Home
Class 12
CHEMISTRY
Obtain a relation between E(1)^(@), E(2)...

Obtain a relation between `E_(1)^(@), E_(2)^(@) and E_(3)^(@)` , where `E_(1)^(@)=E_(Fe^(2+)|Fe)^(@), E_(2)^(@)=E_(Fe^(3+)|Fe(2+))^(@) & E_(3)^(@)=E_(Fe^(3+)|Fe)^(@)`.

Text Solution

Verified by Experts

`Fe^(2+)+2e rarr Fe,DeltaG_(1)^(@)=-nFE_(1)^(@)=-2FE_(1)^(@)" ...[1]"`
`Fe^(3+)+e rarr Fe^(2+), DeltaG_(2)^(@)=-1FE_(2)^(@)" ...[2]"`
`Fe^(3+)+3e rarr Fe, DeltaG_(3)^(@)=-3FE_(3)^(@)" ...[3]"`
Adding equation (1) and (2), we get `Fe^(3)+3e rarr Fe`
and `DeltaG_(1)^(@)+DeltaG_(2)^(@)=DeltaG_(3)^(@)`
`"or, "-2FE_(1)^(@)+(-1FE_(2)^(@))=-3FE_(3)^(@) or, 2E_(1)^(@)+E_(2)^(@)=3E_(3)^(@)" ...[4]"`
Equation (4) gives the relation between `E_(1)^(@), E_(2)^(@) and E_(3)^(@)`.
Promotional Banner

Topper's Solved these Questions

  • ELECTROCHEMISTRY

    CHHAYA PUBLICATION|Exercise ADVANCED LEVEL NUMERICAL BANK|38 Videos
  • ELECTROCHEMISTRY

    CHHAYA PUBLICATION|Exercise ENTRANCE QUESTION BANK (WBJEE)|14 Videos
  • ELECTROCHEMISTRY

    CHHAYA PUBLICATION|Exercise SOLVED NCERT TEXTBOOK PROBLEMS (NCERT EXERCISE QUESTIONS)|35 Videos
  • D - AND F - BLOCK ELEMENTS

    CHHAYA PUBLICATION|Exercise PRACTICE SET 8 (Answer the following questions)|10 Videos
  • ENVIRONMENTAL CHEMISTRY

    CHHAYA PUBLICATION|Exercise PRACTICE SET 14(Answer the following questions)|6 Videos

Similar Questions

Explore conceptually related problems

In which pair of electrodes, will the reductant of the first electrode reduce the oxidant of the second electrode - ["Given : "E_(Mg^(2+)|Mg)^(@) lt E_(Al^(3+)|Al)^(@) lt E_(Zn^(2+)|Zn)^(@) lt E_(Fe^(2+)|Fe)^(@) lt E_(Ni^(2+)|Ni)^(@) lt E_(Cu^(2+)|Cu)^(@) lt E_(Ag^(+)|Ag)^(@)]

Given that : E_(Ni^(2+)|Ni)^(@)=-0.25V, E_(Cd^(2+)|Cd)^(@)=-0.40V, e_(Fe^(2+)|Fe)^(@)=-0.44V, E_(Cu^(2+)|Cu)^(@)=+0.34V, E_(Fe^(3+)|Fe)^(@)=-0.036V, E_(2H^(+)|H_(2))^(@)=0.00V, E_(I_(2))|(2I^(-))^(@)=+0.80V Select appropriate species in each of the following cases : Cu^(2+) is reduced but not Fe^(3+)

Find the stronger oxidising agent in each pair, stating reason. (d) Cr^(3+), Fe^(3+) "Given : "E_((1)/(2)Cl_(2)|Cl^(-))^(@)=+1.36V, E_((1)/(2)Br_(2)|Br^(-))^(@)=+1.07V, E_(Cu^(2+)|Cu)^(@)=+0.34V, E_(Ag^(+)|Ag)^(@)=+0.80V, E_(Pb^(2+)|Pb)^(@)=-0.13V, E_(Fe^(2+)|Fe)^(@)=-0.44V , E_(Fe^(3+)|Fe)^(@)=-0.036V, E_(Cr^(3+)|Cr)^(@)=-0.74V

Given that : E_(Ni^(2+)|Ni)^(@)=-0.25V, E_(Cd^(2+)|Cd)^(@)=-0.40V, e_(Fe^(2+)|Fe)^(@)=-0.44V, E_(Cu^(2+)|Cu)^(@)=+0.34V, E_(Fe^(3+)|Fe)^(@)=-0.036V, E_(2H^(+)|H_(2))^(@)=0.00V, E_(I_(2))|(2I^(-))^(@)=+0.80V Select appropriate species in each of the following cases : Ni^(2+) is reduced but not Fe^(2+)

Given that : E_(Ni^(2+)|Ni)^(@)=-0.25V, E_(Cd^(2+)|Cd)^(@)=-0.40V, e_(Fe^(2+)|Fe)^(@)=-0.44V, E_(Cu^(2+)|Cu)^(@)=+0.34V, E_(Fe^(3+)|Fe)^(@)=-0.036V, E_(2H^(+)|H_(2))^(@)=0.00V, E_(I_(2))|(2I^(-))^(@)=+0.80V Select appropriate species in each of the following cases : I^(-) is oxidised but not Br^(-) .

