)^(-)` to `MnO_(4)^(2-)` electricity required =0.1F

Discussion

1269.	Consider the following cell with hydrogen electrodes at difference pressure `p_(1)` and `p_(2)`. The `EMF` of the cell is given byA. `(RT)/(F)ln.(p_(1))/(p_(2))`B. `(RT)/(2F)ln.(p_(1))/(p_(2))`C. `(RT)/(F)ln.(p_(2))/(p_(1))`D. `(RT)/(2F)ln.(p_(2))/(p_(1))`
Answer» Correct Answer - b `E_(cell)=0.059[pH_(a)-pH_(c)+(1)/(2).((p_(H_(2))))/((p_(H_(2)))_(c)]]` `=0.059[0-0+(1)/(2)log `(p_(1))/(p_(2))]` `=(0.059)/(2)log `(p_(1))/(p_(2))=(RT)/(2F)ln.(p_(1))/(p_(2))`

Discussion

1270.	On electrolysis, which of the following does not give out hydrogen?A. Acidic water using Pt electrodesB. Fused NaOH using Pt electrodesC. Dilute `H_(2)SO_(4)` using Pt electordesD. Dilute `H_(2)SO_(4)` using Cu electordes
Answer» Correct Answer - D On electrolysis of dilute `H_(2)SO_(4)` solution using Cu electrodes Cu will get dissolved at anode and Cu w ill get deposited on cathode No `H_(2)` will be evolved.

Discussion

1271.	`Pt(H_(2))(p_(1))\|H^(o+)(1M)\|(H_(2))(p_(2)),Pt` cell reaction will be exergonic ifA. `p_(1)=p_(2)`B. `p_(1)gtp_(2)`C. `p_(2)gtp_(1)`D. `p_(1)=1 atm`
Answer» Correct Answer - b For a cell to be spontaneous `(E_(cell)=+ve or Delta G =-ve)(` exergonic `)p_(1)p_(2) or c_(2)gtc_(1)`. Factual statement.

Discussion

1272.	`Pt(Cl_(2))(p_(1))\|HCl(0.1M)\|(Cl_(2))(p_(2)),Pt` cell reaction will be endergonic ifA. `p_(1)=p_(2)`B. `p_(1)gtp_(2)`C. `p_(2)gtp_(1)`D. `p_(1)=1 atm`
Answer» Correct Answer - c Cell is spontaneous when `p_(2)(` cathode `)gtp_(1)(` anode `)` At anode `: 2Cl^(c-)(0.1M) rarr Cl_(2)(p_(1))+2e^(-)` At cathode `: Cl_(2)(p_(2))+2e^(-) rarr 2 C l^(c-)(0.1M)` `ulbar(Cell reaction : Cl_(2)(p_(2))rarr Cl_(2)(p_(1)))` `C_(cell)=E^(c-)._(cell)-(0.059)/(2)log.(p_(1))/(p_(2))` `=0+(0.059)/(2)log `(p_(2))/(p_(1))` `E_(cell)gt0` if `p_(2)gtp_(1)`.

Discussion

1273.	For the concentration cell `Cu\|Cu^(2+)(C_(1))\|\|Cu^(2+)(C_(2))\|Cu,DeltaG` will be negative if `Pt(Cl_(2)-P_(1))\|HCl(0.1M)\| Pt(Cl_(2)-P_(2))` the cell reaction is spontaneous ifA. `p_(1)=p_(2)`B. `p_(1)gtp_(2)`C. `p_(2)gtp_(1)`D. None of these.
Answer» Correct Answer - C It is a concentration cell reversible w.r.t. anion `2Cl^(-)(0.1M)rarrCl_(p_(1))+2e^(-)` `Cl_(2)(p_(2))+2e^(-)rarr2Cl^(-)(0.1M)` `Cl_(2)(p_(2))+2e^(-)rarr2Cl^(-)(0.1M)` `Cl_(2)(p_(2))rarrCl_(2)(p_(1))` `E_("Cell")=-(0.0591)/(5)"log"(p_(1))/(p_(2))` `=(0.0591)/(5)"log"(p_(1))/(p_(1))` `E_("cell")gt0` if `p_(2)gtp_(1)`

Discussion

1274.	The specific conductances of four electrolytes in `ohm^(-1)cm^(-1)` are given below. Which one offers higher resistance to passage of electric current?A. `7.0xx10^(-4)`B. `9.2xx10^(-10)`C. `6.0xx10^(-8)`D. `4.0xx10^(-9)`
Answer» Correct Answer - B Lesser the speci fic conductance more is the specific resistance.

Discussion

1275.	The gas`X` at `1 atm` is bubbled through a solution containing a mixture of `1MY^(c-)` and `1MZ^(c-)` at `25^(@)C` . If the order of reduction potential is `ZgtYgtX`, thenA. `Y` will oxidize `X` onlyB. `Y` will oxidize `Z` onlyC. `Z` will oxidize `X` and `Y`D. `Z` will reduce both `X` and `Y`
Answer» Correct Answer - a,c `E^(c-)._(Z)` is highest `implies` Strongest oxidizing agent. `E^(c-)._(X)` is lowest `implies` Wealest pxodozomg agent. `implies Y` will oxidize `X` but not `Z`. Also will oxidize both `X` and `Y`.

