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1801.

Given the data at `25^(@)C` `Ag+I^(-)rarrAgl+e^(-)" "E^(@)=0.153V` `Ar rarrAg^(+)+e^(-)E^(@)=0.800V` What is the value of `log K_(sp)` for AgI?A. `-37.83`B. `-16.13`C. `-8.12`D. `+8.612`

Answer» Correct Answer - B
`Ag+I^(-)rarrAgI+e^(-),E^(@)=0.152V`
`Ag^(+)+e^(-)rarrAg,E^(@)=0.800V`
`Ag^(+)+I^(-)rarrAgI,E_("cell")=0.952V`
`k_(C)=( [AgI])/([Ag^(+)][I^(-)])=(1)/(K_(sp))`
`therefore 0.952=(2.303)/(F)RT"log"k_(C)`
`0.952=0.059"log"(1)/(k_(sp))`
`0.952=-0.059"log" k_(sp)`
or `"log" k_(sp)=-16.13`
Discussion
1802.

Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1M is `100Omega`. The conductivity of this solution is 1.29 `S m^(-1)`. Resistance of the same cell when filled with 0.2M of the same solution is `520Omega`. the molar conductivity of `0.02`M solution of the electrolyte will beA. `1.24xx10^(-4)Sm^(2)"mol"^(-1)`B. `12.4xx10^(-4)S m^(2) "mol"^(-1)`C. `124xx10^(-4)S m^(2)"mol"^(-1)`D. `1240xx10^(-4)S m^(2)"mol"^(-1)`

Answer» Correct Answer - C
Cll constant =`x xxR`
`=1.29S m^(-1)xx100Omega=129m^(-1)`
In the 2nd case `x=("cell constant")/("Resistance")`
`=(129m^(-1))/(520Omega)= 0.248Omega^(-1)m^(-1)`
`^^_(m)=(x)/("molarity")=(0.248Omega^(-1)m^(-1))/(0.02xx10^(3)"mol"m^(-1))`
`=124xx10^(-4)Sm^(2)"mol"^(-1)`
Discussion
1803.

An aqueous solution of X is added slowly to an aqueous solution of Y as shown in List-I. The variation in conductivity of these reactions is given in List-II. Match List-I with List-II and select the correct answer using the code given below the lists: `{:(List-I,List-II),((P)(C_(2)H_(5))_(3)N +CH_(3)COOH,(1)"Conductivity decreases adn then increases"),(" X Y",),((Q)KI(0.1M)+AgNO_(3)(0.01M),(2)"Conductivity decreases and then does not change much"),(" X Y",),((R)CH_(3)COOH+KOH,(3)"Conductivity increases and then does not change much"),(" X Y",),((S)NaOH+Hi,(4)"Conductivity does not change much and then increases"),(" X Y",):}` Codes:A. `{:(P,Q,R,S),(3,4,2,1):}`B. `{:(P,Q,R,S),(4,3,2,1):}`C. `{:(P,Q,R,S),(2,3,4,1):}`D. `{:(P,Q,R,S),(1,4,3,2):}`

Answer» Correct Answer - a
Discussion
1804.

Resistance of a conductvity cell filled with a solution of an electrolyte of concentration 0.1 M is 100 `Omega`. The conductivity of this solution is 1.29 `Sm^(-1)`. Resistance of the same cell when filled with 0.02M of the same solution is `520 Omega`. the molar conductivity of 0.02M solution of the electrolyte will be:A. `124xx10^(-4) S m^(2) mol^(-1)`B. `1240xx10^(-4) S m^(2) mol^(-1)`C. `1.24xx10^(-4) S m^(2) mol^(-1)`D. `12.4xx10^(-4) S m^(2) mol^(-1)`

Answer» Correct Answer - A
`kappa=1/Rxxl/A`
`1.29=1/100xxl/A`
`l/A=129 m^(-1)`
`Lambda_(m)=kappaxx1000/M`
`=(1/Rxxl/A)xx1000/M`
`=(1/520xx129)xx1000/0.02xx10^(-6)`
`=124xx10^(-4) S m^(2) mol^(-1)`
Discussion
1805.

