3195 + Interview Questions in ELECTROCHEMISTRY IN GENERAL AWARENESS Page 46 InterviewSolution

2251.	At what concetration of `Ag^(2+)` ions, will the electrode have a potential of 0.0 V?
Answer» Correct Answer - `2.9xx10^(-14) M` `Ag^(+)+e^(-) to Ag`. `(E_(Ag)^(+))/(Ag) to (E_(Ag^(+))^(@))/(Ag)-(0.0591)/(n)"log"(1)/([Ag^(+)])` `0=0.80-(0.0591)/(1)"log"(1)/([Ag^(+)])=0.80+0.0591 " log "[Ag^(+)]` `[Ag^(+)]=((-)0.80)/(0.0591)=-13.5364=overset(-" ")14.4636`. `[Ag^(+)]="Antilog "log[overset(-" ")14.4636]=2.9xx10^(-14)" M "`.

Discussion

2252.	The EMF of the following cell is found to be 0.20 V at 298 K `Cd\|Cd^(2+)(?)\|\|Ni^(2+)(2.0M)\|Ni` What is the molar concentration of `Cd^(2+)` ions in the solution? `(E_(Cd^(2+)//Cd)^(@)=-0.40V,E_(Ni^(2+)//Ni)^(@)=-0.25V)`
Answer» The cell reaction is: `Cd+Ni^(2+)toCd^(2+)+Ni" "E_(Cell)^(@)=-0.25-(-0.40)==0.15V` Applying nernst equation, `E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"([Cd^(2+)])/([Ni^(2+)])` `0.20=0.15-(0.0591)/(2)"log"([Cd^(2+)])/(2)" or "log[Cd^(2+)]-log2=-1.690` or `log[Cd^(2+)]=-1.690+0.3021=-1.3879" or "[Cd^(2+)]="antilog "overline(2).6121=0.0409`M

Discussion

2253.	At what pH of HCl slutionw ill hydrogen gas electrode show electrode potential of `-0.118V`? `H_(2)` gas is bubbled at 298 K and 1 atm pressure.
Answer» Writing electrode reaction as reduction reaction, `H^(+)+e^(-)to(1)/(2)H_(2)` Applying nernst equation, `E_(H^(+)//H_(2))=E_(H^(+)//H_(2))^(@)-(0.0591)/(1)"log"(1)/([H^(+)])=0+0.0591log[H^(+)]` or `-0.118=-0.0591pH` or `pH=2` (`becausepH=-log[H^(+)])`

Discussion

2254.	Iodine `(I_(2))` and bromine `(Br_(2))` are added to a solution containing iodine and bromide `(Br^(-))` ions. What reaction would occur if the concentration of each species is 1M? The electrode potentials for the reaction are: `E_(I_(2)//I^(-))^(@)=0.54V,E_(Br_(2)//Br^(-))^(@)=1.08V`
Answer» The reaction can be either `Br_(2)+2I^(-)to2Br^(-)+I_(2)` or `I_(2)+2Br^(-)to2I^(-)+Br_(2)` For 1st reaction, E.M.F.`=E_(Br_(2)//Br^(-))^(@)-E_(I_(2)//I^(-))^(@)=1.08-0.54=0.54V` For 2nd reaction, E.M.F.`=E_(I_(2)//I^(-))^(@)-E_(Br_(2)//Br^(-))^(@)=0.54-1.08=-0.54V` As E.M.F. is positive for the 1st reaction, hence the cell reaction is `Br_(2)+2I^(-)to2Br^(-)+I_(2)`.

Discussion

2255.	At what pH of HCl solution, will hydrogen gas electrode show electrode potential of -0.118 V ? `H_(2)` gas is bubbled at 298 K and 1 atm pressure.
Answer» Correct Answer - pH=2 `H^(+)+e^(-) to 1//2H_(2)` `E_(cell)=E_(cell)^(@)-(0.0591)/(1)"log"(1)/([H^(+)])` `-0.188=0-0.0591 " pH" " "(because pH="log "1//[H^(+)])` `pH=(0.118)/(0.0591)=2`.

Discussion

2256.	Calculate the EMF of the following concentration cell at 298K `Zn\|ZnSO_(4)(0.05M)\|\|ZnSO_(4)(0.5M)\|Zn`
Answer» `E_(Cell)=(0.0591)/(n)"log"(C_(2))/(C_(1))=(0.0591)/(2)"log"(0.5)/(0.05)=0.02955V`

Discussion

2257.	A cell consists of two hydrogen electrodes. The negative electrode is in contact with a solution having `10^(-6)MH^(+)` ion concentration. Calculate the concentration of `H^(+)` ions at the positive electrode, if the emf of the cell is found to be 0.118 V at 298 K.
Answer» Here, `C_(1)=10^(-6)M," "C_(2)=`tobe calculated. For given concentration cell, `E_(cell)=(0.0591)/(n)"log"(C_(2))/(C_(1))` `0.118=(0.0591)/(1)"log"(C_(2))/(10^(-6))" or ""log"(C_(2))/(10^(-6))=2" or "(C_(2))/(10^(-6))="Antilog 2"=10^(2) " "thereforeC_(2)=10^(2)xx10^(-6)=10^(-4)M`.

Discussion

2258.	Assertion: Copper liberates hydrogen from a dilute solution of hydrochloric acid. Reason: Hydrogen is below copper in the electrochemical series.A. If both assertion and reason are true, and reason is the true explanation of the assertionB. if both assertion and reason are true, but reason is not the true explanation of the assertion.C. if assertion is true, but reason is false.D. If both assertion and reason are false.
Answer» Correct Answer - D Correct A. Copper does not liberate hydrogen from dilute hydrochloric acid solution. Correct R. Copper is below hydrogen in the electrochemical series.

Discussion

2259.	Assertion: Copper reacts with hydrochloric and liberates hydrogen from the solution of dilute hydrochloric acid. Reason: Hydrogen is below copper in the electrochemical.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.
Answer» Correct Answer - D Copper is present below hydrogen therefore hydrogen from HCl canot be liberated by treating with copper. Hence assertion and reason both false.