Discussion

1276.	The gas`X` at `1 atm` is bubbled through a solution containing a mixture of `1MY^(c-)` and `1MZ^(c-)` at `25^(@)C` . If the order of reduction potential is `ZgtYgtX`, thenA. `Y` will oxidise `X` and not `Z`B. `Y` will oxidise `Z` and not `X`C. `Y` will oxidise both `X` and `Z`D. `Y` will reduce both `X` and `Z`
Answer» Correct Answer - A `E_(RP_(Z//Z^(-)))^(@) gt E_(RP_(Y//Y^(-)))^(@) gt E_(RP_(X//X^(-)))^(@)` Thus order of oxidizing power will be `Z gt Y gt X`

Discussion

1277.	If it is disreed to construct the following voltaic cell to have `E_(cell) = 0.0860 V`, what `[Cl^(-)]` must be present in the cathodic half cell to achieve the desired e.m.f.? Given `K_(SP)` of `AgCl` and `Agl` are `1.8 xx 10^(-10)` and `8.5 xx 10^(-17)` respectively. `Ag_((s))\|Ag^(+)[Sat. Agl_((aq.))]\|\|Ag^(+)[Sat.AgCl.xMCl^(-)\|Ag_((s))]`
Answer» Correct Answer - `6.8 xx 10^(-4)M ;`

Discussion

1278.	`Zn \| Zn^(2+)(C_(1) \|\| Zn^(2+)(C_(2)\| Zn`. For this cell `DeltaG` is negative if:A. `c_(1)=c_(2`B. `c_(1)gtc_(2)`C. `c_(2)gtc_(1)`D. None
Answer» Correct Answer - c `E=E^(c-)-(0.0591)/(2)log((c_(1))/(c_(2)))=(0.0591)/(2)log.(c_(2))/(c_(1))` To make `DeltaG=-ve,E=+ve` Hence, `c_(2)gtc_(1)`

Discussion

1279.	`Zn\|Zn^(2+)(c_(1)) \|\| Zn^(2+)(c_(2))\|Zn`. For this cell `DeltaG` is negative if `:`A. `C_(1) = C_(2)`B. `C_(1) gt C_(2)`C. `C_(2) gt C_(1)`D. None of these
Answer» Correct Answer - C `{:(ZnrarrZn_(C_(1))^(2+)+2e),(Zn_(C_(1))^(2+)+2erarrZn):}/(Zn_(C_(1))^(2+)rarrZn_(C_(1))^(2+))` Thus `E_("cell") = +ve` only `c_(2) gt c_(1)` bacause `E_("cell") = (0.059)/(2)"log"([Zn_(C_(2))^(2+)])/([Zn_(C_(1))^(2+)])`

Discussion

1280.	For the reaction `:` `A+2B^(3+) rarr 2B^(2+)+A^(2+)` `E^(c-)` of the given redox reaction is `:` `A^(2+)+2e^(-) rarr A` `E^(c-)=Xv` `B^(3+)+e^(-) rarr B^(2+)` `E^(c-)=yV`A. `x-2y`B. `x+y//2`C. `x-y`D. `y-x`
Answer» Correct Answer - d It is a redox reaction, thus Cell reaction `:` At anode `: A rarr A^(2+)+2e^(-)(` Oxidation `)` `E^(c-)._(o x i d)=-xV,E^(c-)._(red)=xV` At cathode `: 2B^(3+)2e^(-) rarr 2B^(2+)(` Reduction `)` `E^(c-)._(red)=yV` `ulbar(Cell reaction : A+2B^(3+) rarr A^(2+)+2B^(2+))` `E^(c-)._(cell)=(E^(c-)._(red))_(c)-(E^(c-)._(red))_(a)=(y-x)V`

Discussion

1281.	`Zn\|Zn^(2+)(c_(1)) \|\| Zn^(2+)(c_(2))\|Zn`. For this cell `DeltaG` is negative if `:`A. `C_(1) = C_(2)`B. `C_(1) gt C_(2)`C. `C_(2) gt C_(1)`D. None
Answer» Correct Answer - C

Discussion

1282.	In an electrochemical cell that function as a voltaic cell:-A. electrons move frm the cathode to the cathodeB. electrons move through a salt-bridgeC. electrons can move either form the cathode to the anode or form the anode to the cathodeD. reduction occurs at the cathode
Answer» Correct Answer - D

