Home
Class 12
CHEMISTRY
Distinguish clearly between the meanings...

Distinguish clearly between the meanings of `E_(cell)=0 and E_(cell)^@=0`

Text Solution

Verified by Experts

`E_(cell)=0` means that the concentration ratio is the Nernst equation for the cell reduces the `E_(cell)^@` value to zero while `E_(cell)^@=0` indicates a concentration cell.
Promotional Banner

Topper's Solved these Questions

  • ELECTROMOTIVE FORCE

    BODY BOOKS PUBLICATION|Exercise PROBLEMS |50 Videos

  • ELECTROMOTIVE FORCE

    BODY BOOKS PUBLICATION|Exercise EXERCISE |25 Videos

  • ELECTROCHEMISTRY

    BODY BOOKS PUBLICATION|Exercise EXAMPLES|134 Videos

  • General Principles and Processes of Isolation of Elements

    BODY BOOKS PUBLICATION|Exercise Example|108 Videos

Similar Questions

Explore conceptually related problems

Distinguish between crystal lattice (space lattice) and unit cell.

Distinguish between : Hexagonal and monoclinic unit cells

How will you distinguish between the following pairs of term? b) Crystal lattice and unit cell.

Distinguish between : face- centred and end centred unit cells

60% of angiosperms, pollen grains shed at 2 celled stage, but in some species pollen grains are shed at 3 celled stage. Distinguish between 2 celled and 3 celled stage of pollen grains ?

For the redox reaction: Zn(s)+Cu^(2+) (0.1M) rarr Zn^(2+) (1M) + Cu(s) taking place in a cell, E_(cell)^(@) is 1.10V. E_(cell)^(@) for the cell will be [2.303(RT/F)=0.0591]

Construct galvanic cells from the following pairs of half cells and calculate their emf at 25^@C Sn,SnCl_2(1M) E_(Sn^(2+),Sn)^@=-0.14 volt and Fe|FeSO_4 (1M) E_(Fe^(2+),Fe)^@=0.44 volt

A potentiometer is sued for the comparison of emf of two cells E_(1) and E_(2) . For cell E_(1) the no deflection point is obtained at 20cm and for E_(2) the no deflection point is obtained at 30cm. The ratio of their emf's will be

BODY BOOKS PUBLICATION-ELECTROMOTIVE FORCE -EXERCISE
  1. Distinguish clearly between the meanings of E(cell)=0 and E(cell)^@=0

    Text Solution

    |

  2. Is 1M H^+ solution under hydrogen gas at 1 atm capable of oxidising Ag...

    Text Solution

    |

  3. The potential of hydrogen electrode is -118mV. The concentration of H^...

    Text Solution

    |

  4. E^@ for the half cell Zn^(2+)|Zn is -0.76V emf of the cell Zn|Zn^(2+...

    Text Solution

    |

  5. The standard reduction potentials at 298K for the following half react...

    Text Solution

    |

  6. The standard reduction potentials E^@ for the half reactions are as ...

    Text Solution

    |

  7. From the following E^@ values of half cells, (i) A+e to A^(-) , E^@=...

    Text Solution

    |

  8. From the following E^@ values of half cells (i) A to A^(+)+e, E^@=+1...

    Text Solution

    |

  9. From the following E^@ values of half cells (i) A^(3-)to A^(2-) +e,...

    Text Solution

    |

  10. IF the following half cells have the E^@ values as Fe^(3+)+e to Fe^(...

    Text Solution

    |

  11. E^@ (red.) values of the half cells Mg^(2+)//Mg and Cl2//Cl^- are resp...

    Text Solution

    |

  12. For the cell reaction Zn(s)+Mg^(2+) (1M)=Zn^(2+) (1M) +Mg, the emf has...

    Text Solution

    |

  13. E^@ for F2+2e=2F^(-) is 2.8V, E^@ for 1/2 F2+e=F^(-) is

    Text Solution

    |

  14. Delta G^@ of the cell reaction AgCl(s)+1/2H2(g)=Ag(s)+H^+ +Cl^(-) is ...

    Text Solution

    |

  15. The value of equilibrium constant for the feasible cell reaction is

    Text Solution

    |

  16. E^@ for the reaction Fe+Zn^(2+)=Zn+Fe^(2+) is -0.35 V. The given cell...

    Text Solution

    |

  17. A galvanic cell is composed of two hydrogen electrodes, one of which i...

    Text Solution

    |

  18. 1/2 H2(g)+AgCl(s)=H^+ (aq)+Cl^(-) (aq)+Ag(s) occurs in the galvanic c...

    Text Solution

    |

  19. For the cell Tl|Tl^(+) (0.001M) ||Cu^(2+) (0.1M) |Cu. E(cell) at 25^@C...

    Text Solution

    |

  20. (i) E^@ (Cu^(2+),Cu)=0.34V (ii) E^@(Cu^+,Cu)=+0.52V (iii) E^@[O2(...

    Text Solution

    |

  21. Given below the half cell reactions Mn^(2+)+2e to Mn, E^@=-1.18V 2...

    Text Solution

    |