The standard reduction potential data at `25^(@)C` is given below `E^(@) (Fe^(3+), Fe^(2+)) = +0.77V`, `E^(@) (Fe^(2+), Fe) = -0.44V`, `E^(@) (Cu^(2+),Cu) = +0.34V`, `E^(@)(Cu^(+),Cu) = +0.52 V`, `E^(@) (O_(2)(g) +4H^(+) +4e^(-) rarr 2H_(2)O] = +1.23V` `E^(@) [(O_(2)(g) +2H_(2)O +4e^(-) rarr 4OH^(-))] = +0.40V`, `E^(@) (Cr^(3+), Cr) =- 0.74V`, `E^(@) (Cr^(2+),Cr) = - 0.91V`, Match `E^(@)` of the redox pair in List-I with the values given in List-II and select the correct answer using the code given below teh lists: `{:(List-I,List-II),((P)E^(@)(Fe^(3+),Fe),(1)-0.18V),((Q)E^(@)(4H_(2)O hArr 4H^(+)+4OH^(+)),(2)-0.4V),((R)E^(@)(Cu^(2+)+Curarr2Cu^(+)),(3)-0.04V),((S)E^(@)(Cr^(3+),Cr^(2+)),(4)-0.83V):}` Codes:A. `{:(P,Q,R,S),(4,1,2,3):}`B. `{:(P,Q,R,S),(2,3,4,1):}`C. `{:(P,Q,R,S),(1,2,3,4):}`D. `{:(P,Q,R,S),(3,4,1,2):}`

Answer» Correct Answer - D
`(P) e^(-) + Fe^(3+) rarr Fe^(+2)`
`DeltaG_(1)^(ul(@)) =- 1 xx F xx 0.77`
`2e^(-) +Fe^(2+) rarr Fe`
`DeltaG_(2)^(ul(@))=- 2 xx F xx (-0.44)`
Required equation
`Fe^(3+) +3e^(-) rarr Fe`
`DeltaG_(3)^(ul(@)) =- 3 xx F xx E^(@)`
`DeltaG_(3)^(ul(@)) = DeltaG_(2)^(ul(@)) +DeltaG_(1)^(ul(@))`
`- 3 xx F xx E^(ul(@)) =- 1 xx F xx 0.77 - 2 xx F xx (-0.44)`
`E^(ul(@)) =- (0.11)/(3) =- 0.04V`
`(Q) O_(@)(g) +4H^(+) +4e^(-) rarr 2H_(2)O`
`E_(1)^(ul(@)) = 1.23` ...(i)
`O_(2)(g) +2H_(2)O +4e^(-) rarr 40H^(-) E_(2)^(ul(@)) = 0.40V` ...(ii)
Required equation
`4H_(2)O hArr 4H^(+) + 4OH^(-) (E_(2)^(ul(@))-E_(1)^(ul(@))) =- 0.83V`
`(R) {:(Cu^(+2)+e^(-),rarr,Cu^(+)),(E_(1)^(ul(@))=0.34V,,),(Cu,rarr,Cu^(+)+e^(-)),(E_(2)^(ul(@))=-0.52,,):}`
`bar(Cu^(+2) +Cu rarr 2Cu^(+1)`
`-0.18 V`
`(S) Cr^(+3) +3e^(-) rarr Cr`
`DeltaG_(1)^(ul(@)) =- 3 xx F xx (-0.74)`
`Cr^(+2) +2e^(-) rarr Cr`
`DeltaG_(2)^(ul(@)) = - 2xx F xx (-0.91)`
Required equation
`Cr^(3+) +e^(-) rarr Cr^(+2)`
`DeltaG_(3)^(ul(@)) =- 1 xx F xx (E^(@))`
`DeltaG_(1)^(ul(@)) = DeltaG_(2)^(ul(@)) +DeltaG_(3)^(ul(@))`
`- 3 xx F xx (-0.74) =- 2 xx F xx (-0.91) - 1 xx F xx E^(@)`
`E^(@) = 1.82 - 2.22 =- 0.4V`
Discussion
1806.

The standard reduction potential data at `25^(@)C` is given below `E^(@) (Fe^(3+), Fe^(2+)) = +0.77V`, `E^(@) (Fe^(2+), Fe) = -0.44V`, `E^(@) (Cu^(2+),Cu) = +0.34V`, `E^(@)(Cu^(+),Cu) = +0.52 V`, `E^(@) (O_(2)(g) +4H^(+) +4e^(-) rarr 2H_(2)O] = +1.23V` `E^(@) [(O_(2)(g) +2H_(2)O +4e^(-) rarr 4OH^(-))] = +0.40V`, `E^(@) (Cr^(3+), Cr) =- 0.74V`, `E^(@) (Cr^(2+),Cr) = - 0.91V`, Match `E^(@)` of the redox pair in List-I with the values given in List-II and select the correct answer using the code given below teh lists: `{:(List-I,List-II),((P)E^(@)(Fe^(3+),Fe),(1)-0.18V),((Q)E^(@)(4H_(2)O hArr 4H^(+)+4OH^(+)),(2)-0.4V),((R)E^(@)(Cu^(2+)+Curarr2Cu^(+)),(3)-0.04V),((S)E^(@)(Cr^(3+),Cr^(2+)),(4)-0.83V):}` Codes:A. `{:(P,Q,R,S),(4,1,2,3):}`B. `{:(P,Q,R,S),(1,2,3,4):}`C. `{:(P,Q,R,S),(2,3,4,1):}`D. `{:(P,Q,R,S),(3,4,1,2):}`

