3195 + Interview Questions in ELECTROCHEMISTRY IN GENERAL AWARENESS Page 29 InterviewSolution

1401.	Write the cell formulation and calculate the standard cell potential of the galvanic cell in operation of which the reaction taking place is: `2Cr(s)+3Cd^(2+)(aq)to2Cr^(3+)(aq)+3Cd(s)` Calculate `Delta_(r)G^(@)` for the above reaction. (Given: `E_(Cr^(3+)//Cr)^(@)=-0.74V,E_(Cd^(2+)//Cd)^(@)=-0.40V,F=96500" "C" "mol^(-1)`)
Answer» Correct Answer - `+0.34V,-196.86kJ" "mol^(-1)` `Cr(s)\|Cr^(3+)(aq)\|\|Cd^(2+)(aq)\|Cd(s)` `2Crto2Cr^(3+)+6e^(-)` or `3Cd^(2+)+6e^(-)to3Cd`. Hence, n=6 `E_(cell)^(@)=-0.40-(-0.74)=+0.34V,Delta_(r)G^(@)=-nFE_(cell)^(@)`

Discussion

1402.	Which allotropic form of carbon is used for making eletrodes ?
Answer» Graphite is used for making electrodes as it is a good conductor of electricity.

Discussion

1403.	Calculate the cell e.m.f. at `25^(@)C` for the cell: `Mg(s)\|Mg^(2+)(0.01M)\|\|Sn^(2+)(0.1M)\|Sn(s)` Given `E_(Mg^(2+)//Mg)^(@)=-2.34V,E_(Sn^(2+)//Sn)^(@)=-0.136V,1F=96,500" C "mol^(-1)` Calculate the maximum work that can be accomplished by the operation of this cell.
Answer» Correct Answer - `E_(cell)^(@)=2.50V,w_(max)=425.372kJ` `Mg+Sn^(2+)toMg^(2+)+Sn,E_(cell)^(@)=-0.136-(-2.34)=2.204V` `E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"([Mg^(2+)])/([Sn^(2+_)]), thereforeE_(cell)=2.204-(0.0591)/(2)"log"(0.01)/(0.1)=2.500V` `w_(max)=-DeltaG^(@)=nFE_(cell)^(@)=2xx96500xx2.204=425372J` (Note that `w_(max)` is related to `E_(cell)^(@)` and not `E_(cell)`).

Discussion

1404.	Which of the following solutions is used as an anti-rusting solutionA. `Na_(2)SO_(4)`B. `Na_(3)PO_(4)`C. `Na_(3)BO_(3)`D. `Na_(2)S`
Answer» Correct Answer - B

Discussion

1405.	Calculate the e.m.f. of the following cell at `25^(@)C` `Mg(s)//Mg^(2+)(0.01 M)\|\|Sn^(2+)(0.1 M)//Sn(s)` Given `" " E_(Mg^(2+)//Mg)^(@)=-2.34 V, E^(@)Sn^(2+)//Sn=-0.136 V` Also calculate the maximum work that can be accomplished by the operation of the cell.
Answer» Calculate of the cell e.m.f. According to Nernst equation, `E_(cell)=E_(cell)^(@)-(0.0591)/(n)"log"([Mg^(2+)(aq)])/([Sn^(2+)(aq)])` `E_(cell)^(@)=E_(cathode)^(@)-E_(anode)^(@)=-0.136-(-2.34)=2.204 V` `E_(cell)=2.204 V-(0.0591)/(n)"log"(0.01)/(0.1)` `=2.204V-(0.0591)/(2)"log"(10^(-1))` =2.204 V+0.02955 V=2.23 V Calculate of maximum work `(DeltaG^(@))` `(-DeltaG^(@))=nFE_(cell)^(@)=(2"mol")xx(96500" C mol"^(-1))xx(2.204 V)` `425372" CV"=425372 J=425.372 kJ`

