The standard reduction potentials of some half cell reactions are given below: `PbO_2+4H^++2e^(-)hArrPb^(2+)+2H_(2)O , E^@=1.455 V` `MnO_4^(-)+8H^++5e

Monjur Alahi

The standard reduction potentials of some half cell reactions are given below:
`PbO_2+4H^++2e^(-)hArrPb^(2+)+2H_(2)O , E^@=1.455 V`
`MnO_4^(-)+8H^++5e^(-)hArrMn^(2+)+4H_(2)O , E^@=1.51 V`
`Ce^(4+)+e^(-)hArrCe^(3+), E^@=1.61 V`
`H_2O_2+2H^++2e^(-)hArr 2H_2O, E^@=1.71 V`
Pick out the correct statement:
A. `Ce^(+4)` will oxidise `Pb^(2+)` to `PbO_2`
B. `MnO_(4)^(-)` will oxidise `Pb^(2+)` to `PbO_2`
C. `H_2O_2` will oxidise `Mn^(+2)` to `MnO_4^(-)`
D. `PbO_2` will oxidise `Mn^(+2)` to `MnO_4^(-)`


0 like 0 dislike
1 views
Share with your friends

1 Answers

Salim Ahmed Kawsar

Call

Correct Answer - A,B,C
On basis of given SRP values

0 like 0 dislike
1 views
Talk Doctor Online in Bissoy App
Following are the values of standard potentials of two half cells, which of the following will be the value of standard cell potential `(E_("cell")^(@

Correct Answer - B `E_("Cell")^(@)=E_("cathode")^(@)-E_("anode")^(@)` `=-0.25-(-0.77)=0.520`

2 Answers 1 views
The standard rectuion potentials for two half-cell reactions are given below `Cd^(2+)(aq)+2e^(-)rarrCd(s),E^(@)=-0.40V` `Ag^(+)(aq)+e^(-)rarrAg(s),E^(

Correct Answer - C ` 2Ag^(+)(aq)+2e^(-)rarr2Ag(s) :E^(@)= 0.80V` `Cd(s)rarrCd^(2+)(aq)+2e^(-):E^(@)=0.400` `2Ag^(+)(aq)+Cd(s)rarrCd^(2+)(aq)+2Ag(s)` `E^(@)=1.20V` `DeltaG=-nE^(@)F` `=-2xx1 .20xx96500` `=-231.6kJ`

2 Answers 1 views
A cell is prepared by dipping a copper rod in 1M `CuSO_(4)` solution and a nickel rod in 1M `NiSO_(4)` solution. The standard reduction potentials of

(i) The nickel electrode with smaller `E^(@)` value (-0.25 V) will work as anode while copper with more `E^(@)` value (+0.34) will work as cathode. (ii) The cell reaction may...

2 Answers 1 views
The standard reduction potentials of `Cu^(2+)//Cu` and `Ag^(+)//Ag` electrodes are 0.337 volt and 0.799 volt respectively. Construct a galvanic cell u

`E_(cell)^(@)=E_(cathode)^(@)-E_(anode)^(@)` =0.799-0.337=0.462 V Applying Nernst equation , `E_(cell)=E_(cell)^(@)-(0.0591)/(n)"log"([Anode])/([Cathode])` Given `" " E_(cell)^(@)=0.462 V , E_(cell)=zero , n=2, [Cu^(2+)]=0.01 M` `0=0.462-(0.0591)/(2)"log"([Cu^(2+)])/([Ag^(+)]^(2))=0.462-0.02955" log"([Cu^(2+)])/([Ag^(+)]^(2)` `"log"([Cu^(2+)])/([Ag^(+)]^(2))=(0.462)/(0.02955)=15.6345` `([Cu^(2+)])/([Ag^(+)]^(2)) ="Antilog" 15.6345=4.3102xx10^(15)` `[Ag^(+)]^(2)=(0.01)/(4.3102xx10^(15))=0.2320xx10^(-17)` `[Ag^(+)]=(0.2320xx10^(-17))^(1//2)=(2.320xx10^(-18))^(1//2)=1.523xx10^(-9) M`.

2 Answers 1 views
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),

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

2 Answers 1 views
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^(-)

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...

2 Answers 1 views
Redox reactions play a pivotal role in chemistry and biology. The value of standard reduction potetials `(E^(@))` of the two half cells reactions deci

Correct Answer - C (c ) `2I^(-)+Cl_(2) to 2Cl^(-)+l_(2)` `I^(-)` ions is oxidised by `Cl_(2)` because e.m.f. of the cell comes out to be positive. `E_(cell)^(@)=E_(Cl_(2)//2Cl^(-))^(@)-E_(2I^(-)//I_(2))^(@)` `=1.36-0.54=0.82" V"`

2 Answers 1 views
The standard redox potentials `E^@` of the following systems are `{:("System",E^@("volts")),((i)MnO_4^(-)+BH^(+)+5e^(-)toMn^(2+)+4H_2O,1.51),((ii)Sr^(

Correct Answer - B,C,D `S.R.P uarr O.A uarr " " S.R.P=-S.O.P` `E_(Ce^(+4)|Ce^(+3))^@gtE_(MnO_4^(-)|Mn^(+2))^@gtE_(Cr_2O_7^(-2)|Cr^(+3))^@gtE_(Sn^(+4)|Sn^(+2))^@` `1. 61` `1.51` `1.33` `0.15`

2 Answers 1 views
Heat of formation of `CH_(4)` are: If given heat: `C(s)+ O_(2)(g) rarr CO_(2) (g) " "DeltaH =-394 KJ` `2H_(2) (g)+ O_(2)(g) rarr 2H_(2)O(l) rarr 2H_(2

Correct Answer - A::B `(1) + (2)-(3)` `DeltaH_("reaction") =-394 -568+892=-70 KJ` `" "=-16.607Kcal.`

2 Answers 1 views
Prove that for two half reactions having potentials `E_(1)` and `E_(2)` which are combined to yield a third half reaction, having a potential `E_(3)`,

`DeltaG_(3)=DeltaG_(1)+DeltaG_(2)` `-n_(3)FE_(3)=-n_(1)FE_(1)-n_(2)FE_(2)` or `n_(3)E_(3)=n_(1)E_(1)+n_(2)E_(2)` or `E_(3)=(n_(1)E_(1)+n_(2)E_(2))/n_(3)`

2 Answers 1 views
X