Answer» Correct Answer - d
Discussion
1807.

The conductivity of 0.1 n NaOH solution is 0.022 S`cm^(-1)`.When equal volume of 0.1 N HCl solution is added, the conductivity of resultant solution is decreased to 0.0055 S `cm^(-1)` . The equivalent conductivity of NaCl solution is :A. 0.0055B. 0.11C. 110D. None of these

Answer» Correct Answer - C
Discussion
1808.

The pressure of `H_(2)` required to make the potential of `H_(2)`- electrode zero in pure water at 298 K isA. `10^(-10)m`B. `10^(-4)atm`C. `10^(-14)atm`D. `10^(-12)atm`

Answer» Correct Answer - C
The reduction reaction for `H_(2)`-electrode is
`2H^(+)(aq)+2^(-)toH_(2)(g)`
`H_(H^(+)//H_(2))^(@)=E_(H^(+)//H_(2))^(@)-(0.0591)/(2)"log"(p_(H_(2)))/([H^(+)]^(2))`
In pure water at 298K, `[H^(+)]=10^(-7)M`
`therefore0=0-(0.0591)/(2)"log"(p_(H_(2)))/((10^(-7))^(2))`
or `"log"(p_(H_(2)))/(10^(-14))=0` or `(p_(H_(2)))/(10^(-14))=1" "(becauselog1=0)`
or `p_(H_(2))=10^(-14)atm`
Discussion
1809.

The pressure of `H_(2)` required to make the potential of `H_(2)-`electrode zero in pure water at 298K isA. `10^(-14)atm`B. `10^(-12)atm`C. `10^(-10)atm`D. `10^(-4)atm`

Answer» Correct Answer - A
`2H^(+)(aq)=2e^(-)toH_(2)(g)`
`thereforeE=E^(0)-(0.0591)/(2)"log"(P_(H_(2)))/([H^(+)]^(2))`
`0=0-0.295"log"(P_(H_(2)))/((10^(-7))^(2))`
`(P_(H_(2)))/((10^(-7))^(2))=1`
`P_(H_(2))=10^(-14)atm`
Discussion
1810.

What pressure of `H_(2)` would be required to make emf of hydrogen electrode zero in pure water at `25^(@)C` ?A. `10^(-7) atm`B. `10^(-14) atm`C. 1 atmD. 0.5 atm

Answer» Correct Answer - C
(c ) It is the correct answer,
Discussion
1811.

The pressure of `H_(2)` required to make the potential of `H_(2)-`electrode zero in pure water at 289K is :A. `10^(-12) atm `B. `10^(-10) atm `C. `10^(-4) atm `D. `10^(-14)atm`

Answer» Correct Answer - D
(d) From the question, we have an equation
`2H^(+)+2e^(-) to H_(2)(g)`
According to Nernst equation,
`E=E^(@)-(0.059)/(2)log""(p_(Hg))/([H^(+)]^(2))`
`=0-(0.059)/(2) log""(p_(Hg))/((10^(-7))^(2))" " [:.[H^(+)]=10^(-7)]`
`:.` For potential of `H_(2)` electrode to be zero, `p_(H_(2))` should be equal to `[H^(+)]^(2)`,i.e `10^(-14)` atm.
`:." " log""(10^(-14))/((10^(-7))^(2))=0`
Discussion
1812.

Calculate `E_(cell)^(@)` for the following reaction at 298 K: `2AI(s) + 3Cu^(2+)(0.01M) to 2A1^(3+)(0.01M) + 3Cu(s)` `"Given" : E_(cell) = 1.98V` (b) Using the `E^(@)` values A and B, predict which is better for coating the surface of iron `[E^(@)(Fe^(2+)//Fe) = -0.44V]` to prevent corrosion and why? `"Given" : E^(@)(A^(2+)//A) = 2.37V : E^(@)(B^(2+) //B) = 0.14V`

Answer» A is better since its `E^(@)` value is more negative.
Discussion
1813.