Discussion

1406.	The given figure shows the corrosion of iron in atmosphere. Fill in the blanks by choosing an appropriate option. At a particular spot of an object made of iron, ___(i)___ of iron to ferrous ion takes place and that spot behaves as ___(ii)___. Electrons released at anodic spot move through the metal and go to another spot on the metal and reduce oxygen in presence of `H^+` . This spot behaves as ___(iii)___. The ferrous ions are further oxidised by atmospheric oxygen to ferric ions which come out as rust, __(iv)____ and with further production of ____(v)__ ions.A. i-oxidation , ii-anode , iii-cathode , iv-`Fe_2O_3.xH_2O` , v-hydrogenB. i-reduction, ii-cathode, iii-anode , `Fe_3O_4` , v-hydroxideC. i-oxidation, ii-cathode, iii-anode , iv-`Fe_2O_3.xH_2O` , v-hydrogenD. i-oxidation , ii-anode, iii-cathode , iv-`Fe_2O_3.H_2O`, v-ferrous
Answer» Correct Answer - A

Discussion

1407.	During rusting the anode isA. CuB. `H_(2)`C. FeD. Carbon
Answer» Correct Answer - C Fe sheet itself acts as an anode .

Discussion

1408.	`Zn` acts as sacrifical or cathodic protect iont to prevent rusting of iron becauseA. `E_(OP)^(@)` of `Zn lt E_(OP)^(@)` of FeB. `E_(OP)^(@) ` of ` Zn gt E_(OP)^(@)` of FeC. `E_(OP)^(@) ` of `Z n =E_(OP)^(@)` of FeD. Zn is cheaper than iron
Answer» Correct Answer - B Zn is oxidised in place of iron and iron acts as cathode if Zn plating or galvanisation of iron is made and thus called cathodic protection . Since Zn is oxidised and thus also called sacrifical protection .

Discussion

1409.	Define electrolysis.
Answer» Electrolysis : The process of a nonspontaneous chemical decomposition of an electrolyte by the passage of an electric current through its aqueous solution or fused mass and in which electrical energy is converted into chemical energy is called electrolysis. E.g. Electrolysis of fused NaCl.

1409.

Define electrolysis.

Answer»

Electrolysis : The process of a nonspontaneous chemical decomposition of an electrolyte by the passage of an electric current through its aqueous solution or fused mass and in which electrical energy is converted into chemical energy is called electrolysis.

E.g. Electrolysis of fused NaCl.

Discussion

1410.	The corrosion of iron object is not favoured byA. presence of `H^(+)` ionB. presence of moisture in airC. presence of impurities in iron objectD. presence of vacuum
Answer» Correct Answer - D Follow theory of corrosion .

Discussion

1411.	Determine the values of equilibrium constant `(K_(c))` and `DeltaG^(@)` for the reaction `Ni(s)+2Ag^(+)(aq)rarrNi^(2+)(aq)+2Ag(s), E^(@)=1.05 V.` (`"Given " 1F=96500"C mol"^(-1))`
Answer» (i) Calculate of `DeltaG^(@)` for the reaction `Ni(s)+2Ag^(+)(aq)rarr Ni^(2+)(aq)+2Ag(s)` `DeltaG^(@)=-nFE_(cell)^(@)` `=(-2)xx(96500" C mol"^(-1))xx(1.05 V)=-202650 "CV mol"^(-1)` =-202650 J `mol^(-1)`=-202.65 kJ `mol^(-1)` (ii) Calculate of equilibrium constant `(K_(c))` for the reaction `E^(@)=(2.303RT)/(nF)"log "K_(c)=(0.0591)/(n)"log"K_(c)` `logK_(c)=(nxxE^(@))/(0.0591 V)=(2xx(1.05 V))/((0.0591 V))=35.53` `K_(c)="Antilog " 35.53=3.41xx10^(35)`

Discussion

1412.	What is an electrochemical cell? What does it consist of?
Answer» Electrochemical cell : It consists of two electronic conductors such as metal plates dipping into an electrolytic or ionic conductor which is an aqueous electrolytic solution or a pure liquid of a molten electrolyte.

Discussion

1413.	What would happen if the protective tin coating over an iron bucket is broken in some places ?
Answer» Iron will get rusted because iron has lesser reduction potential than tin.

Discussion

1414.	What are electrochemical reactions?
Answer» 1. Electrochemical reactions : The chemical reactions occurring in electrochemical cells which involve transfer of electrons from one species to other are called electrochemical reactions. They are redox reactions. 2. These reactions are made of two half reactions namely oxidation at one electrode (anode) and reduction at another electrode (cathode) of the electrochemical cell. 3. The net reaction is the sum of the above two half reactions.