Discussion

1283.	`E^(c-)._(red)` of different half cell are given as `:` `E^(c-)._(Cu^(2+)\|Cu)=0.34V,E^(c-)._(Zn^(2+)\|Zn)=-0.76V`. `E^(c-)._(Ag^(o+)\|Ag)=0.80 V, E^(c-)._(Mg^(2+)\|Mg)=-2.37V.` In which cell `DeltaG^(c-)` is most negative ?A. `Zn\|Zn^(2+)(1M)\|\|Mg^(2+)(1M)\|Mg`B. `Zn\|Zn^(2+)(1M)\|\|Ag^(o+)(1M)\|Ag`C. `Cu\|Cu^(2+)(1M)\|\|Ag^(o+)(1M)\|Ag`D. `Ag\|Ag^(o+)(1M)\|\|Mg^(2+)(1M)\|Mg`
Answer» Correct Answer - b `DeltaG^(c-)=-nFE^(c-)._(cell)` For `DeltaG^(c-)` to be most negative, check for maximum value of `E^(c-)._(cell)`. `a.` `E^(c-)._(cell)=(E^(c-)._(red))_(c)-(E^(c-)._(red))_(a)=-2.37-(-0.76)` `=-1.61V` `b.` `E^(c-)._(cell)=0.80V-(0.76)=1.56V` `c.` `E^(c-)._(cell)=0.80V-0.34=0.46V` `d.` `E^(c-)._(cell)=-2.37-0.80=-3.17`

Discussion

1284.	`100mL` of buffer of `1 M NH_(3)(aq)` and `1 M NH_(4)^(o+)(aq)` are placed in two compartments of a voltaic cell separately. A current of `1.5A` is passed through both cells for `20 m i n`. If only electrolysis of water takes place, thenA. `L.H.S.` will increaseB. `R.H.S.` will increaseC. Both side will increaseD. Both side will decrease
Answer» Correct Answer - B In `R.H.S.` formation of `OH^(-)` takes palce which reacts with buffer as `NH_(4)^(+) + OH^(-) rarr NH_(3) + H_(2)O` Thus `[NH_(4)^(+)]` decrease and `[NH_(3)]` increasee. Also `pOH = pK_(b) + "log"([NH_(4)^(+)])/([NH_(3)])` Thus, `pOH` decreases or `pH` increases. Similarly in `L.H.S.` formation of `H^(+)` shows. `NH_(3) rarr H^(+) rarr NH_(4)^(+)` or `[NH_(4)^(+)]` Increases and `[NH_(3)]` decreases. Thus `pOH` increases or `pH` decreases.

Discussion

1285.	`Cu^(+) + e rarr Cu, E^(@) = X_(1)` volt, `Cu^(2+) + 2e rarr Cu, E^(@) = X_(2)`X_(2) volt For `Cu^(2+) + e rarr Cu^(+), E^(@)` will be :A. `X_(1) - 2X_(2)`B. `X_(1) + 2X_(2)`C. `X_(1) - X_(2)`D. `2X_(2) - X_(1)`
Answer» Correct Answer - D `{:(Cu^(+)+erarrCu,-Delta G_(1)^(@)=1xxX_(1)F),(underset(-)(Cu)^(2+)+2erarr underset(-)(Cu),-Delta G_(2)^(2)=underset(-)(2)X_(2)F),(bar(Cu^(+)rarr Cu^(2+)+e, :. -Delta G_(3)^(@)=(-2X_(2)+X_(1))F)),(orCu^(+)+e rarr Cu^(1+), :. Delta G^(@)=(-X_(1)+2X_(2))F):}` or `1xxE_(3)^(@)xxF=(2X_(2)-X_(1))F` or `E_(3)^(@)=(2X_(2)-X_(1))V`.

Discussion

1286.	`Cu^(+) + e rarr Cu, E^(@) = X_(1)` volt, `Cu^(2+) + 2e rarr Cu, E^(@) = X_(2)`X_(2) volt For `Cu^(2+) + e rarr Cu^(+), E^(@)` will be :A. `x_(1) -2x_(2)`B. `x_(1)+2x_(2)`C. `x_(1)-x_(2)`D. `2x_(2)-x_(1)`
Answer» Correct Answer - D `Cu^(+)+e^(-)rarr Cu Delta G_(1)= -F E^(@)= -FX_(1)` `Cu^(2+)+2e^(-)rarr Cu Delta G_(2)= -2F E^(@)= -2 FX_(2)` `Cu^(2+)+e^(-)rarr Cu^(+) Delta G_(3)= -2F E^(@)= -2 FX_(2)` `Delta G_(3) = Delta G_(2)-Delta G_(1) -F E^(@)= -2 FX_(2)+FX_(1)` `E^(@)=2X_(2)-X_(1)`

Discussion

1287.	At what concentration of `[overset(c-)(O)H]` does the following half reaction has a potential of `0V` when other species are at `1M ?` `NO_(3)^(c-)+H_(2)O+2e^(-) rarr NO_(2)^(c-)+2overset(c-)(O)H,E^(c-)._(cell)=-0.06V`A. `2.0M`B. `1.0M`C. `0.1M`D. `0.01M`
Answer» Correct Answer - c `E_(cell(red))=E^(c-)._(cell(red))-(0.059)/(2)log [overset(c-)(O)H]^(2)` `=-0.06-0.059log[overset(c-)(O)H]=0` `[overset(c-)(O)H]=0.14`