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+)` and `Fe^(2+)`C. `Fe^(3+)` oxidises `H_(2)O` to `O_(2)`D. `Mn^(3+)` oxidises `H_(2)O` to `O_(2)`

Answer» Correct Answer - d
`Mn^(3+)` oxidises `H_(2)O` to `O_(2)` because the standard reduction potential of `(Mn^(3+) rarr Mn^(2+))` is greater than that of `(O_(2) rarr H_(2)O)`.
Discussion
1814.

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)` to `O_(2)`D. `Mn^(3+)` oxidises `H_(2)O` to `O_(2)`

Answer» Correct Answer - D
Discussion
1815.

The standard redox potentials for the ractions `Mn^(2+)+2e^(-)toMn` and `Mn^(3+)+e^(-)toMn^(2+)` are -1.18V and 1.51 V respectively. What is the redox potential for the reaction `Mn^(3+)+3e^(-)toMn`A. 0.33VB. 1.69VC. `-0.28`VD. `-0.85V`

Answer» Correct Answer - C
`Mn^(2+)+2e^(-)toMn` . . . .(i)
`DeltaG^(o)=-nFE^(o)=2.36F`
`Mn^(3+)+e^(-)toMn^(2+)` . . .(ii)
`DeltaG^(o)=-nFE^(o)=-1.51F`
Add eqns. (i) and (ii),
`Mn^(3+)+3e^(-)toMn`
`Delta^(0)=2.36F+(-1.51F)=0.85F`
`DeltaG^(0)=-nFE^(o)`
`0.85F=-3xxFxxE^(@)impliesE^(@)=-0.28V`
Discussion
1816.

For the cell, `Pt|Cl_2(g,0.4"bar")|Cl^(-)(aq,0.1M)"||"Cl^(-)(aq),0.01M)|Cl_2(g,0.2"bar")|pt`A. 0.051VB. -0.051C. 0.102VD. 0.0255V

Answer» Correct Answer - A
Discussion
1817.

Fe is reacted with `1.0M HCI.E^(@)` for `Fe//Fe^(2+) = +0.34` volt. The correct observation (s) regarding this reaction is/are:A. `Fe` will be oxidised to `Fe^(2+)`B. `Fe^(2+)` will be reduced to `Fe`C. since e.m.f is positive, the reaction shall occurD. since e.m.f. is positive the reaction shall not occur

Answer» Correct Answer - A::C
`2H^(+) +Fe rarr Fe^(2+) +H_(2)`
`E_(cell)^(@) = E_("cathod") -E_("anode")^(@) = 0 - (-0.34) = 0.34V`
Fe will oxidise, emf is +ve, the reaction shall occur.
Discussion
1818.

Which one of the following will increase the voltage of the cell ? `(T = 298 K)` ` Sn+ 2Ag^+ rarr Sn^(2+) +2 Ag`.A. increases in the size of silver rodB. increases in the concentration of `Sn^(+2)` ionsC. increases in the concentration of `Ag^(+)` ionsD. none of the above

Answer» Correct Answer - C
Due to reduction of `Ag^(+)`
`Ag^(+)+e^(-)rarr Ag : Ag` will deposit on cathode
Ag rod will increase in size.
Discussion
1819.

An acidic solution of copper (II) sulphate containing some contaiminations of zinc and iron (II) ions was electrolysed till all the copper is deposited. If electrolysis is further continued for sometime, the product liberated at cathode isA. `Fe`B. `Zn`C. `H_(2)`D. Alloy of `Zn` and `Fe`

Answer» Correct Answer - C
On further electrlysed the solution, there will be `Fe^(2+) & H^(+)` ions and electrode potential of `H^(+)` is greater. Hence it will reduce
`2H^(+) +2e^(-) rarr H_(2)`
Discussion
1820.

A current of `9.65` ampere is passed through the aqueous solution `NaCI` using suitable electrodes for `1000s`. The amount of `NaOH` formed during electrolysis isA. `2.0g`B. `4.0g`C. `6.0g`D. `8.0g`

Answer» Correct Answer - B
`W =Z I t = (40)/(96500) xx 9.65 xx 10^(3) = 4g`
Discussion
1821.