Find the stronger oxidising agent in each pair, stating reason. (a) Cl_(2), Br_(2) "Given : "E_((1)/(2)Cl_(2)|Cl^(-))^(@)=+1.36V, E_((1)/(2)Br_(2)|Br^(-))^(@)=+1.07V, E_(Cu^(2+)|Cu)^(@)=+0.34V, E_(Ag^(+)|Ag)^(@)=+0.80V, E_(Pb^(2+)|Pb)^(@)=-0.13V, E_(Fe^(2+)|Fe)^(@)=-0.44V , E_(Fe^(3+)|Fe)^(@)=-0.036V, E_(Cr^(3+)|Cr)^(@)=-0.74V

Find the stronger oxidising agent in each pair, stating reason. (c) Ag^(+), Cu^(2+) "Given : "E_((1)/(2)Cl_(2)|Cl^(-))^(@)=+1.36V, E_((1)/(2)Br_(2)|Br^(-))^(@)=+1.07V, E_(Cu^(2+)|Cu)^(@)=+0.34V, E_(Ag^(+)|Ag)^(@)=+0.80V, E_(Pb^(2+)|Pb)^(@)=-0.13V, E_(Fe^(2+)|Fe)^(@)=-0.44V , E_(Fe^(3+)|Fe)^(@)=-0.036V, E_(Cr^(3+)|Cr)^(@)=-0.74V

Fe has a much lower tendency to get oxidised than Mn and Cr - explain. [Given : E_(Cr^(2+)|Cr)^(0)=-0.9V, E_(Cr^(3+)|Cr^(2+))^(0)=-0.4V E_(Mn^(2+)|Mn)^(0)=-1.2V, E_(Mn^(3+)|Mn^(2+))^(0)=+1.5V E_(Fe^(2+)|Fe)^(0)=-0.4V, E_(Fe^(3+)|Fe^(2+))^(0)=+0.8V ]

Stability of Fe^(3+) ion is greater than Mn^(3+) ion but less than Cr^(3+) - explain. [Given : E_(Cr^(2+)|Cr)^(0)=-0.9V, E_(Cr^(3+)|Cr^(2+))^(0)=-0.4V E_(Mn^(2+)|Mn)^(0)=-1.2V, E_(Mn^(3+)|Mn^(2+))^(0)=+1.5V E_(Fe^(2+)|Fe)^(0)=-0.4V, E_(Fe^(3+)|Fe^(2+))^(0)=+0.8V ]

CHHAYA PUBLICATION-ELECTROCHEMISTRY-HIGHER ORDER THINKING SKILL (HOTS) QUESTIONS
  1. Given : (5) D+H(2)SO(4)rarr X. Arrange the following elements in asc...

    Text Solution

    |

  2. Between (a) 0.01 (M) H(2)SO(4) and (b) 0.01 (M) H(3)PO(4) solution, wh...

    Text Solution

    |

  3. Represent symbolically the galvanic cells in which the following react...

    Text Solution

    |

  4. Represent symbolically the galvanic cells in which the following react...

    Text Solution

    |

  5. Represent symbolically the galvanic cells in which the following react...

    Text Solution

    |

  6. Given that, E(Zn^(2+)|Zn)^(@)=-0.76V, E(Cu^(2+)|Cu)^(@)=+0.34V, E(...

    Text Solution

    |

  7. Given that, E(Zn^(2+)|Zn)^(@)=-0.76V, E(Cu^(2+)|Cu)^(@)=+0.34V, E(...

    Text Solution

    |

  8. Given that, E(Zn^(2+)|Zn)^(@)=-0.76V, E(Cu^(2+)|Cu)^(@)=+0.34V, E(...

    Text Solution

    |

  9. Given that, E(Zn^(2+)|Zn)^(@)=-0.76V, E(Cu^(2+)|Cu)^(@)=+0.34V, E(...

    Text Solution

    |

  10. Given that, E(Zn^(2+)|Zn)^(@)=-0.76V, E(Cu^(2+)|Cu)^(@)=+0.34V, E(...

    Text Solution

    |

  11. Given that, E(Zn^(2+)|Zn)^(@)=-0.76V, E(Cu^(2+)|Cu)^(@)=+0.34V, E(...

    Text Solution

    |

  12. Given that : E(Ni^(2+)|Ni)^(@)=-0.25V, E(Cd^(2+)|Cd)^(@)=-0.40V, e(F...

    Text Solution

    |

  13. Given that : E(Ni^(2+)|Ni)^(@)=-0.25V, E(Cd^(2+)|Cd)^(@)=-0.40V, e(F...

    Text Solution

    |

  14. Given that : E(Ni^(2+)|Ni)^(@)=-0.25V, E(Cd^(2+)|Cd)^(@)=-0.40V, e(F...

    Text Solution

    |

  15. See whether the cell representation given below is in accordance with ...

    Text Solution

    |

  16. Obtain a relation between E(1)^(@), E(2)^(@) and E(3)^(@) , where E(1)...

    Text Solution

    |

  17. Why is the conductivity of 0.1 (N) HCl solution greater than 0.1 (N) ...

    Text Solution

    |

  18. Between 0.1 (M) NH(4)OH and 0.1 (M) NaOH solutions, which one has a gr...

    Text Solution

    |

  19. On the basis of following data, explain why Co(III) is not stable in a...

    Text Solution

    |

  20. Calomel electrode is represented as - Hg(l)|Hg(2)Cl(2)(s)|HCl(aq), It...

    Text Solution

    |