Discussion

2260.	Assertion: The resistivity for a substance is its resistance when its is one meter long and its area of cross section is one square meter. Reason: The SI units of resistivity are ohm metre `(Omega" "m)` and ohm centimeter `(Omega" "cm)`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.
Answer» Correct Answer - B We know, `Rprop(1)/(A)` orr `R=rho((l)/(A))`, where proportionality constant `rho` is called resistivity. If l=1m and A=`1m^(2)`, then `R=rho` i.e., resistance=Resistivity.

Discussion

2261.	A current of 9.65 A is passed through three different electrolytes NaCl, `AgNO_(3) and CuSO_(4)` for 30 min separately. Calculate the ratio of the metals deposited at the respective electrodes. Also find out the weights of various metals deposited at the respective electrodes
Answer» Quantity of electricity passed through the electrolyte = current strength `xx` time of flow = `9.65 xx 30 xx 60` coulombs `m_(Na) : m_(Ag) : m_(Cu) = E_(Na) : E_(Ag) : E_(Cu)` `E_(Na) = (23)/(1) = 23 " " (Na^(+) + e^(-) rarr Na)` `E_(Ag) = (108)/(1) = 108 (Ag^(+) + e^(-) rarr Ag)` `E_(Cu) = (63.5)/(2) 31.75 " " (Cu^(+2) + 2e^(-) rarr Cu)` `m_(Na) : m_(Ag) : m_(Cu) = 23 : 108 : 31.75` Weight of sodium deposited `= (23 xx 9.65 xx 30 xx 60)/(96500) = 4.14g` Weight of silver deposite `= (108 xx 9.65 xx 30 xx 60)/(96500) = 19.44 g` Weight of copper deposited `= (31.75 xx 9.65 xx 30 xx 60)/(96500) = 5.715g`

Discussion

2262.	Depict the galvanic cell in which the reaction Zn(s) + 2Ag+(aq) → Zn2+ + 2Ag(s) takes place. Further show :1. Which of the electrodes is negatively charged?2. The carriers of current in the cell.3. Individual reaction at each electrode.
Answer» The electrochemical cell which will involve the given cell reaction is depicted as Zn(s) \| Zn²⁺(aq) \|\| Ag⁺(aq) \| Ag(s). 1. In this cell, the electrode Zn(s) \| Zn²⁺ (aq) will act as negative terminal and Ag \| Ag⁺(aq) electrode will act as positive terminal. 2. The conventional current will flow from silver to zinc electrode in the external circuit. 3. Individual reaction at each electrode. Cathodic reaction : 2Ag⁺(aq) + 2e^- → 2Ag(s) Anodic reaction : Zn(s) → Zn²⁺(aq) + 2e^-

2262.

Depict the galvanic cell in which the reaction Zn(s) + 2Ag+(aq) → Zn2+ + 2Ag(s) takes place. Further show :1. Which of the electrodes is negatively charged?2. The carriers of current in the cell.3. Individual reaction at each electrode.

Answer»

The electrochemical cell which will involve the given cell reaction is depicted as Zn(s) | Zn²⁺(aq) || Ag⁺(aq) | Ag(s).

1. In this cell, the electrode Zn(s) | Zn²⁺ (aq) will act as negative terminal and Ag | Ag⁺(aq) electrode will act as positive terminal.

2. The conventional current will flow from silver to zinc electrode in the external circuit.

3. Individual reaction at each electrode.

Cathodic reaction : 2Ag⁺(aq) + 2e^- → 2Ag(s)

Anodic reaction : Zn(s) → Zn²⁺(aq) + 2e^-

Discussion

2263.	∧0 for CH3COOH is 390.7 Ω-1 cm2 mol-1. If ∧0 for CH3COOK, and HBr in Ω-1 cm2 mol-1 are 115 and 430.4 respectively, then ∧0 for KBr is :(a) 74.6 Ω-1 cm2 mol-1(b) 180.6 Ω-1 cm2 mol-1(c) 154.7 Ω-1 cm2 mol-1(d) 706.1 Ω-1 cm2 mol-1
Answer» Option : (c) 154.7 Ω^-1 cm² mol^-1

Discussion

2264.	Formulate the galvanic cell in which the following reaction takes place: `Zn(s)+2Ag^(+)(aq)toZn^(2+)(aq)+2Ag(s)` State (i) Which one of the electrodes is negatively charged? (ii) The reaction taking place at each of its electrode. ltBrgt (iii) The carriers of current within this cell.
Answer» The cell reaction is : `Zn(s)+2Ag^(+)(aq)toZn^(2+)(aq)+2Ag(s)` The cell is represented as: `Az(s)\|Zn^(2+)(aq)\|\|Ag^(+)(aq)+Ag(s)` (i) Anode i.e., zinc electrode is negatively charged. (ii) At anode: `Zn(s)toZn^(2+)(aq)+2e^(-)` At cathode: `Ag^(+)(aq)+e^(-)toAg(s)` (iii) Within the cell, the current is carried by the cations and the anions through the salt bridge. In the external circuit, electrons flow from zinc to silver and hence conventional current flows from silver to zinc.

Discussion

2265.	Why is it necessary to use a salt bridge in a Gavanic cell?
Answer» To complete the inner circuit and to maintain electrical neutrality of the electrolytic solutions of the half cells.

Discussion

2266.	What would happen if no sal bridge were used in an electrochemical cell (like `Zn-Cu` cell)?
Answer» The metal `(Zn^(2+))` formed by the loss of electrons will accumulate in one electrode and the negative ions `(SO_(4)^(2-))` will accumulate in the other. Thus, the solutions will develop charges and the current stops flowing. Moreover, inner circuit is not completed.