Zn gives `H_(2)` gas with `H_(2)SO_(4)` and HCl but not with `HNO_(3)` becauseA. `Zn` acts as oxidizing agent when reacts with `HNO_(3)`B. `HNO_(3)` is weaker acid than `H_(2)SO_(4)` and HClC. In electrochemical series, Zn is above hydrogenD. `NO_(3)^(-)` is reduced in preference to hydronium ion

Answer» Correct Answer - B
In the given reaction, `I^(-)` has been oxidized to `I_(2)` and `Cr_(2)O_(7)^(2-)` ions have been reduced to `Cr^(3+)`
`thereforeE_(cell)^(@)=E_(Cr_(2)O_(7)^(2-))^(@)-E_(I_(2))^(@)`
i.e., `0.79=1.33-E_(I_(2))^(@)` or `E_(I_(2))^(@)=0.54V`
Discussion
1822.

`Cr_(2)O_(7)^(2-)+I^(-)toI_(2)+Cr^(3+)` `E_(cell)^(@)=0.79V,E_(Cr_(2)O_(7)^(2-))^(@)=1.33V,E_(I_(2))^(@)=?`A. `0.54V`B. `-0.054V`C. `+0.18V`D. `-0.18V`

Answer» Correct Answer - D
In the given raction, `I^(-)` has been oxidized to `I_(2)` and `Cr_(2)O_(7)^(2-)` ions have been reduced to `Cr^(3+)`
`thereforeE_(cell)^(@)=E_(Cr_(2)O_(7)^(2-))^(@)-E_(I_(2))^(@)`
i.e., `0.79=1.33-E_(I_(2))^(@)` or `E_(I_(2))^(@)=0.54V`
Discussion
1823.

Standard electrode potential for `Sn^(4+)//Sn^(2+)` couple is +0.15V and that for the `Cr^(2+)//Cr` couple is -0.74V. These two couples in their standard state are connected to make a cell. The cell potential will beA. `+1.83V`B. `+1.19V`C. `+0.89V`D. `+0.18V`

Answer» Correct Answer - C
`E_(cell)^(@)=E_("cathode")^(@)-E_("anode")^(@)`
`=+0.15-(-0.74)+0.89V`
Discussion
1824.

Standard electrode potential for `Sn^(4+)//Sn^(2+)` couple is +0.15V and that for the `Cr^(3+)//Cr` couple is -0.74V. These two couples in their standard state are connected to make a cell. The cell potential will beA. `+1.83V`B. `+1.19V`C. `+0.89V`D. `+0.18V`

Answer» Correct Answer - C
`E_(cell)^(@)=E_("cathode (RP)")^(@)-E_("anode (RP)")^(@)`
`=0.15-(-0.74)=+0.89V`
Discussion
1825.

Two half cell reactions are given as : (i) `Fe_(aq)^(2+)+2e^(-) to Fe_((s)),E^(@)=-0.44" V "` (ii) `2H_((aq))^(+)+(1)/(2)O_(2)(g)+2e^(-) to H_(2)O(l),E^(@)=+1.23" V "` The `E^(@)` for the reaction `Fe_(s)+2H^(+)+(1)/(2)O_(2)(g) to Fe_(aq)^(2+)+H_(2)O_(l)` isA. `+1.67" V "`B. `-1.67" V "`C. `+0.79" V "`D. `-0.79" V "`

Answer» Correct Answer - A
(a) Subtract equation (i) from equation (ii) to get equation (iii) `E^(@)=(+1.23)-(-0.44)=+1.67" V "`.
Discussion
1826.

A solution contains `Fe^(2+), Fe^(3+)` and `T^(-)` ions. This solution was treated with iodine at `35^(@)C. E^(@)` for `Fe^(3+), Fe^(2+)` is `0.77 V` and `E^(@)` for `I_(2)//2I^(-)` = 0.536 V. The favourable redox reaction is:A. `I_(2)` will be reduced to `I^(-)`B. there will be no redox reactionC. `I^(-)` will be oxidised to `I_(2)`D. `Fe^(2+)` will be oxidised to `Fe^(3+)`

Answer» Correct Answer - C
Since the reduction potential of `Fe^(3+) | Fe^(2+)` is greater than that of `I_(2) | I^(-) , Fe^(3+)` will be reduced and `I^(-)` will be oxidised .
Discussion
1827.