Discussion

1288.	Consider the cell reaction `:` `Mg(s)+Cu^(2+)(aq) rarr Cu(s) +Mg^(2+)(aq)` If `E^(c-)._(Mg^(2+)\|Mg(s))` and `E^(c-)._(Cu^(2+)\|Cu(s))` are `-2.37` and `0.34V`, respectively. `E^(c-)._(cell)` isA. `2.03V`B. `-2.03V`C. `-2.17 V`D. `2.71 V`
Answer» Correct Answer - d The cell is represented as `:` `Mg(s)\|Mg^(2+)\|\|Cu^(2+)\|Cu(s)` `E^(c-)._(cell)=(E^(c-)._(red))_(c)-(E^(c-)._(red))_(a)=0.34-(-2.37)=2.71V`

Discussion

1289.	If hydrogen electrodes dipped in two solutions of `pH=3` and `pH=6` are connected by a salt bridge, the `EMF_(cell)` isA. `0.052V`B. `0.104V`C. `0.177V`D. `0.3V`
Answer» Correct Answer - C For the given concentration cell `:` `H_(2)\|H^(o+)(c_(1))\|\|H^(o+)(c_(2))\|H_(2)` `E_(cell)=-0.059(pH_(c)-pH_(a))` `=-0.059(3-6)=0.1773V` Cell will be feasible only when `[c_(2)]_(c)gt[c_(1)]_(a)` or `pH_(c)ltpH_(a)`

Discussion

1290.	Two electrochemical cells, `Zn^(2+)\|Zn^(2+)\|\|Cu^(2+)\|Cu` and `Fe\|Fe^(2+)\|\|Cu^(2+)\|Cu` are connected in series. What will be the net e.m.f. of the cell at `25^(@)C` ? Given : `E^(@)` of `Zn^(2+)\|Zn=-0.76" V"`, `Cu^(2+)\|Cu=+0.34" V",Fe^(2+)\|Fe=-0.41" V"`A. `+1.85" V"`B. `-1.85" V"`C. `+0.83" V"`D. `-0.83" V"`
Answer» Correct Answer - A (a) emf of the cell,`Zn(s)\|Zn^(2+)(aq)\|\|Cu^(2+)(aq)\|Cu(s)` is : `E_(Cu^(2+)//Cu)^(@)-E_(Zn^(2+)//Zn)^(@)=+0.34-(-0.76)=1.10" V"` emf of the cell, `Fe(s)\|Fe^(2+)(aq)\|\|Cu^(2+)(aq)\|Cu(s)` is : `E_(Cu^(2+)//Cu)^(@)-E_(Fe^(2+)//Fe)^(@)=+0.34-(0.41)=0.75" V"` Since the two electrodes are connected in series, the net emf of cell is : `=1.10+0.75=+1.85" V"`

Discussion

1291.	Which of the following statements regarding rusting of iron is `//` are correct ?A. It takes place is moist air.B. It is stopped in `CO_(2)` atmosphereC. If produces `Fe(III)` oxide.D. It is an electrochemical process.
Answer» Correct Answer - a,c,d Refer section

Discussion

1292.	Given `:` `A^(2+)+2e^(-) rarr A(s)" "E^(c-)=0.08V` `B^(o+)+e^(-)rarr B(s)" "E^(c-)=-0.64V` `X_(2)(g)+2e^(-) rarr 2X^(c-)" "E^(c-)=1.03V` Which of the following statements is `//` are correct ?A. `X_(2)(g)` will oxidize both `(A)` and `(B)`.B. `A^(2+)` will oxidize both `(A)` and `(B)`C. The reaction `2X^(c-)(1.0M)+A^(2+)(1.0M) rarr X_(2)(1 atm)+A(s)` will be spontaneous.D. The oxidizing power of `A^(2+),B^(o+),` and `X_(2)(g)` is in the order `X_(2)gtA^(2+)gtB^(o+)`
Answer» Correct Answer - a,b,d `E^(c-)._(cell)` corresponding to the reaction is `:` `2X^(c-)(1M)+A^(2+)(1.0M) rarr X_(2)(1atm)+A(s)` `E^(c-)._(cell)E^(c-)._(A^(2+)\|A)-(E^(c-)._(X2\|2X^(c-))=0.08-1.30=-1.22V` `E^(c-)._(cell)` is negative, so `DeltaG^(c-)` will be `(+ve)` and hence reaction `(c)` will not be feasible. So, statement `(c)` is wrong. Statements `(a),(b),` and `(d)` are correct.