1414.

What are electrochemical reactions?

Answer»

1. Electrochemical reactions : The chemical reactions occurring in electrochemical cells which involve transfer of electrons from one species to other are called electrochemical reactions. They are redox reactions.

2. These reactions are made of two half reactions namely oxidation at one electrode (anode) and reduction at another electrode (cathode) of the electrochemical cell.

3. The net reaction is the sum of the above two half reactions.

Discussion

1415.	Can absolute electrode potential of an electrode be measured?
Answer» The correct answer is no.

Discussion

1416.	Can absolute value of electrode potential be measured ?
Answer» No, Absolute value of electrode potentials cannot be measured as electrode reaction cannot take place by its own.

Discussion

1417.	What is the difference between cell potential and standard cell potential ?
Answer» A cell potential becomes standard cell potential if the temperature and pressure are `25^(@)C` and 1 atmosphere respectively and the concentration of the electrolytes in the respective half cells is 1 M.

Discussion

1418.	The oxidation potential of Mg and Al are +2.37 and +1.66 volts respectively . The Mg in chemical reactionsA. will be replaced by AlB. will replace AlC. will not be able to replace AlD. none of these
Answer» Correct Answer - B Mg is more reaction than Al.

Discussion

1419.	When `E_(Ag)^(@)+_(//Ag)=0.8V` and ` E_(Zn^(@))2^(+)._(//Zn)` `=-0.76V`. Which of the following is correct?A. `Ag^(+)` can be reduced by `H_(2)`B. `Ag` can oxidise `H_(2)` into `H^(+)`C. `Zn^(2+)` can be reduced by `H_(2)`D. Ag can reduce `Zn^(2+)` ion.
Answer» Correct Answer - A Zinc has lower red pot, than hydrogen and hence can be reduced. Ag has more reduction potential so it is reduced easily.

Discussion

1420.	The oxidation potentials of A and B are +2.37 and +1.66 V respectively . In chemical reactionsA. A will be replaced by BB. A will replace BC. A will not replace BD. A and B will not replace each other .
Answer» Correct Answer - B Greater the oxidation potential , more is the reactivity .

Discussion

1421.	When `E_(Ag)^(@)+_(//Ag)=0.8V` and ` E_Zn^2+_(//Zn)` `=-0.76V`. Which of the following is correct?A. `Ag^(+)` can be reduced by `H_(2)`B. Ag can oxidise `H_(2)` into `H^(+)` ionC. `Zn^(2+)` can be reduced by `H_(2)`D. Ag can reduce `Zn^(2+)` ion .
Answer» Correct Answer - A Check the emf in each case . It should be positive .

Discussion

1422.	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 The given equation of y = mx + c. i.e. equation of straight line . So graph of `E_(Mg^(2+)) \| Mg `(y) Vs log `[Mg^(2+)]` (x) is a straight line with slope `(0.0592)/(2)` and intercept `E_(Mg^(2+) \| Mg)^(@)`.

Discussion

1423.	Which of the following statement is not correct about an inert electrode in a cell?A. It does not participate in the cell reaction.B. It provides surface either for oxidation or for reduction reaction.C. It provides surface for conduction of electrons.D. It provide surface for redox reaction
Answer» Correct Answer - D Inert electrode does not participate in the chemical reaction and acts only as source or sink for electrons and provides surface either for oxidation or for reduction reaction.

Discussion

1424.	In the electrolysis of aqueous solution of `CuSO_4` using copper electrodes , the process tha takes place at the anode is
Answer» Correct Answer - D `Cu^(+)to2e^(-)toCu` (cathode) `CutoCu^(2+)+2e^(-)` (anode) In the electrolysis of aq.. `CuSO_(4)`, as the current flows, copper from anode dissolves while pure copper is deposited on cathode.

Discussion

1425.	Which of the following statements is `//` are correct ?A. `F_(2)` is the strongest oxidizing agent. FB. `Li` is the strongest reducing agent.C. `Li^(o+)` is the weakest oxidizing agent.D. `F_(2)` has a highest reduction potential.
Answer» Correct Answer - a,b,c,d Reduction potential of `F_(2)` is highest. Oxidation potential of `Li` is highest.