If the half cell reactions are given as (i) `Fe_((aq))^(2+)+2etoFe_((s)),E^(o)=0.44V` (ii) `2H_((aq))^(+)+(1)/(2)O_(2(g))+2etoH_(2)O_((l)),E^(o)=+1.23V` The `E^(o)` for the reaction `Fe_((s))+2H^(+)+(1)/(2)O_(2(g))toFe_((aq))^(2+)+H_(2)O_((l))`A. `+1.67V`B. `-1.67V`C. `+0.79V`D. `-0.79V`

Answer» Correct Answer - A
Subtracting equation (i) from (ii) we will get equation (iii)
`Fe_((s))+2H^(+)+(1)/(2)O_(2)toFe_((aq))^(2+)+H_(2)O_((l))` . . .(iii)
So, `E^(o)=E_(2)^(o)--E_(1)^(0)`
`=+1.23-(-0.44)=+1.6V`
Discussion
1828.

If the e.m.f of a galvanic cell is neagative, it implies that:A. the cell reaction is spontaneousB. the cell reaction is non-spontaneousC. the cell reaction is exothermicD. the cell is working in reverse direction

Answer» Correct Answer - A::B
Discussion
1829.

A solution contains `Fe^(2+),Fe^(3+) and I^(-)` ions. This solution was treated with iodine at `35^(@)C. E^(@)` for `Fe^(3+)//Fe^(2+)` is +0.77V and `E^(@)` for `I_(2)//2I^(-)=0.536V`. The favourable redox reaction isA. `I^(-)` will oxidised to `I_(2)`B. `Fe^(2+)` will be oxidised to `Fe^(3+)`C. `I_(2)` will be reduced to `I^(-)`D. There will be no redox reaction

Answer» Correct Answer - A
`2(e^(-)+Fe^(+3)toFe^(+2)" "E^(@)=0.77V`
`underline(2I^(-)toI_(2)+2e^(-)" "E^(@)=0.536V" ")`
`2Fe^(+3)+2e^(-)to2Fe^(+2)+I_(2)" "E^(o)=E_(o x)^(o)+E_(red)^(o)`
`=0.77=0.536=0.164V`
So, Reaction will taken place.
Discussion
1830.

If the `E_(cell)^(o)` for a given reaction has a negative value, then whih of the following gives the correct relationships for the value of `DeltaG^(@) and K_(eq)`A. `DeltaG^(@)gt0,K_(eq)lt1`B. `Delta^(@)gt0,K_(eq)gt1`C. `DeltaG^(+)lt0,K_(eq)gt1`D. `DeltaG^(@)lt0,K_(eq)lt1`

Answer» Correct Answer - A
`ThereforeDeltaG^(@)=-nFE^(@) and DeltaG^(@)=-Rtlog_(e)K_(eq)`
Discussion
1831.

The potential of standard hydrogen electrode is zero. This implies thatA. `DeltaG_(f)^(o)(H^(+),aq)=0`B. `DeltaH_(f)^(o)(H^(+),aq)=0`C. `DeltaG_(f)^(o)(H^(+),aq)lt0`D. `DeltaG_(f)^(o)(H^(+),aq)gt0`

Answer» Correct Answer - A
As `DeltaG_(f)^(o)=-nFE^(o)`
if `E^(o)=0`
then `DeltaG_(f)^(o)=0`.
Discussion
1832.

Which of the following statements (or equation) is correctA. The unit of cell e.m.f. is V `cm^(-1)`B. `DeltaG=-(nF)/(E_(cell))`C. In galvanic cell, cchemical energy is transformed into electrical energyD. Oxidation state of Mn in potassium permanganate is +6

Answer» Correct Answer - C
The unit of cell e.m.f. is volt
`DeltaG=-nFE_(cell)`
In `KMnO_(4)`,oxidation state of `Mn` is +7.
Discussion
1833.

Which one of the following reaction is not possible ?A. `Fe + H_(2)SO_(4) to FeSO_(4) + H_(2)`B. `Cu + 2 AgNO_(3) to Cu (NO_(3))_(2) +2 Ag`C. `2 KBr + I_(2) to 2 KI + Br_(2)`D. `Cu O + H_(2) to Cu + H_(2)O`

Answer» Correct Answer - C
`I_(2)` is less reactive than `Br_(2)` .
Discussion
1834.

`Mg_((s))+Cu^(2+)(aq)toCu_((s))+Mg^(2+)(aq)` If the standard reduction potential of Mg and Cu are -2.37 and +0.34V respectively. The emf of the cell isA. 2.03VB. `-2.03V`C. `+2.71V`D. `-2.71V`

Answer» Correct Answer - C
`E_(Cell)^(@)=E_("cathode")-E_("anode")^(@)=0.34-(2.37)=+2.71V`
Discussion
1835.