Discussion

1293.	The `EMF` of the cell `Ag\|0.1(N)AgNO_(3)\|\|1(N) KBr, AgBr(s)\|Ag` was found to be `-0.64V` at `298 K. 0.1 N Ag NO_(3)` is `81.3%` dissociated and` 1N KBr` is `75.5%` dissociated. Calculate `:` `a.` Solubility `b.` Solubility product of `AgBr ` at `298 K`
Answer» `Ag\|0.1N AgNO_(3)\|\|1N KBr, AgBr(s)\|Ag` The above cell is concentration cell `:. E^(c-)._(cell)=0` `Ag^(o+)._((anode))rarr Ag^(o+)(0.1N)+e^(-)` `Ag^(o+)._((cathode))+e^(-)rarr Ag_((cathode))` `ulbar(Ag^(o+)._((cathode))rarr Ag^(o+)(0.1N))` Since `AgNO_(3)` is `81.3%` ionized `:. [Ag^(o+)]_(a)=0.1xx0.813=0.0813N` `-0.64=0-0.059log .(0.0813)/([Ag^(o+)._((cathode))])` `:. [Ag^(o+)]_((cathode))=3.2xx10^(-12)M` Since `1 N KBr` is `75.5%` dissociated `:. [Br^(c-)]=1xx0.755=0.755 M` `K_(sp)[Ag^(o+)][Br^(c-)]=3.2xx10^(-12)xx0.755` `2.416xx10^(-12)` `S=sqrt(2.416xx10^(-12))` `=1.544xx10^(-6)M`

Discussion

1294.	Construct the cells in which the following reactions are taking place. Which of the electrodes shall act as anode (negative electrode) and which one as cathode (positive electrode) ? (a) `Zn+CuSO_(4)=ZnSO_(4)+Cu` (b) `Cu+2AgNO_(3)=Cu(NO_(3))_(3)+2Ag` (c) `Zn+H_(2)SO_(4)=ZnSO_(4)+H_(2)` (d) `Fe+SnCl_(2)=FeCl_(2)+Sn`
Answer» It should always be kept in mind that the metal which goes into solution in the form of its ions undergoes oxidation and thus acts as negative electrode (anode) and the element which comes into the free state undergoes reduction and acts as positive electrode (cathode) : (a) In this case `Zn` is oxidised to `Zn^(2+)` and thus acts as anode (negative electrode) while `Cu^(2+)` is reduced to copper and thus acts as cathode (positive electrode). the cell can be represented as `Zn\|ZnSO_(4)\|\|CuSO_(4)\|Cu` or `underset("Anode (-)")(Zn\|Zn^(2+))\|\|underset("Cathode (+)")(Cu^(2+)\|Cu)` (b) In this case Cu is oxidised to `Cu^(2+)` and `Ag^(+)` is reduced to Ag. The cell can be represented as `Cu\|Cu(NO_(3))_(2)\|\|AgNO_(3)\|Ag` or `underset("Anode (-)")(Cu\|Cu^(2+))\|\|underset("Cathode (+)")(Ag^(+)\|Ag)` (c) In this case, Zn is oxidised to `Zn^(2+)` and `H^(+)` is reduced to `H_(2)`. The cell can be represented as : `Zn\|ZnSO_(4)\|\|H_(2)SO_(4)\|H_(2)(Pt)` or `underset("Anode (-)")(Zn\|Zn^(2+))\|\|underset("Cathode (+)")(2H^(+)\|H_(2) (Pt))` (d) Here, Fe is oxidised to `Fe^(2+)` and `Sn^(2+)` is reduced to Sn. the cell can be represented as : `Fe\|FeCl_(2)\|\|SnCl_(2)\|Sn` or `underset("Anode (-)")(Fe\|Fe^(2+))\|\|underset("Cathode (+)")(Sn^(2+)\|Sn)`

Discussion

1295.	Resistance of lead is greater than that of gold.1. Do you agree? 2. What do you mean by resistance? 3. What do you mean by resistivity?
Answer» 1. Yes. 2. Resistance is the hindrance offered by a substance to the flow of electricity through it. 3. Resistivity is the resistance of a conductor of length 1 cm and area of cross section 1 cm². R ∝ \(\frac{l}{A}\) Or, R = ρ \(\frac{l}{A}\) where R → Resistance and ρ → resistivity. When l = 1 cm and A = 1 cm then, R = ?

1295.

Resistance of lead is greater than that of gold.1. Do you agree? 2. What do you mean by resistance? 3. What do you mean by resistivity?

Answer»

1. Yes.

2. Resistance is the hindrance offered by a substance to the flow of electricity through it.

3. Resistivity is the resistance of a conductor of length 1 cm and area of cross section 1 cm².

R ∝ \(\frac{l}{A}\)

Or, R = ρ \(\frac{l}{A}\)

where R → Resistance and ρ → resistivity.

When l = 1 cm and A = 1 cm then, R = ?

Discussion

1296.	How many hours does it take to reduce 3 mol of `Fe^(3+)` to `Fe^(2+)` with 2.00 A current (1 Faraday=96,500 C `mol^(-1),R=8.314JK^(-1)mol^(-1)`).
Answer» Correct Answer - 40.2 hrs `Fe^(3+)+e^(-)toFe^(2+)` i.e., 1 mol requires 1F. `therefore3` moles will require=3F=`3xx96500C` `t=Q//I=(3xx96500)//2.0=144750s=40.2`hrs.

