Home
Class 12
CHEMISTRY
Calculate the reduction potential of the...

Calculate the reduction potential of the following electrode at 298 K :
Pt `Cl_(2) (2.5` atm) HCl `(0.01 M), E^(Theta) Cl_(2) | 2 Cl^(-) = 1.36` V .

Text Solution

Verified by Experts

The electrode reaction is :
`Cl_(2) (g) + 2e^(-) to 2 Cl^(-) (aq)`
`E (Cl_(2)|2Cl^(-)) = E^(Theta) (Cl|2Cl^(-)) - ((0.059))/(2) "log" ([Cl^(-)]^(2))/(p (Cl_(2)))`
`p(Cl_(2)) = 2.5` atm , `[Cl^(-)] = 0.1 M` (same as the concentration of HCl)
`E^(Theta) (Cl_(2) | 2Cl^(-)) = 1.36` V
`therefore E(Cl_(2) |Cl^(-)) = 1.36 - ((0.059))/(2) "log" ((0.1)^(2))/((2.5))`
`= 1.36 + 0.0295 xx 4.398 `
`= 1.36 + 0.13 = 1.49` V
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • ELECTROCHEMISTRY

    MODERN PUBLICATION|Exercise PROBLEMS|11 Videos

  • ELECTROCHEMISTRY

    MODERN PUBLICATION|Exercise PRACTICE PROBLEMS|57 Videos

  • D AND F-BLOCK ELEMENTS

    MODERN PUBLICATION|Exercise UNIT PRACTICE TEST|12 Videos

  • GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS

    MODERN PUBLICATION|Exercise COMPETION FILE (Integer Type Numerical Value Type Questions)|5 Videos

Similar Questions

Explore conceptually related problems

Calculate the electrode potential of given electrode Pt, Cl_(2) (1.5 "bar") | 2Cl^(-) (0.01 M) , Solu E_(Cl_(2)//2Cl^(-))^(@) = 1.36 V tion:

Find the electrode potentials of the following electrodes. (a) Pt, H_2 (1 atm)/HCl (0.1 M), (b) Pt, H_2 (2 atm)/ H_2SO_4 (0.01 M)

Calculate the electrode potential of given electrode Pt, Cl_(2) (1.5 bar) 2Cl^(-) (0.01M), E_(Cl_(2)//2Cl^(-))^(@) = 1.36V

Calculate the reduction potential for the following half cells at 25^@C Cl_2|Cl^(-)(2 times 10^-5 M), E_(Cl_2,Cl^-)^@=+1.36V

Calculate EMF of the following half cells : a. Pt, H_(2)(2 atm)|HCl(0.02M)" "E^(c-)=0V b. Pt , Cl_(2)(10 atm)|HCl(0.1 M)" "E^(-c)=1.36V

Calculate the equilibrium constant for the following reaction at 298 K: Cu(s) + Cl_(2) (g) to CuCl_(2) (aq) Given: R =8.314 J K^(-1) mol^(-1) , E_(Cu^(2+)//Cu)^(@) = 0.34 V, E_(1/2Cl_(2)//Cl^(-))^(@) =1.36 V, 1 F = 96500 C mol^(-1)

Calculate the equilibrium constant for the following reaction at 298K. Cu(s)+Cl_(2)(g)toCuCl_(2)(aq) R=8.314JK^(-1)mol^(-1),E_(Cu^(2+)//Cu)^(@)=0.34V,E_(1//2" "Cl_(2)//Cl^(-))^(@)=1.36V,F=96500" C "mol^(-1)

Calculate the electrode potential for underset(1 atm)((Pt)H_2) | H^(+) (c = 0.1)

Derive expression to calculate emf (reduction) of the following half cells at 25^@C Cl_2 (g)|2Cl^(-) (Pt) (1 atm)

Calculate the potential of the following half - cells | cells : a. Cr|Cr^(3+)(0.1 M)||Fe^(2+)(0.01M)|Fe Given : E^(c-)._(Cr^(3+)|Cr)=-0.74V E^(c-)._(Fe^(2+)|Fe)=-0.44V b. 6e^(c-)+BrO_(3)^(c-)(aq)+3H_(2)OrarrBr^(c-)(aq)+6OH(aq) Given :E^(c-)._((BrO_(3)^(c-)|Br^(c-)))=0.61V, [BrO_(3)^(c-)]=2.5xx10^(-3)M,[Br^(c-)]=5.0xx10^(-3)M,pH=9.0 c. Ag|Ag^(o+)(0.1M)||Cl^(c-)(0.02M)|Cl_(2)(g)(0.5atm)|Pt Given E_((Ag^(o+)|Ag))=0.80V,E^(c-)((Cl_(2)|2Cl^(c-))=1.36V d. NO_(3) ^(c-)(aq)+2H^(o+)(aq)+e^(-)rarrNO_(2)+H_(2)O Given : E^(c-)._(NO_(3)^(c-)|NO_(2)=0.78V What will be the reductino potential of the half cell in neutral solution ? Assuming all the other species to be at unit concentration.

Home
Profile