What is `E^(@)` for electrode represented by `Pt,O_(2)(1atm)//2H^(+)(1m)`A. UnpredictableB. ZeroC. 0.018VD. 0.118V

Answer» Correct Answer - B
`E^(o)=0` because hydrogen have zero potential.
Discussion
1836.

The element that is easiest to be reduced isA. FeB. CuC. AgD. Sn

Answer» Correct Answer - C
More is `E_(RP)^(@)` , more is the tendency to get reduced . `E_(RP)^(@)` for Ag is maximum .
Discussion
1837.

Use the data given in Q.8 and find out the most stable ion in its reduced form.A. `Cr^(3+)`B. `MnO_(4)^(-)`C. `Cr_(2)O_(7)^(2-)`D. `Mn^(2+)`

Answer» Correct Answer - A
Since `Cr^(3+)` ion is the strongest reducing agant, it has been oxidised to the maximum. It is therefore, the most stable oxidised species.
Discussion
1838.

Use the data given in Q.8 and find out the most stable ion in its reduced form.A. `Cl^(-)`B. `Cl^(3+)`C. CrD. `Mn^(2+)`

Answer» Correct Answer - D
`Mn^(2+)` ion is the most stable in its reduced form because `MnO_(4)^(-)` is the strongest oxidising agent.
Discussion
1839.

Which of the statements about solutions of electrolytes is not correct?A. Conductivity of solution depends upon size of ions.B. Conductivity depends upon viscosity of solution.C. Conductivity does not depend upon solvation of ions present in solution.D. Conductivity of solution increases with temperature.

Answer» Correct Answer - C
Greater the solvation of the ions, less is the conductivity.
Discussion
1840.

Which of the statements about solution of electrolytes is not correct?A. Conductivity so solution depends upon size of ionsB. conductivity depends upon viscosity of solutionC. Conductivity does not depend upon solvation of ions present in solution.D. Conductivity of solution increases with temperature

Answer» Solution consists of electrolytes is known as electrolytic solution and conductivity of electrolytic solution depends upon the following fac tors.
(i) Size of ions As ion size increases, ion mobility decreases and conductivity decreases.
(ii) Viscosity of solution Greater the viscosity of the solvent lesser will be the conductivity of the solution.
(iii) Solvation of ions Greater the solvation of ions of an electrolyte lesser will b e the electrical conductivity of the solution.
(iv) Temperature of medium conductivity of solution inc reases with increase in temperature.
Discussion
1841.

Find out which of the following is the strongest oxidizing agent.A. `Cl^(-)`B. `Mn^(2+)`C. `MnO_(4)^(-)`D. `Cr^(3+)`

Answer» Correct Answer - C
Strongest oxidizing agent is one which is reduced most easily, i.e., which has highest reduction potential viz. `MnO_(4)^(-)` (reduced to `Mn^(2+)`).
Discussion
1842.

Using the data given below find out the strongest reducing agent. `E_(Cr_(2)O_(7)^(2-)//Cr^(3+))^(Theta)=1.33V,E_(Cl_(2)//Cl^(-))^(Theta)=1.36V` `E_(MnO_(4)^(-)//Mn^(2+))^(Theta)=1.51V,E_(Cr^(3+)//Cr)^(Theta)=-0.74V`A. `Cl^(-)`B. `Cr`C. `Cr^(3+)`D. `Mn^(2+)`

Answer» Correct Answer - B
Strongest reducing agent is one which is oxidized most easiliy, i.e., which has highest oxidation potential. Given values are reduction potentials. Reversing the sign will give oxidation potentials.thus, `Cr//Cr^(3+)` will have highest oxidation potential (+0.74V). hence, Cr is oxidized most easiliy `(CrtoCr^(3+)+3e^(-))` and, therefore, is the strongest reducing agent.
Discussion
1843.

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 provides surface for redox reaction.

Answer» Correct Answer - D
Inert electrode provides surface for oxidation or reduction reaction but not for redox reaction. Hence, (d) is not correct
Discussion
1844.

Determine the equilibrium constant of the following reaction at 298 K : `2Fe^(3+)+Sn^(2+) rarr 2Fe^(2+) +Sn^(4+)` `("Given: "E_(Sn^(4+)//Sn^(2+))^(@)=0.15" volt, "E_(Fe^(3+)//Fe^(2+))^(@)=0.771" volt"` )

Answer» Correct Answer - `K=1.0xx10^(21)`
Calculate `E_(cell)^(@)`. The value is 0.21 volt.
Apply `E_(cell)^(@)=0.0591/2 log K`
Discussion
1845.