Discussion

2267.	Write expressions for equivalent conductivity and molar conductivity of `Al_(2)(SO_(4))_(3)` at infinite dilutio in terms of their ionic conductivitiies.
Answer» `wedge_(eq)(Al_(2)SO_(4))=lamda_(Al^(3+))^(@)+lamda_(SO_(4)^(2-))^(@),wedge_(m)(Al_(2)SO_(4))_(3))=2lamda_(Al^(3+))^(@)+3lamda_(SO_(4)^(2-))^(@)`

Discussion

2268.	The charge required for the reduction of 1 mol of `MnO_(4)^(-)` to `MnO_(2)` isA. 1FB. 3FC. 5FD. 6F
Answer» Correct Answer - B `overset(+7)(MnO_(4)^(-)rarroverset(+4)(MnO_(2))` `i.e. MnO_(4)^(-)+3e^(-)rarrMnO_(2)` `MnO_(4)^(-)+2H_(2)O+3e^(-) rarrMnO_(2)+4OH^(-)` Thus for reduction of 1 mol of `MnO_(4)^(-)` to `MnO_(2)` charge required =3F

Discussion

2269.	The products formed when an aqueous solution of `NaBr` is electrolysed in a cell having inert electrodes are `:`A. Na and `Br_(2)`B. Na and `O_(2)`C. `H_(2),Br_(2)` and NaOHD. `H_(2)` and `O_(2)`
Answer» Correct Answer - C `NaBr(aq)hArrNa^(+)(aq)+Br^(-)(aq)` `Na^(+)+e^(-)rarrNa` `2H_(2)O+ 2e^(-)rarrH_(2)=2OH^(-)` As R.P. of 2nd reactions is more `H_(2)` is produced at cathode `2Br^(-)rarrBr_(2)+2e^(-)` `2H_(2)OrarrO_(2)+4 H^(+)+4e^(-)` As O.P. of 1st reaction is more, `Br_(2)` is produced at anode `Na^(+)` from ionisation of NaBr and `OH^(-)` ions produced due to reduction of water at cathode combine to give NaOH.

Discussion

2270.	The products formed when an aqueous solution of `NaBr` is electrolysed in a cell having inert electrodes are `:`
Answer» Correct Answer - C At cathode `H^(+)` reduces to `H_(2)` and at anode `Br^(-)` oxidises to `Br_(2)`, leaving NaOH in the solution.

Discussion

2271.	Molar conductivity of ionic solution depends on ___________.(i) temperature.(ii) distance between electrodes.(iii) concentration of electrolytes in solution.(iv) surface area of electrodes.
Answer» (i) temperature. (iii) concentration of electrolytes in solution.

Discussion

2272.	Conductivity κ , is equal to ____________.(i) 1l/RA(ii) G*/R(iii) Λm(iv) 1/A
Answer» (i) 1l/RA (ii) G^*/R

2272.

Conductivity κ , is equal to ____________.(i) 1l/RA(ii) G*/R(iii) Λm(iv) 1/A

Answer»

(i) 1l/RA

(ii) G^*/R

Discussion

2273.	What will happen during the electrolysis of aqueous solution of CuSO4 in the presence of Cu electrodes?(i) Copper will deposit at cathode.(ii) Copper will dissolve at anode.(iii) Oxygen will be released at anode.(iv) Copper will deposit at anode.
Answer» (i) Copper will deposit at cathode. (ii) Copper will dissolve at anode.

Discussion

2274.	Aqueous solution of which electrolyte produces H2 gas at cathode, when electrolysed among inert electrodes? (1) NaCl (2) MgCl2 (3) CuCl2 (4) AgCl
Answer» Correct option (1, 2) Explanation: H⁺ has lower discharge potential as compared to Na⁺ and Mg²⁺ Hence, in case of NaCl and MgCl₂ reaction is 2H⁺ + 2e^– →H₂.

2274.

Aqueous solution of which electrolyte produces H2 gas at cathode, when electrolysed among inert electrodes? (1) NaCl (2) MgCl2 (3) CuCl2 (4) AgCl

Answer»

Correct option (1, 2)

Explanation:

H⁺ has lower discharge potential as compared to Na⁺ and Mg²⁺

Hence, in case of NaCl and MgCl₂ reaction is 2H⁺ + 2e^– →H₂.

Discussion

2275.	Which is/are correct statements about salt bridge? (1) Velocity of ions of salt bridge are almost equal (2) Salt bridge completes the electric circuit (3) Ions of salt bridge discharge at electrode (4) Ions of salt bridge do not discharge at electrode
Answer» Correct option (1, 2, 4) Explanation: Salt Bridge contains electrolyte which do not participate in the electrochemical change, completes the cell circuit and it is also necessary that velocity of ions of salt bridge are almost equal.

2275.

Which is/are correct statements about salt bridge? (1) Velocity of ions of salt bridge are almost equal (2) Salt bridge completes the electric circuit (3) Ions of salt bridge discharge at electrode (4) Ions of salt bridge do not discharge at electrode

Answer»

Correct option (1, 2, 4)

Explanation:

Salt Bridge contains electrolyte which do not participate in the electrochemical change, completes the cell circuit and it is also necessary that velocity of ions of salt bridge are almost equal.

Discussion

2276.	Which statement is/are correct? (1) In voltaic cell electrons flow from anode to cathode (2) In voltaic cell, anode is negative electrode and cathode is positive electrode (3) Oxidation take place at anode and reduction take place at cathode in electrochemical cell (4) In electrolytic cell oxidation take place at cathode and reduction take place at anode
Answer» Correct option (1, 2, 3) Explanation: At cathode always reduction takes place and at anode always oxidation takes place. Hence (4) will not the correct statement.

2276.

Which statement is/are correct? (1) In voltaic cell electrons flow from anode to cathode (2) In voltaic cell, anode is negative electrode and cathode is positive electrode (3) Oxidation take place at anode and reduction take place at cathode in electrochemical cell (4) In electrolytic cell oxidation take place at cathode and reduction take place at anode

Answer»

Correct option (1, 2, 3)

Explanation:

At cathode always reduction takes place and at anode always oxidation takes place. Hence (4) will not the correct statement.