Discussion

1297.	The position of some metals in the electrochemical series in the decreasing order of electropositive character is given : Mg > Al > Zn > Cu > Ag1. What would happen if a copper spoon is used to stir a solution of aluminium?2. Is there any reaction?3. Comment on your response and justify.
Answer» 1. Nothing will happen. The Al solution can be safely stirred using a Cu spoon. 2. No. 3. The Cu spoon will not react with the solution of Al. Greater the value of reduction potential of a metal, more easily it is reduced. The element having low reduction potential is more reactive and is oxidised. Cu has greater reduction potential than Al and is more active metal than Cu.

1297.

The position of some metals in the electrochemical series in the decreasing order of electropositive character is given : Mg > Al > Zn > Cu > Ag1. What would happen if a copper spoon is used to stir a solution of aluminium?2. Is there any reaction?3. Comment on your response and justify.

Answer»

1. Nothing will happen. The Al solution can be safely stirred using a Cu spoon.

2. No.

3. The Cu spoon will not react with the solution of Al. Greater the value of reduction potential of a metal, more easily it is reduced. The element having low reduction potential is more reactive and is oxidised. Cu has greater reduction potential than Al and is more active metal than Cu.

Discussion

1298.	In the chemistry lab, a student immersed two platinum electrodes in a solution of CuSO4 and electric current was passed through the solution. After sometime he noticed that the colour of CuSO4 disappeares with the evolution of a gas at the electrode. In his Practical Log, he recorded that the colourless solution contains H2SO4 .1. Is it true? 2. Comment on his findings and justify your answer.
Answer» 1. Yes. 2. When aqueous CuSO₄ solution is electrolysed using Pt electrodes, Cu is deposited at the cathode and O₂ is liberated at the anode. CuSO₄ (aq) → Cu²⁺(aq) + SO₄^2-(aq) Cu²⁺ (aq) + 2\(\bar e\) → Cu(s) (at cathode) H₂O(l) → 2H+(aq) + 1/2 O₂(g) + 2\(\bar e\) (at anode) Thus, H₂SO₄ is formed as the secondary product.

1298.

In the chemistry lab, a student immersed two platinum electrodes in a solution of CuSO4 and electric current was passed through the solution. After sometime he noticed that the colour of CuSO4 disappeares with the evolution of a gas at the electrode. In his Practical Log, he recorded that the colourless solution contains H2SO4 .1. Is it true? 2. Comment on his findings and justify your answer.

Answer»

1. Yes.

2. When aqueous CuSO₄ solution is electrolysed using Pt electrodes, Cu is deposited at the cathode and O₂ is liberated at the anode.

CuSO₄ (aq) → Cu²⁺(aq) + SO₄^2-(aq)

Cu²⁺ (aq) + 2\(\bar e\) → Cu(s) (at cathode)

H₂O(l) → 2H+(aq) + 1/2 O₂(g) + 2\(\bar e\) (at anode)

Thus, H₂SO₄ is formed as the secondary product.

Discussion

1299.	How much electricity in Faraday is required to produce1. 20.0 g of Ca from molten CaCl2 ?2. 45.0 g of Al from molten Al2O3 ?(Hint: Atomic masses – Ca: 40 u, Al: 27 u)
Answer» 1. The electrode reaction is Ca²⁺(aq) + 2e^- → Ca(s) Electricity required to produce 40 g Ca = 2 F Electricity required to produce 20 g Ca = 1 F 2. The electrode reaction is. Al³⁺ (aq) + 3e^- → Al(s) Electricity required to produce 27 g of Al = 3 F Electricity required to produce 45 g Al = 5 F

1299.

How much electricity in Faraday is required to produce1. 20.0 g of Ca from molten CaCl2 ?2. 45.0 g of Al from molten Al2O3 ?(Hint: Atomic masses – Ca: 40 u, Al: 27 u)

Answer»

1. The electrode reaction is Ca²⁺(aq) + 2e^- → Ca(s)

Electricity required to produce 40 g Ca = 2 F

Electricity required to produce 20 g Ca = 1 F

2. The electrode reaction is.

Al³⁺ (aq) + 3e^- → Al(s)

Electricity required to produce 27 g of Al = 3 F

Electricity required to produce 45 g Al = 5 F

Discussion

1300.	Identify the false statement among the followingA. Oxidation takes place at positive electrode in electrolytic cell, whereas reduction at negative electrode in voltaic cellB. Deposition of metals takes place in both electrolytic and galvanic cellsC. Direction of flow of current is from cathode to anode in electrolytic cellsD. Both a and b
Answer» In an electrolytic cell, the direction of flow of electrons is from cathode to anode and hence the flow of current is taken in opposite direction from anode to cathode.

Discussion

Explore topic-wise InterviewSolutions in .