An electrochemical cell can behave like an electrolytic cell when_____A. `E_(cell)=0`B. `E_(cell) gt E_(ext)`C. `E_(ext)gtE_(cell)`D. `E_(cell)=E_(ext)`

Answer» Correct Answer - C
When external EMF applied `(E_(ext))` is greater than `E_(cell)`, electrons flow from cathode to anode, i.e., like an electrolytic cell.
Discussion
1846.

An electrochemical cell an behave like an electrolytic cell whenA. `E_("cell")=0`B. `E_("cell")gtE_("ext")`C. `E_("ext")gtE_("cell")`D. `E_("cell")=E_("ext")`

Answer» If an external opposite potential is applied on the galvanic cell and increased reaction continues to take place till the opposing voltage reaches the value 11V.
At this sstage no current flow through the cell and if there is any further increase in the external potential then reaction starts functioning in oppsoite direction.
Hence, this works as an electrolytic cell.
Discussion
1847.

Calculate the value of equilibrium constant for the reaction taking place between Cu(II) and Sn (II) ions in aqueous solution at 298 K. Given : `E_(cu^(2+)//cu)^(@)=0.34" V" , E_(Sn^(2+)//Sn^(4+))^(@)=-0.154" V"`.

Answer» Calculation of `E^(@)` for the redox reaction
The redox reaction that takes place in aqueous solution is :
`(Sn^(2+)(aq) to Sn^(4+)(aq)+2e^(-) , E^(@)=-0.154" V".)/(Cu^(2+)(aq)+2e^(-) to Cu(s) ," " E^(@)=+0.34" V"`.
Add : `Sn^(2+)(aq)+Cu^(2+)(aq) to Sn^(4+)(aq)+Cu(s) , E_(cell)^(@)=+0.186" V"`.
Step II. Calculation of equilibrium constant
`E_(cell)^(@)=(0.0591)/(n)logK`
`E_(cell)^(@)=0.186" V"`, n=2,
`:. " " logK=(2xx(0.186))/(0.0591)=6.294" or "K="Antilog " 6.294=1.97xx10^(6)`
The `E_(cell)` comes out to negative. Therefore, the reaction given above will process spontaneously.
Discussion
1848.

The standard potential of `Cu | Cu^(2+)` electrode is `-0.34` V . It corresponds to the reactionA. `Cu to Cu^(2+) + 2 e^(-)`B. `Cu^(2+) + 2e^(-) to Cu`C. `Cu^(+) to Cu^(2+) + e^(-)`D. `Cu to Cu^(+) - e^(-)`

Answer» Correct Answer - A
`-ve` SRP , oxidation take place .
Discussion
1849.

If `E^(c-)._(Fe^(3+)|Fe)` and `E^(c-)._(Fe^(2+)|Fe)` are `=-0.36 V` and `-0.439V`, respectively, then the value of `E^(c-)._(Fe^(3+)|Fe^(2+))`A. `3x_(2)-2x_(1)`B. `x_(2)-x_(1)`C. `x_(2)+x_(1)`D. `2x_(2)+3x_(2)`

Answer» Correct Answer - a
`E^(c-)._(Fe^(3+)|Fe)=?=x`
`i. Fe^(2+)+2e^(-)rarr Fe` `E^(c-)._(1(Fe^(2+)|Fe))=x_(1)`
`DeltaG^(c-)._(1)=-nFE^(c-)._(1)=-2xxFxxx_(1)`
`ii. Fe^(3+)+3e^(-)rarr Fe` `E^(c-)._(2(Fe^(3+)|Fe))=x_(2)`
`DeltaG^(c-)._(2)=-nFE^(c-)._(2)=-3Fx_(2)`
`iii. Fe^(3+)+e^(-) rarr Fe^(2+)` `DeltaG^(c-)._(3)=-nFE^(c-)._(3)=-Fx`
`DeltaG^(c-)._(3)=DeltaG^(c-)._(2)-DeltaG^(c-)._(1)`
`-Fx=-3Fx_(2)-(-2Fx_(1))`
`:. x=3x_(2)-2x_(1)`
Discussion
1850.

For a cell containing copper and silver electrodes , which of the following statements is correct ?A. copper accepts electrodes and gets reducedB. silver electrode is the negative electrodeC. oxidation occurs at the copper electrodeD. reduction occurs at the copper electrode

Answer» Correct Answer - C
Given half-cell is showing oxidation of copper .
`E_(Cu)^(@) = 0.337` V and `E_(Ag)^(@) = 0.799` V
Discussion