Discussion

1426.	Assertion: The electrolysis of auqueous NaCl solution gives hydrogen at the cathode and chlorine at the anode. Reason: Chlorine has higher oxidation potential than water.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 - C Correct R. chlorine has higher reduction potential than water but lower discharge potential.

Discussion

1427.	Identify the correct statements `(s):`A. `wedge_(m)` increases with increase in temperature.B. `wedge_(m)` decreases with increase in concentration.C. Specific conductance increase with increase in concentration.D. Specific conductance decreases with increse in temperature.
Answer» Correct Answer - a,b,c As temperature increases solubility decreases `:. K darr,wedge_(m)uarr` Specific conductance `(kuarr)` increases as concentration increases.

Discussion

1428.	Statement-1: Electrolysis of an aqueous solution fo KI gives `I_(2)` at the anode but that of KF gives `O_(2)` at the anode and not `F_(2)`. Statement-2: `F_(2)` is more reactive than `I_(2)`.A. Statement-1 is True, statement-2 is true, statement-2 is a correct explanation of statement-1B. Statement-1 is true, statement-2 is true, statement-2 is not a correct explanation of statement-1.C. Statement-1 is true, statemet-2 is falseD. Statement-1 is false, statement-2 is true
Answer» Correct Answer - B Correct explanation. `I^(-)` ions have much higher oxidation potential than that of water whiel `F^(-)` ions have much lower oxidation potential than that of water.

Discussion

1429.	Which of the following cells is `//` are rechargeable or secondary cell `(s)` ?A. `Ni-Cd` cellB. Mercury cellC. Lead storage cellD. Lithium battery
Answer» Correct Answer - a,c,d Refer Section

Discussion

1430.	Electrolysis of KBr(aq) gives Br2 at anode but KF(aq) does not give F2. Give reason.
Answer» Oxidation takes place at anode. Now higher the oxidation Potential, easier to oxidize. Oxidation potential of Br^-, H₂O, F^- are in the following order. Br^- > H₂O > F^- Therefore in aq. Solution of KBr. Br^- ions are oxidized to Br₂ in preference to H₂O. On the other hand, in aq. Solution of KF, H₂O is oxidized in preference to F^-. Thus in this case oxidation of H₂O at anode gives O₂ and no F² is produced.

1430.

Electrolysis of KBr(aq) gives Br2 at anode but KF(aq) does not give F2. Give reason.

Answer»

Oxidation takes place at anode. Now higher the oxidation Potential, easier to oxidize. Oxidation potential of Br^-, H₂O, F^- are in the following order.

Br^- > H₂O > F^-

Therefore in aq. Solution of KBr. Br^- ions are oxidized to Br₂ in preference to H₂O. On the other hand, in aq. Solution of KF, H₂O is oxidized in preference to F^-. Thus in this case oxidation of H₂O at anode gives O₂ and no F² is produced.

Discussion

1431.	What happens when a piece of copper is added to (a) an aq solution of FeSO4 (b) an Aq solution of FeCl3?
Answer» a. Nothing will happen when the piece of copper is added to FeSO₄ because reduction potential E⁰Cu₂/Cu(0.34) is more than the reduction potential E⁰(Fe²⁺/Fe) (0.44 V). b. Copper will dissolve in an aq solution of FeCl₃ because reduction potential E⁰ Fe³⁺/Fe²⁺(0.77 V) is more than the reduction potential of E⁰ Cu²/Cu(0.34) Cu(s) + 2FeCl₃ (aq) → Cu₂(aq) + 2FeCl₂(aq)

1431.

What happens when a piece of copper is added to (a) an aq solution of FeSO4 (b) an Aq solution of FeCl3?