Discussion

2277.	Which is/are correct about corrosion? (1) Due to corrosion FeO.xH2O formed (2) Due to corrosion Fe2O3.xH2O formed (3) Presence of air and moisture increases the rate of corrosion (4) Magnesium is used as sacrificial anode
Answer» Correct option (2, 3, 4) Explanation: In corrosion Fe →Fe²⁺ + 2e is formed and the formation of oxide i.e., Fe₂O₃.xH₂O takes place and presence of air and moisture is must. Mg can be used as sacrificial anode.

2277.

Which is/are correct about corrosion? (1) Due to corrosion FeO.xH2O formed (2) Due to corrosion Fe2O3.xH2O formed (3) Presence of air and moisture increases the rate of corrosion (4) Magnesium is used as sacrificial anode

Answer»

Correct option (2, 3, 4)

Explanation:

In corrosion

Fe →Fe²⁺ + 2e is formed and the formation of oxide i.e., Fe₂O₃.xH₂O takes place and presence of air and moisture is must.

Mg can be used as sacrificial anode.

Discussion

2278.	Electrode potential for `Mg` electrode varies according to the equation `E_(Mg^(2+)\|Mg)=E_(Mg^(2+)\|Mg)^(ϴ) -(0.059)/2 "log" 1/([Mg^(2+)])` The graph of `E_(Mg^(2+)\|Mg) vs log [Mg^(2+)]` isA. B. C. D.
Answer» Correct Answer - B `E=E^@+0.059/2"log"[Mg^(2+)]` hence, plot of E vs log `[Mg^(2+)]` will be linear with positive slope and intercept =`E^@`

Discussion

2279.	The correct order of `E_(M^(2+)//M)^(@)` Values with negative sign for the four successive elements `Cr, Mn, Fe` and `Co` is:A. `Fe gt Mn gt Cr gt Co`B. `Cr gt Mn gt Fe gt Co`C. `Mn gt Cr gt Fe gt Co`D. `Cr gt Fe gt Mn gt Co`
Answer» Correct Answer - C `E^(@)` values of for `M^(2+)\|M` with negative sign are `Cr = -0.9 V ,Mn = -1.18 V , Fe = -0.44 V, ` Co = -28 V Thus order is `Mn gt Cr gt Fe gt Co`

Discussion

2280.	The standard e.m.f. for the cell reaction. `2Cu^(+)(aq)rarrCu(s)+Cu^(2+)(aq)` is +0.36 V at 298K. The equilibrium constant of the reaction isA. `5xx10^(6)`B. `1.4xx10^(12)`C. `7.4xx10^(12)`D. `1.3xx10^(6)`
Answer» Correct Answer - D Cell reactions are `Cu^(+)rarrCu^(2+)+e^(-)` `Cu^(2+)+e^(-)rarrCu` here `n=1,E_("cell")^(@)=+0.36V` `E_("Cell")^(@)=(0.0591V)/(n)"log" k_(c)` `0.36=(0.0591)/(1)"log"k_(c)` `"log" k_(c)=(0.36)/(0.591)=(36)/(5.91)= 6.09` `k_(c)=1.3xx10^(6)`

Discussion

2281.	Standard reduction potentials for the half reactions are given below : `F_(2)(g)+2e^(-) to 2F^(-)(aq) , E^(@)=+2.85" V "` `Cl_(2)(g)+2e^(-) to 2Cl^(-)(aq) , E^(@)=+1.36" V "` `Br_(2)(g)+2e^(-) to 2Br^(-)(aq) , E^(@)=+1.06" V "` `I_(2)(g)+2e^(-) to 2I^(-)(aq) , E^(@)=+0.53" V "` The strongest oxidising and reducing agents respectivity are :A. `F_(2)` and `I^(-)`B. `Br_(2)` and `Cl^(-)`C. `Cl_(2)` and `Br^(-)`D. `Cl_(2)` and `I_(2)`.
Answer» Correct Answer - A (a) More negative (or less positive) the value of reduction potential, stranger will be the reducing agent. Thus `l^(-)` is the strongest reducing agent. More positive is the values of reduction potential, stronger will be the oxidising agent. Thus, `F_(2)` is the strongest oxidising agent.

Discussion

2282.	Which of the following expressing is correctA. `DeltaG^(@)=-nFE_(cell)^(@)`B. `DeltaG^(@)=+nFE_(cell)^(@)`C. `DeltaG^(@)=-2.303RTnFE_(cell)^(@)`D. `DeltaG^(@)=-nF" log "K_(C)`
Answer» Correct Answer - A `DeltaG^(@)=-2.303RTlogK_(eq)` or `DeltaG^(o)=-nFE_(cell)^(o)`

Discussion

2283.	The standard reduction potential for `Zn^(2+)//Zn, Ni^(2+)//Ni` and `Fe^(2+)//Fe` are `-0.76, -0.23` and `0.44V` respectively. The reaction `X + Y^(2) rarr X^(2+) + Y` will be spontaneous when:A. `X = Ni, Y = Zn`B. `X = Fe, Y = Zn`C. `X = Zn, Y = Ni`D. `X = Ni, Y = Fe`
Answer» Correct Answer - C The elements with high negative value of standard reduction potential are good reducing agent and can be easily oxidised. ltbRgt Thus `X` should have high negative value of standard potential then `Y`, so that it will be oxidised to `X^(2+)` by reducing `Y^(2+)` to `Y`. `X = Zn, Y = Ni` `Zn + Ni^(2+) rarr Zn^(2+) + Ni` Alternatively, for a spontaneous reaction `E^(@)` must be positive. `E^(@) = E_("reduced")^(@) - E_("oxidised")^(@)` `= -0.23 - (0.76)` implies `E^(@) = +0.53 V`