`1F` of electricity is passed through `10 L` of a solution of aqueous solution of `NaCl` . Calculate the `pH` of the solution.

The density of copper is `8.95 g mL^(-1)`. Find out the number of coulombs needed to plate an area of `100cm^(2)` to a thickness of `10^(-2)cm` using `CuSO_(4)` solution as electrolyte. `(` Atomic weight of `Cu=63.5g)`

In an electrolysis of AgNO3 solution, 0.7 g of Ag is deposited after a certain period of time. Calculate the quantity of electricity required in coulomb. (Molar mass of Ag is 107.9 g mol-1.)

The amount of sodium produced by passage of 2 moles of electrons in the electrolysis of fused NaCl is ______A. 23 gB. 46 gC. `11.5 g`D. `58.5g`

In a voltaic cell, if iron and silver electrodes are connected with each other then current flowA. from iron to silver outside the cellB. from silver to iron within the cellC. from silver to iron outside the cellD. current does not flow in this cell

An external voltage opposing the cell potential is applied to a Daniel cell with zinc and copper electrodes. What happen when the external emf is increased from 0.8V to 1.1 V and then 1.54 V ?

A galvanic cell has `E_(cell)^(@)=1.5" V"`. If an opposing potential of `1.5" V"` is applied to cell, what will happen to the cell reaction and current flowing through the cell ?

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Assume that during the electrolysis of `AgNO+(3)`, only `H_(2)O` is electrolyzed and `O_(2)` is formed as `2H_(2)O rarr 4H^(o+)+O_(2)+4e(-)` `O_(2)` formed at `NTP` due to passage of 2 amperes of current for 96 second isA. `0.112 L`B. `0.224L`C. `11.2L`D. `22.4L`

Given:`Pt(s)"|"underset(P_1atm)Cl_2(g)"|"Cl^(-)(C_1)"||"Cl^(-)(Cl_2)"|"underset(P_2atm)(Cl)_(2)(g)"|"Pt(s)` identify in which of following condition working of cell takes place:A. `C_1gtC_2"and"P_1=P_2`B. `P_1gtP_2"and"C_1=C_2`C.D. `P_1ltP_2"and"C_1=C_2`

How will the pH of brine (aq. NaCl solution) be affected when it is electrolysed?

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

Which one of the following will increase the voltage of the cell ? `(T = 298 K)` ` Sn+ 2Ag^+ rarr Sn^(2+) +2 Ag`.A. increase in concentration of ` Sn^(2+)`B. Increase in size of silver rodC. Increase in the concentration of `Ag^+`D. none of these

`Pt | Cl_2 (P_2) |HCl (0.M) |Pt|Cl_2 (P_2)` , cell reaction will be spontaneous if :A. ` P_1 =P_2`B. ` P_1 gt P_2`C. ` P_2 gt P_1`D. ` P_1 = P_2 =1 ` atm

The half-cell reduction potential of a hudrogen electrode at `pH= 10` will be.A. ` 0. 95 V`B. ` -0. 59 V`C. ` 0.059 V`D. ` -0 .059 `

In an electrolytic cell, one litre of a 1 M aqueous solution of `MnO_(4)^(-)` is reduced at the cathode the quantity of electricity required so that the final solution is `0.1M MnO_(4)^(2-)` will beA. 0.1FB. 1FC. 10FD. 0.01F

Consider the following cell with hydrogen electrodes at difference pressure `p_(1)` and `p_(2)`. The `EMF` of the cell is given byA. `(RT)/(F)ln.(p_(1))/(p_(2))`B. `(RT)/(2F)ln.(p_(1))/(p_(2))`C. `(RT)/(F)ln.(p_(2))/(p_(1))`D. `(RT)/(2F)ln.(p_(2))/(p_(1))`

On electrolysis, which of the following does not give out hydrogen?A. Acidic water using Pt electrodesB. Fused NaOH using Pt electrodesC. Dilute `H_(2)SO_(4)` using Pt electordesD. Dilute `H_(2)SO_(4)` using Cu electordes

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

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

For the concentration cell `Cu|Cu^(2+)(C_(1))||Cu^(2+)(C_(2))|Cu,DeltaG` will be negative if `Pt(Cl_(2)-P_(1))|HCl(0.1M)| Pt(Cl_(2)-P_(2))` the cell reaction is spontaneous ifA. `p_(1)=p_(2)`B. `p_(1)gtp_(2)`C. `p_(2)gtp_(1)`D. None of these.