Answer»

a. Nothing will happen when the piece of copper is added to FeSO₄ because reduction potential E⁰Cu₂/Cu(0.34) is more than the reduction potential E⁰(Fe²⁺/Fe) (0.44 V).

b. Copper will dissolve in an aq solution of FeCl₃ because reduction potential E⁰ Fe³⁺/Fe²⁺(0.77 V) is more than the reduction potential of E⁰ Cu²/Cu(0.34)

Cu(s) + 2FeCl₃ (aq) → Cu₂(aq) + 2FeCl₂(aq)

Discussion

1432.	What will happen when chloride is passed through an aqueous solution of potassium bromide ?
Answer» The solution will acquire an orange colour due to the vapours of bromine that are evolted. Actually , chlorine is a stronger oxidising agent than bromine and oxidises `Br^(-)`ions (in KBr) to `Br_(2)` and itself is reducted to `Cl^(-)` ions. `2KBr+Cl_(2) to 2KCl+Br_(2)`

Discussion

1433.	Which of the following is not used to determine cell constant ?A. `10^(-2)` M KClB. `10^(-1)`M KClC. 1 M KClD. Saturated KCl
Answer» Correct Answer - D (d) Saturated solution of KCl cannot be used to determine the value of cell constant.

Discussion

1434.	Galvanization is applying a coating of :A. PbB. CrC. CuD. Zn
Answer» Correct Answer - D (d) Galvanisation implies applying a coating of zinc on the surface of iron which is prone to rusting. It checks the rusting of iron.

Discussion

1435.	Galvanization is applying a coating of :A. `Pb`B. `Cr`C. `Cu`D. `Zn`
Answer» Correct Answer - D

Discussion

1436.	The number of Faradays needed to reduce 4 gm equivalents of `Cu^(2+)` to Cu metal will beA. 1B. 2C. `1//2`D. 4
Answer» Correct Answer - D 1 g eq.require 1 F .

Discussion

1437.	Which of the following statements are correct concerning redox properties? (i) A metal M for which `E^(@)` for the half cell reaction `M^(n+)+"ne"^(-)hArrM` is very negative will be a good reducing agent. (ii) The oxidizing power of the halogen decreases from chlorine to iodine. (iii) The reducing power of hydrogen halides increases from hydrogen chloride to hydrogen iodide.A. i,ii and iiiB. i and iiC. i onlyD. ii and iii only
Answer» Correct Answer - A (i) Highly negative `E^(@)` for `M^(n+)+"ne"^(-)hArrM` means high positive value for the reverse reaction `MhArrM^(n+)+"ne"^(-)`. This means M is oxidized very easily. Hence it is a good reducing agent. (ii) Reduction potentials of halogens are in order `Cl^(2)gtBr_(2)gtI_(2)`. thus `Cl_(2)` is reduced most easily and hence is the best oxidizing agent. (iii) the size of the halide ions is in order `Cl^(-)ltBr^(-)ltI^(-)`. Greater the size of the halide ions. more easily it can lose electrons and get oxidized. Thus `I^(-)` can be oxidized most easily an d hence have the greatest reducing power.

Discussion

1438.	Which of the following statements are correct concerning redox properties? (i) A metal M for which `E^(@)` for the half life reaction `M^(n+)+n e^(-)hArrM` is very negative will be a good reducing agent. (ii) The oxidizing power of the halogen decrease from chlorine to iodine. (iii) The reducing power of hydrogen halides increases from hydrogen chloride to hydrogen iodide.A. (i),(ii) and (iii)B. (i) and (ii)C. (i) onlyD. (ii) and (iii) only
Answer» Correct Answer - A (i) High negative `E^(@)` for `M^(n+)+n e^(-)hArrM` means a high positive value for the reverse reaction, `M hArrM^(n+)+n e^(-)`. This means M is oxidized very easily. Hence, it is a good reducing agent. (ii) Reduction potentials of halogens are in the order: `Cl_(2) gt Br_(2) gt I_(2)`. thus, `Cl_(2)` is reduced most easily and hence is the best oxidizing agent. (iii) The size of the halide ions is in the order `Cl^(-) lt Br^(-) lt I^(-)`. greater the size of the halide ion, more easily it can lose electrons and get oxidized. Thus, `I^(-)` ions can be oxidized most easily and hene have the greatest reducing power.

Discussion

1439.	Small quantities of solutions of compounds TX, TY and TZ are put into separate test tubes containing X, Y and Z solutions. TX does not react with any of these. TY reacts with both X and Z. TZ reacts with X. the decreasing order of oxidation of the anions `X^(-),Y^(-),Z^(-)` is:A. `Y^(-),Z^(-),X^(-)`B. `Z^(-),X^(-),Y^(-)`C. `Y^(-),X^(-),Z^(-)`D. `X^(-),Z^(-),Y^(-)`
Answer» Correct Answer - A TX does ot ract with any solution. This means that `X^(-)` is least easily oxidized. TY reacts with both X and Z. this means `Y^(-)` is oxidized by both. `(Y^(-)+XtoY+X^(-),Y^(-)+ZtoY+Z^(-))` TZ reacts with X only `(Z^(-)+XtoZ+X^(-))` `Y^(-)`, thus, decreasing order of oxidation of anions `X^(-),Y^(-),Z^(-)` will be : `Y^(-),Z^(-),X^(-)`.