Discussion

2284.	The standard reduction potentials for `Zn^(2+)//Zn,Ni^(2+) and Fe^(2+)//Fe` are -0.76,-0.23 and -0.44V respectively. The reaction `X+Y^(2+)toX^(2+)+Y` will be spontaneous whenA. `X=Ni,Y=Fe`B. `X=Ni,Y=Zn`C. `X=Fe,Y=Zn`D. `X=Zn,Y=Ni`
Answer» Correct Answer - D `X+Y^(2+)toX^(2+)+Y` for reaction to be spontaneous `E^(o)` must be positive `underset(("Positive"))(E_(zn//zn^(+2))^(o)+E_(Ni^(2+)//Ni)^(o)=0.76+(-0.23)=+0.53)`

Discussion

2285.	Standard reduction potentails of the half reactions are given below: `F_(2)(g)+2e^(-) rarr 2F^(-)(aq.),, E^(ɵ)= +2.87` `Cl_(2)(g)+2e^(-) rarr 2Cl^(-)(aq.),, E^(ɵ)= +1.36 V` `Br_(2)(g)+2e^(-) rarr 2Br^(-)(aq.),, E^(ɵ)= +1.09 V` `I_(2)(s)+2e^(-) rarr 2l^(-)(aq.),, E^(ɵ)= +0.54 V` The strongest oxidizing and reducing agents respectively are:A. `Cl_(2)` and `Br^(-)`B. `Cl_(2)` and `I_(2)`C. `F_(2)` and `I^(-)`D. `Br_(2)` and `Cl^(-)`
Answer» Correct Answer - C Higher the reduction potential more easily it is reduced and hence stronger is the oxidising agent. `F_(2)` has highest reduction potential. `therefore F_(2)` is the strongest oxidising agent. Considering the reverse reaction (oxidiation reactions). higher the oxidiation potential more easily it is oxidised and hence stronger is the reducing agent. Oxidation potential `1(-0.53V)` is highest out of 2.85,-1.36,-1.36 and 0. 53, Hence I is the strongest reducing agent.

Discussion

2286.	Which electrolyte is used in dry cell ?A. Connecting iron to more electropositive metal cathodic protectionB. Commecting iron to more electropositive metal anodic protectionC. Connecting iron to less electropositive metal anodic protectionD. Connecting iron to less electropositive metal cahtodic protection
Answer» Correct Answer - B `MnO_2` in Leclanche cell.

Discussion

2287.	The standard reduction potential for `Zn^(2+)//Zn, Ni^(2+)//Ni` and `Fe^(2+)//Fe` are `-0.76, -0.23` and `-0.44V` respectively. The reaction `X + Y^(2) rarr X^(2+) + Y` will be spontaneous when:A. `x=Zn,y=Ni`B. `x=Ni,y=Fe`C. `x=Mn,y=Zn`D. `x=Fe,y=Zn`.
Answer» Correct Answer - A The reaction `x+y^(2+)rarrx^(2+)+y` will be spontaneous if `E_("cell")^(@)` is +ve (A) `Zn+Ni^(2+)rarrZn^(2+)+Ni` `E_("cell")^(@)=E_(cu^(2+)//Ni)^(@)=E_(Zn^(+2)//Zn)^(@)` `=-0.23-(-0.76)V=0.53V` (B) `Ni+Fe^(2+)rarrNi^(2+)+Fe` `E_("cell")^(@)=E_(Fe^(2+)..Fe)^(@)-E_(Ni^(2+)//Ni)^(@)=-0.44-(-.23)V` =-0.21V (C) `Ni+Zn^(2+)rarrNi^(2+)+Zn` `E_("cell")^(@)+Zn ^(2+)rarrNi^(2+)+Zn` `E_("cell")^(@)=0.53V` from (A) (D) `Fe+Zn^(2+)rarrFe^(2+)+Zn` `E_("cell")^(@)=E_(Zn^(2+)//Zn)^(@)-E_(Fe^(2+)//Fe)^(@)` `=-0.76-(-0.44)=-0.32V`

Discussion

2288.	Give reason: (i) Rusting of iron pipe can be prevented by joining it with a piece of magnesium. (ii) Conductivity of an electrolyte solution decreases with the decrease in concentration.
Answer» (i) it is due to cathodic protection in which magnesium metal is oxidized in preference to iron and acts as the anode. (ii) It is due to the fact that on dilution, number of ions per unit volume decreases.

Discussion

2289.	Metals can be prevented from rusting by .A. They are more efficientB. The are free from pollutionC. They rum till reactanis are activeD. All of these
Answer» Correct Answer - A The process is called cathodic protection where iron acts as cathode and thus not oxidised.

Discussion

2290.	Iron can be prevented from rusting byA. Connecting iron to more electropositive metal `-` a case of cathodic protectionB. Connecting iron to more electropositive metal`-` a case of anodic protection.C. Connecting iron to less electropositive metal `-` a case of anodic protectionD. Connecting iron to less electropositive metal`-` a case of cathodic protection.
Answer» Correct Answer - a,c Cathodic protection `:` A techniquie to control corrosion of a metal surface by making it work as a cathode of an electrochemical cell by placing in contact with a the metal to be protected another more easily corroded matal to act as the anode of the electrochemical cell. Most commonly used to protect steel, water pipelines, and storage tanks. Anodic protection `:` A technique to control corrosion of a matal by making it work as anode developing a passive film on the metal. it used in extremely corrosive conditions and most extensively used to store and hangle sulphuric acid container.