The specific conductances of four electrolytes in `ohm^(-1)cm^(-1)` are given below. Which one offers higher resistance to passage of electric current?A. `7.0xx10^(-4)`B. `9.2xx10^(-10)`C. `6.0xx10^(-8)`D. `4.0xx10^(-9)`

The gas`X` at `1 atm` is bubbled through a solution containing a mixture of `1MY^(c-)` and `1MZ^(c-)` at `25^(@)C` . If the order of reduction potential is `ZgtYgtX`, thenA. `Y` will oxidize `X` onlyB. `Y` will oxidize `Z` onlyC. `Z` will oxidize `X` and `Y`D. `Z` will reduce both `X` and `Y`

`Zn | Zn^(2+)(C_(1) || Zn^(2+)(C_(2)| Zn`. For this cell `DeltaG` is negative if:A. `c_(1)=c_(2`B. `c_(1)gtc_(2)`C. `c_(2)gtc_(1)`D. None

`Zn|Zn^(2+)(c_(1)) || Zn^(2+)(c_(2))|Zn`. For this cell `DeltaG` is negative if `:`A. `C_(1) = C_(2)`B. `C_(1) gt C_(2)`C. `C_(2) gt C_(1)`D. None of these

For the reaction `:` `A+2B^(3+) rarr 2B^(2+)+A^(2+)` `E^(c-)` of the given redox reaction is `:` `A^(2+)+2e^(-) rarr A` `E^(c-)=Xv` `B^(3+)+e^(-) rarr B^(2+)` `E^(c-)=yV`A. `x-2y`B. `x+y//2`C. `x-y`D. `y-x`

`Zn|Zn^(2+)(c_(1)) || Zn^(2+)(c_(2))|Zn`. For this cell `DeltaG` is negative if `:`A. `C_(1) = C_(2)`B. `C_(1) gt C_(2)`C. `C_(2) gt C_(1)`D. None

In an electrochemical cell that function as a voltaic cell:-A. electrons move frm the cathode to the cathodeB. electrons move through a salt-bridgeC. electrons can move either form the cathode to the anode or form the anode to the cathodeD. reduction occurs at the cathode

`Cu^(+) + e rarr Cu, E^(@) = X_(1)` volt, `Cu^(2+) + 2e rarr Cu, E^(@) = X_(2)`X_(2) volt For `Cu^(2+) + e rarr Cu^(+), E^(@)` will be :A. `X_(1) - 2X_(2)`B. `X_(1) + 2X_(2)`C. `X_(1) - X_(2)`D. `2X_(2) - X_(1)`

`Cu^(+) + e rarr Cu, E^(@) = X_(1)` volt, `Cu^(2+) + 2e rarr Cu, E^(@) = X_(2)`X_(2) volt For `Cu^(2+) + e rarr Cu^(+), E^(@)` will be :A. `x_(1) -2x_(2)`B. `x_(1)+2x_(2)`C. `x_(1)-x_(2)`D. `2x_(2)-x_(1)`

At what concentration of `[overset(c-)(O)H]` does the following half reaction has a potential of `0V` when other species are at `1M ?` `NO_(3)^(c-)+H_(2)O+2e^(-) rarr NO_(2)^(c-)+2overset(c-)(O)H,E^(c-)._(cell)=-0.06V`A. `2.0M`B. `1.0M`C. `0.1M`D. `0.01M`

Consider the cell reaction `:` `Mg(s)+Cu^(2+)(aq) rarr Cu(s) +Mg^(2+)(aq)` If `E^(c-)._(Mg^(2+)|Mg(s))` and `E^(c-)._(Cu^(2+)|Cu(s))` are `-2.37` and `0.34V`, respectively. `E^(c-)._(cell)` isA. `2.03V`B. `-2.03V`C. `-2.17 V`D. `2.71 V`

If hydrogen electrodes dipped in two solutions of `pH=3` and `pH=6` are connected by a salt bridge, the `EMF_(cell)` isA. `0.052V`B. `0.104V`C. `0.177V`D. `0.3V`

Which of the following statements regarding rusting of iron is `//` are correct ?A. It takes place is moist air.B. It is stopped in `CO_(2)` atmosphereC. If produces `Fe(III)` oxide.D. It is an electrochemical process.

The `EMF` of the cell `Ag|0.1(N)AgNO_(3)||1(N) KBr, AgBr(s)|Ag` was found to be `-0.64V` at `298 K. 0.1 N Ag NO_(3)` is `81.3%` dissociated and` 1N KBr` is `75.5%` dissociated. Calculate `:` `a.` Solubility `b.` Solubility product of `AgBr ` at `298 K`

Resistance of lead is greater than that of gold.1. Do you agree? 2. What do you mean by resistance? 3. What do you mean by resistivity?

How many hours does it take to reduce 3 mol of `Fe^(3+)` to `Fe^(2+)` with 2.00 A current (1 Faraday=96,500 C `mol^(-1),R=8.314JK^(-1)mol^(-1)`).

The position of some metals in the electrochemical series in the decreasing order of electropositive character is given : Mg > Al > Zn > Cu > Ag1. What would happen if a copper spoon is used to stir a solution of aluminium?2. Is there any reaction?3. Comment on your response and justify.

How much electricity in Faraday is required to produce1. 20.0 g of Ca from molten CaCl2 ?2. 45.0 g of Al from molten Al2O3 ?(Hint: Atomic masses – Ca: 40 u, Al: 27 u)