Discussion

1440.	The standard potential of hydrogen half cell is zero at all conditions of temperature and pressure.
Answer» Correct Answer - F Pressure is fixed `(1 atm)` . Temperature can have any value.

Discussion

1441.	The conventional value of zero of the standard hydrogen half cell holds good at all temperature.
Answer» Correct Answer - F The conventinal value is at `298K`.

Discussion

1442.	Redox reactions play a pivotyal role in chemistry and biology. The values of standard redox potential `(E^(@))` of two half-cell reactions decide which way the reaciton is expected to proced. A simple exampler is a Daniel cell in which zinc goes into solution and copper gerts deposited. Given below are a set of half-cell reactions (acidic medium) along with their `E^(@)` (`V` with respect to normal hydrofgen electrode) values. Using this data : `I_(2) + 2e^(-) rarr 2I^(-)" " E^(@) = 0.54`, `CI_(2) + 2e^(-) rarr 2CI^(-) " "E^(@) = 1.36`,. `Mn^(3+) + e^(-) rarr Mn^(2+) " "E^(@) = 1.50`, `Fe^(3+) + e^(-) rarr Fe^(2+)" " E^(@) = 0.77`, `O_(2) + 4H^(+) + 4e^(-) rarr 2H_(2)O " "E^(@) = 1.23`, Soldium fusion extract, obtained from aniline, on treatment with iron `(II)` sulphatge and `H_(2)SO_(4)` in presence of air gives a Prussion bule precipitate. The blue colour is due to the formation of :A. `Fe_(4)[Fe(CN)_(6)]_(3)`B. `Fe_(3)[Fe(CN)_(6)]_(2)`C. `Fe_(4)[Fe(CN)_(6)]_(2)`D. `Fe_(3)[Fe(CN)_(6)]_(3)`
Answer» Correct Answer - A `Na + C + N rarr NaCN` `Fe^(2+) + 6CN^(-) rarr [Fe(CN)_(6)]^(4-)` `4Fe^(3+) + 3[Fe(CN)_(6)]^(4-) rarr underset("Prussion blue")(Fe_(4)[Fe(CN)_(6)]_(3darr)`

Discussion

1443.	The current `-` carrying ions in an electrolytic cell are not necessarily discharged at the electrodes.
Answer» Correct Answer - T They can also be discharged in the solution.

Discussion

1444.	Redox reactions play a pivotal role in chemistry and biology. The values standard redox potential `(E^(c-))` of two half cell reactions decided which way the reaction is expected to preceed. A simple example is a Daniell cell in which zinc goes into solution and copper sets deposited. Given below are a set of half cell reactions `(` acidic medium `)` along with their `E^(c-)(V` with respect to normal hydrogen electrode `)` values. Using this data, obtain correct explanations for Question. `I_(2)+2e^(-) rarr 2I^(c-)," "E^(c-)=0.54` `Cl_(2)+2e^(-) rarr 2Cl^(c-), " "E^(c-)=1.36` `Mn^(3+)+e^(-) rarr Mn^(2+), " "E^(c-)=1.50` `Fe^(3+)+e^(-) rarr Fe^(2+)," "E^(c-)=0.77` `O_(2)+4H^(o+)+4e^(-) rarr 2H_(2)O, " "E^(c-)=1.23` While `Fe^(3+)` is stable, `Mn^(3+)` is not stable in acid solution becauseA. `O_(2)` oxidises `Mn^(2+)` to `Mn^(3+)`B. `O_(2)` oxidises both `Mn^(2+)` to `Mn^(3+)` and `Fe^(2+)` to `Fe^(3+)`C. `Fe^(3+)` oxidises `H_(2)O` to `O_(2)`D. `Mn^(3+)` oxidises `H_(2)O` to `O_(2)`
Answer» Correct Answer - D Reaction of `Mn^(3+)` with `H_(2)O` is spontaneous.