Discussion

2291.	`E_(1),E_(2) and E_(3)` the emf values of the three galvanic cells respectively. (i) `Zn\|Zn_(0.1M)^(2+)\|\|Cu_(0.1M)^(2+)\|Cu`, (ii) `Zn\|Zn_(0.1M)^(2+)\|\|Cu_(0.1M)^(2+)\|Cu` ltBrgt (iii) `Zn\|Zn_(0.1M)^(2+)\|\|Cu_(0.1M)^(2+)\|Cu` Which one of the following is trueA. `E_(2) gt E_(3) gt E_(1)`B. `E_(3) gt E_(2) gt E_(1)`C. `E_(1) gt E_(2) gt E_(3)`D. `E_(1) gt E_(3) gt E_(2)`
Answer» Correct Answer - B E.m.f. of the cell`=E_(cell)^(o)-(0.0591)/(n)log_(10)[(Zn^(2+))/(Cu^(2+))]` Higher the `([Zn^(2+)])/([Cu^(2+)])` ratio lower is the e.m.f. of cell. (i) `"log"[(Zn^(2+)])/([Cu^(2+)])=10," "(ii)" log"([Zn^(2+)])/([Cu^(2+)])=1` (iii) `"log"([Zn^(2+)])/([Cu^(2+)])=0.1` So is `E_(3) gt E_(2) gt E_(1)`.

Discussion

2292.	Define emf. Calculat the emf of the following galvanic cell : `Zn_((s))+Cu_((aq))^(+2)to Zn_((aq))^(+2)+Cu_((s))` `E_(zn^(+2//Zn))^(0) =0.76V(anode) ,E_(cu^(+3//Cu))^(0)=+0.34` (Cathode)
Answer» EMF : The difference in electrode potentials between radiction potential of cathode and reduction potential of anode is called emf. emf = reduction potential of cathode - reduction potential of anode `emf = + 0.34 -(-0.76) implies+ 0.34 +0.76 =1.1V`

Discussion

2293.	Given `E_("Cr_(2)O_(7)^(2-)//Cr^(3+))^(@)=1.33V,E_(MnO_(4)^(-)//Mn^(2+))^(@)=1.51V` Among the following, the strongest reducing agent is `E_(Cr^(3+)//Cr)^(@)=-0.74V^(x),E_(MnO_(4)^(-)//Mn^(2+))^(@)=1.51V` `E_(Cr_(2)O_(7)^(2-)//Cr^(3+))^(@)=1.33V,E_(Cl//Cl^(-))^(@)=1.36V` Based on the data given above strongest oxidising agent will beA. `MnO_(4)^(-)`B. `Cl^(-)`C. `Cr^(2+)`D. `Mn^(2+)`
Answer» Correct Answer - A In the given reaction, i has been oxidised to 1 and `Cr_(2)O_(7)^(2-)` has been reduced to `Cr^(3+)` `therefore E_("cell")^(@)=E_(Cr_(2)O_(7)^(2-))^(@)-E_(I_(2))^(@)` `therefore 0.79=1.33-E_(I_(2))^(@)` or `E_(I_(2))^(@)=1.33-0.79=0.54`

Discussion

2294.	For a Ag-Zn button cell, the net reaction is `Zn (s)+Ag_(2)O(s)rarrZnO(s)+2Ag(s)` `DeltaG_(f)^(@)(Ag_(2)O)=-11.21kJ "mol"^(-1)` and `DeltaG_(f)^(@)(ZnO)=-318.3kJ "mol"^(-1)` The `E_("cell")^ (@)` of this button cell isA. 1.71VB. 1.591VC. 3.182VD. 3.07V
Answer» Correct Answer - b `DeltaG_("reaction")^(@)=DeltaG_(f)^(@)(ZnO)-DeltaG^(@)(Ag_(2)O)` `=-3.18.3+11.21` `=-30.09xx10^(2)J` `DeltaG^(@)= -nE^(@)F` `therefore E^(@)=(-DeltaG^(@))/(nF)` Here n=2 `therefore E^(@)=(+3.07.09xx10^(3))/(2xx96500)=1.591V`

Discussion

2295.	Local action in an electrochemical action can be prevented byA. using by pure electrolytes in tw half cellsB. using very pure metal for anodeC. coating zinc anode with mercuryD. using pure graphite for cathode.
Answer» Correct Answer - B,C Local action leads to the eating away of the metal used as anode in an electrochemical cell. It is due to the presence of impurities in the metal used for making anode. These impurities help in setting up of tiny electrochemical reaction cells on the surface of anode which results in the eating away of the metal . it can be prevented by using very pure metals for anode or by coaing the zinc anode with mercury.

Discussion

2296.	`E^(o)` for the cell `Zn\|Zn_((aq))^(2+)\|\|Cu_((aq))^(2+)\|Cu` is 1.10 V at `25^(@)C`, the equilibrium constant for the reaction `Zn+Cu_((aq))^(2+)hArrCu+Zn_((aq))^(2+)` is of the order ofA. `10^(-28)`B. `10^(-37)`C. `10^(+18)`D. `10^(+17)`
Answer» Correct Answer - B `E_(cell)^(@)=(0.059)/(n)log" "K` `log" "K=(1.10xx2)/(0.059)=37.2881` or `K=10^(-37)`

Discussion

2297.	Calculate `K_(c)` for the reaction at 298 K `Zn_((s))+Cu_((aq))^(+2)hArrZn _((aq))^(2+)+Cu_((s))` `E _(Zn^(2+)//Zn)^(0)=-0.76V, E_(Cu^(2+)//Cu)^(0) =+0.34V.`
Answer» Given `Zn_((s))+Cu_((aq))^(+2)hArrZn _((aq))^(2+)+Cu_((s))` ` E^(0)of Zn ^(+2)//Zn=-0.76V` `E^(0)of Cu^(+2) //Cu =-0.34V` `E^(0)of cell =E_(RHS)-E_(LHS ) =0.34=(-0.76)=1.1V` `DeltaG^(0) =-RT//ln K_(c)=-2.303RT log K_(c)` `-212300=-2.303xx8.314xx298xxlog K _(c)` `log K_(c) =(212300)/(2.303xx8.314xx298)=37.207` `K_(c) =1.6xx10^(37)`

Discussion

2298.	The standard cell potential of `Zn\|Zn_((aq))^(2+)\|\|Cu_((aq))^(2+)\|Cu` cell is 1.10V. The maximum work obtained by this cell will beA. 106.15 kJB. `-212.30kJ`C. `-318.45kJ`D. `-424.60kJ`
Answer» Correct Answer - B

Discussion

2299.	The standard emf of Deniell cell is `1.1` V. Calculate the standard Gibbs energy for the cell reactions: `Zn_((s))+Cu_((aq))^(2+)to Zn _((aq))^(2+)+Cu_((s))`
Answer» `Delta_(r) G^(Θ)=-nFE_((cell))^(Θ)` nin the above equation is `2,F =96487 C mol ^(-1)and E_((cell))^(Θ)=1.1V` Therefore `Delta_(r)G^(Θ)=-2xx1.1V xx96487C mol ^(-1)` `=-21227J mol ^(-1) =-212.27kJ mol ^(-1)`

Discussion

2300.	A fuel cell is supplied 1 mole of `H_(2)` gas and 10 moles `O_(2)`gas. If the fuel cell is operated at 96.5 mA currecnt, how long will it deliver power?
Answer» In `H_(2)-O_(2)` fuel cell, the net reaction is : ` underset((2" mol"))(2H_(2)(g))+underset((1 " mol"))(O_(2)(g)) to 2H_(2)O(l)` In the present case, the cell is supplied with 1 mole of `H_(2)`gas and 10 moles of `O_(2)`gas . This means that `H_(2)`(g) is the limiting reactant. Thus, 1 mole of `H_(2)`reacts with 0.5 mole of `O_(2)`. `H_(2)(g)+(1)/(2)O_(2)(g) to H_(2)O(l),n=2e^(-)` Now, `Q=nF=2 molxx(96500 C mol^(-1))=2xx(96500 AS)` Also `Q=Ixxt=96.5" mA "=(96.5xx10^(-3)" A")xxt` upon equating , `t=( 2xx(96500" AS"))/((96.5xx10^(-3)" A"))=2xx10^(6)s`

Discussion

Explore topic-wise InterviewSolutions in .

At what concetration of `Ag^(2+)` ions, will the electrode have a potential of 0.0 V?

The EMF of the following cell is found to be 0.20 V at 298 K `Cd|Cd^(2+)(?)||Ni^(2+)(2.0M)|Ni` What is the molar concentration of `Cd^(2+)` ions in the solution? `(E_(Cd^(2+)//Cd)^(@)=-0.40V,E_(Ni^(2+)//Ni)^(@)=-0.25V)`

At what pH of HCl slutionw ill hydrogen gas electrode show electrode potential of `-0.118V`? `H_(2)` gas is bubbled at 298 K and 1 atm pressure.

Iodine `(I_(2))` and bromine `(Br_(2))` are added to a solution containing iodine and bromide `(Br^(-))` ions. What reaction would occur if the concentration of each species is 1M? The electrode potentials for the reaction are: `E_(I_(2)//I^(-))^(@)=0.54V,E_(Br_(2)//Br^(-))^(@)=1.08V`

At what pH of HCl solution, will hydrogen gas electrode show electrode potential of -0.118 V ? `H_(2)` gas is bubbled at 298 K and 1 atm pressure.

Calculate the EMF of the following concentration cell at 298K `Zn|ZnSO_(4)(0.05M)||ZnSO_(4)(0.5M)|Zn`

A cell consists of two hydrogen electrodes. The negative electrode is in contact with a solution having `10^(-6)MH^(+)` ion concentration. Calculate the concentration of `H^(+)` ions at the positive electrode, if the emf of the cell is found to be 0.118 V at 298 K.

A current of 9.65 A is passed through three different electrolytes NaCl, `AgNO_(3) and CuSO_(4)` for 30 min separately. Calculate the ratio of the metals deposited at the respective electrodes. Also find out the weights of various metals deposited at the respective electrodes

Depict the galvanic cell in which the reaction Zn(s) + 2Ag+(aq) → Zn2+ + 2Ag(s) takes place. Further show :1. Which of the electrodes is negatively charged?2. The carriers of current in the cell.3. Individual reaction at each electrode.

∧0 for CH3COOH is 390.7 Ω-1 cm2 mol-1. If ∧0 for CH3COOK, and HBr in Ω-1 cm2 mol-1 are 115 and 430.4 respectively, then ∧0 for KBr is :(a) 74.6 Ω-1 cm2 mol-1(b) 180.6 Ω-1 cm2 mol-1(c) 154.7 Ω-1 cm2 mol-1(d) 706.1 Ω-1 cm2 mol-1

Formulate the galvanic cell in which the following reaction takes place: `Zn(s)+2Ag^(+)(aq)toZn^(2+)(aq)+2Ag(s)` State (i) Which one of the electrodes is negatively charged? (ii) The reaction taking place at each of its electrode. ltBrgt (iii) The carriers of current within this cell.

Why is it necessary to use a salt bridge in a Gavanic cell?

What would happen if no sal bridge were used in an electrochemical cell (like `Zn-Cu` cell)?

Write expressions for equivalent conductivity and molar conductivity of `Al_(2)(SO_(4))_(3)` at infinite dilutio in terms of their ionic conductivitiies.

The charge required for the reduction of 1 mol of `MnO_(4)^(-)` to `MnO_(2)` isA. 1FB. 3FC. 5FD. 6F

The products formed when an aqueous solution of `NaBr` is electrolysed in a cell having inert electrodes are `:`A. Na and `Br_(2)`B. Na and `O_(2)`C. `H_(2),Br_(2)` and NaOHD. `H_(2)` and `O_(2)`

The products formed when an aqueous solution of `NaBr` is electrolysed in a cell having inert electrodes are `:`

Molar conductivity of ionic solution depends on ___________.(i) temperature.(ii) distance between electrodes.(iii) concentration of electrolytes in solution.(iv) surface area of electrodes.

Conductivity κ , is equal to ____________.(i) 1l/RA(ii) G*/R(iii) Λm(iv) 1/A

What will happen during the electrolysis of aqueous solution of CuSO4 in the presence of Cu electrodes?(i) Copper will deposit at cathode.(ii) Copper will dissolve at anode.(iii) Oxygen will be released at anode.(iv) Copper will deposit at anode.