Discussion

1445.	Redox reactions play a pivotal role in chemistry and biology. The values standard redox potential `(E^(c-))` of two half cell reactions decided which way the reaction is expected to preceed. A simple example is a Daniell cell in which zinc goes into solution and copper sets deposited. Given below are a set of half cell reactions `(` acidic medium `)` along with their `E^(c-)(V` with respect to normal hydrogen electrode `)` values. Using this data, obtain correct explanations for Question. `I_(2)+2e^(-) rarr 2I^(c-)," "E^(c-)=0.54` `Cl_(2)+2e^(-) rarr 2Cl^(c-), " "E^(c-)=1.36` `Mn^(3+)+e^(-) rarr Mn^(2+), " "E^(c-)=1.50` `Fe^(3+)+e^(-) rarr Fe^(2+)," "E^(c-)=0.77` `O_(2)+4H^(o+)+4e^(-) rarr 2H_(2)O, " "E^(c-)=1.23` Among the following, identify the correct statement.A. Chloride ion is oxidised by `O_(2)`B. `Fe^(2+)` is oxidised by iodineC. Iodide ion is oxidised by chlorineD. `Mn^(2+)` is oxidised by chlorine
Answer» Correct Answer - C Reducation potential of `I_(2)` is less than `Cl_(2)`

Discussion

1446.	Cations having more negative potential than `-0.828V` are reduced in preference to water.
Answer» Correct Answer - T Cations having more value than `-0.828V` are reduced in preference to `H_(2)O`. Hence, the reduction potential of cations would be greater than that of `H_(2)O`. Reduction potential of `H_(2)O=-8.28V`.

Discussion

1447.	Write the chemical equation corresponding ot the oxidation of `H_(2)O_((i))` at the platinum anode.
Answer» The chemical equation corresponding to the oxidation of `H_(2)O_((i))` at the platinum anode is `2H_(2(l))to O_(2(g))+4H_((aq))^(+)+4e^(-)`

Discussion

1448.	Tollen reagent is used for the detection of aldehydes. When a solution of `AgNO_(3)` is added to glucose with `NH_(4)OH`, then gluconic acid is formed. `Ag^(o+)+e^(-) rarr Ag," "E^(c-)._(red)=0.8V` `C_(6)H_(12)O_(6)rarr underset(Gluconic aci d)(C_(6)H_(12)O_(7)+)2H^(o+)+2e^(-) , " "E^(c-)._(o x i d ) =-0.05V` `[Ag(NH_(3))_(2)]^(o+)+e^(-) rarr Ag(s)+2NH_(3), " "E^(c-)._(red)=0.337V` `[Use2.303xx(RT)/(F)=0.0592` and `(F)/(RT)=38.92at 298 K ]` Ammonia is always added in this reaction. Which of the followijng must be wrong ?A. `NH_(3)` combines with `Ag^(+)` to form a complexB. `[Ag(NH_(3))_(2)]^(+)` is a worker oxidising reagent than `Ag^(+)`C. In absence of `NH_(3)` silver salt of gluconic acid is fomedD. `NH_(3)` has affected tge standard reduction potential of glucose`//`glucose acid electrode
Answer» Correct Answer - D Addition of `NH_(3)` affects the value of `E_("oxd")` for glucose`//`gloconic acid and not `E^(@)`. Thus, this statement is wrong. Rest all statements are correct `E_(RP_(Ag^(+)))^(@) gt E_(RP_(Ag(NH_(3))_(2)^(+)))^(@)`

Discussion

1449.	In above question no. `10`, if a current of `0.862` ampere was passed for a period of `5` hr, calculate the amount of metal that would be deposited at cathode.
Answer» Correct Answer - 9 According to above solution `10` Force electrolysis, `w = (E xx i xx t)/(96500)" " ( :. t = 5 xx 60 xx 60 sec)` `= (56 xx 0.862 xx 5 xx 60 xx 60)/(96500) = 9.0 g`

Discussion

1450.	when an aqueous solution of `H_2SO_4` is electrolysed the product at anodes is :A. `H^-`B. `OH^-`C. `SO_4^(2-)`D. `O_2`
Answer» Correct Answer - D

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