Aqueous solution of which electrolyte produces H2 gas at cathode, when electrolysed among inert electrodes? (1) NaCl (2) MgCl2 (3) CuCl2 (4) AgCl

Which is/are correct statements about salt bridge? (1) Velocity of ions of salt bridge are almost equal (2) Salt bridge completes the electric circuit (3) Ions of salt bridge discharge at electrode (4) Ions of salt bridge do not discharge at electrode

Which is/are correct about corrosion? (1) Due to corrosion FeO.xH2O formed (2) Due to corrosion Fe2O3.xH2O formed (3) Presence of air and moisture increases the rate of corrosion (4) Magnesium is used as sacrificial anode

Electrode potential for `Mg` electrode varies according to the equation `E_(Mg^(2+)|Mg)=E_(Mg^(2+)|Mg)^(ϴ) -(0.059)/2 "log" 1/([Mg^(2+)])` The graph of `E_(Mg^(2+)|Mg) vs log [Mg^(2+)]` isA. B. C. D.

The correct order of `E_(M^(2+)//M)^(@)` Values with negative sign for the four successive elements `Cr, Mn, Fe` and `Co` is:A. `Fe gt Mn gt Cr gt Co`B. `Cr gt Mn gt Fe gt Co`C. `Mn gt Cr gt Fe gt Co`D. `Cr gt Fe gt Mn gt Co`

The standard e.m.f. for the cell reaction. `2Cu^(+)(aq)rarrCu(s)+Cu^(2+)(aq)` is +0.36 V at 298K. The equilibrium constant of the reaction isA. `5xx10^(6)`B. `1.4xx10^(12)`C. `7.4xx10^(12)`D. `1.3xx10^(6)`

Which of the following expressing is correctA. `DeltaG^(@)=-nFE_(cell)^(@)`B. `DeltaG^(@)=+nFE_(cell)^(@)`C. `DeltaG^(@)=-2.303RTnFE_(cell)^(@)`D. `DeltaG^(@)=-nF" log "K_(C)`

The standard reduction potential for `Zn^(2+)//Zn, Ni^(2+)//Ni` and `Fe^(2+)//Fe` are `-0.76, -0.23` and `0.44V` respectively. The reaction `X + Y^(2) rarr X^(2+) + Y` will be spontaneous when:A. `X = Ni, Y = Zn`B. `X = Fe, Y = Zn`C. `X = Zn, Y = Ni`D. `X = Ni, Y = Fe`

The standard reduction potentials for `Zn^(2+)//Zn,Ni^(2+) and Fe^(2+)//Fe` are -0.76,-0.23 and -0.44V respectively. The reaction `X+Y^(2+)toX^(2+)+Y` will be spontaneous whenA. `X=Ni,Y=Fe`B. `X=Ni,Y=Zn`C. `X=Fe,Y=Zn`D. `X=Zn,Y=Ni`

The standard reduction potential for `Zn^(2+)//Zn, Ni^(2+)//Ni` and `Fe^(2+)//Fe` are `-0.76, -0.23` and `-0.44V` respectively. The reaction `X + Y^(2) rarr X^(2+) + Y` will be spontaneous when:A. `x=Zn,y=Ni`B. `x=Ni,y=Fe`C. `x=Mn,y=Zn`D. `x=Fe,y=Zn`.

Give reason: (i) Rusting of iron pipe can be prevented by joining it with a piece of magnesium. (ii) Conductivity of an electrolyte solution decreases with the decrease in concentration.

Metals can be prevented from rusting by .A. They are more efficientB. The are free from pollutionC. They rum till reactanis are activeD. All of these

Define emf. Calculat the emf of the following galvanic cell : `Zn_((s))+Cu_((aq))^(+2)to Zn_((aq))^(+2)+Cu_((s))` `E_(zn^(+2//Zn))^(0) =0.76V(anode) ,E_(cu^(+3//Cu))^(0)=+0.34` (Cathode)

For a Ag-Zn button cell, the net reaction is `Zn (s)+Ag_(2)O(s)rarrZnO(s)+2Ag(s)` `DeltaG_(f)^(@)(Ag_(2)O)=-11.21kJ "mol"^(-1)` and `DeltaG_(f)^(@)(ZnO)=-318.3kJ "mol"^(-1)` The `E_("cell")^ (@)` of this button cell isA. 1.71VB. 1.591VC. 3.182VD. 3.07V

Local action in an electrochemical action can be prevented byA. using by pure electrolytes in tw half cellsB. using very pure metal for anodeC. coating zinc anode with mercuryD. using pure graphite for cathode.

`E^(o)` for the cell `Zn|Zn_((aq))^(2+)||Cu_((aq))^(2+)|Cu` is 1.10 V at `25^(@)C`, the equilibrium constant for the reaction `Zn+Cu_((aq))^(2+)hArrCu+Zn_((aq))^(2+)` is of the order ofA. `10^(-28)`B. `10^(-37)`C. `10^(+18)`D. `10^(+17)`

Calculate `K_(c)` for the reaction at 298 K `Zn_((s))+Cu_((aq))^(+2)hArrZn _((aq))^(2+)+Cu_((s))` `E _(Zn^(2+)//Zn)^(0)=-0.76V, E_(Cu^(2+)//Cu)^(0) =+0.34V.`

The standard cell potential of `Zn|Zn_((aq))^(2+)||Cu_((aq))^(2+)|Cu` cell is 1.10V. The maximum work obtained by this cell will beA. 106.15 kJB. `-212.30kJ`C. `-318.45kJ`D. `-424.60kJ`

The standard emf of Deniell cell is `1.1` V. Calculate the standard Gibbs energy for the cell reactions: `Zn_((s))+Cu_((aq))^(2+)to Zn _((aq))^(2+)+Cu_((s))`

A fuel cell is supplied 1 mole of `H_(2)` gas and 10 moles `O_(2)`gas. If the fuel cell is operated at 96.5 mA currecnt, how long will it deliver power?