Home
Class 12
CHEMISTRY
The pK(sp) of AgI is 16.07 if the E^(@) ...

The `pK_(sp)` of `AgI` is `16.07` if the `E^(@)` value for `Ag^(+)//Ag` is `0.7991V`, find the `E^(@)` for the half cell reaction `AgI(s) +e^(-) rarr Ag+I^(-)`

Text Solution

Verified by Experts

The correct Answer is:
`E^(@)=-0.149V`.

`E^(@)=0.7991+(0.0591)/(1)logK_(SP)=0.0791-0.0591xxpK_(SP)=-0.149"volt"`
Promotional Banner

Topper's Solved these Questions

  • ELECTROCHEMISRY

    RESONANCE ENGLISH|Exercise CBSE Problems|34 Videos
  • ELECTRO CHEMISTRY

    RESONANCE ENGLISH|Exercise PHYSICAL CHEMITRY (ELECTROCHEMISTRY)|53 Videos
  • EQUIVALENT CONCEPT & TITRATIONS

    RESONANCE ENGLISH|Exercise Part -IV|22 Videos

Similar Questions

Explore conceptually related problems

Dissociation constant for Ag(NH_(3))_(2)^(+) into Ag^(+) and NH_(3) is 6 xx 10^(-14) . Calculate E^(@) for the half reaction. Ag(NH_(3))_(2)^(+) + e rarr Ag + 2NH_(3) Given, Ag^(+) + e rarr Ag has E^(@) = 0.799 V

The cell Pt(H_(2))(1atm) |H^(+) (pH =?) T^(-) (a=1)AgI(s), Ag has emf, E_(298KK) =0 . The electrode potential for the reaction AgI +e^(-) rarr Ag + I^(Theta) is -0.151 volt. Calculate the pH value:-

K_(d) for dissociation of [Ag(NH_(3))_(2)]^(+) into Ag^(+) and NH_(3) is 6 xx 10^(-8) . Calculae E^(@) for the following half reaction. Ag(NH_(3))_(2)^(+) +e^(-) rarr Ag +2NH_(3) Given Ag^(+) +e^(-) rarr Ag, E^(@) = 0.799V

For Zn^(2+) //Zn, E^(@) =- 0.76 , for Ag^(+)//Ag, E^(@) = -0.799V . The correct statement is

Excess of solid AgCl is added to a 0.1 M solution of Br^(c-) ions. E^(c-) for half cell is : AgBr+e^(-) rarr Ag+Br^(c-),E^(c-)=0.095V AgCl+e^(-) rarr Ag+Cl^(c-), E^(c-)=0.222V The value of [Br^(c-)] ion at equilibrium is : [ Given : Antilog (2.152)=142]

Ag^(+) + e^(-) rarr Ag, E^(0) = + 0.8 V and Zn^(2+) + 2e^(-) rarr Zn, E^(0) = -076 V . Calculate the cell potential for the reaction, 2Ag + Zn^(2+) rarr Zn + 2Ag^(+)

The metal-insoluble salt electrode consists of a metal M coated with a porous insoluble salt MX In a solution of X^(-), A good example is Ihe silver, silver-chloride electrode for which the half-cell reaction is AgCI(s)+e Leftrightarrow Ag(s) + CI (aq), where, the reduction of sold silver chloride produces solid silver and releases chloride ion into solute for the cell M(s)|M^(2+) (aq)||Ag^(+) (aq)|Ag(s)" "E_(M//M^(2+))^(@)=+2.37V and E_(Ag//Ag)^(@)=+0.80V Emf of given cell, when concentration of M^(2+) ion and concentration of Ag^(+) ion both are 0.01 M

The metal-insoluble salt electrode consists of a metal M coated with a porous insoluble salt MX In a solution of X^(-), A good example is Ihe silver, silver-chloride electrode for which the half-cell reaction is AgCI(s)+e Leftrightarrow Ag(s) + CI (aq), where, the reduction of sold silver chloride produces solid silver and releases chloride ion into solute for the cell M(s)|M^(2+) (aq)||Ag^(+) (aq)|Ag(s)" "E_(M//M^(2+))^(@)=+2.37V and E_(Ag//Ag)^(@)=+0.80V If cell reaction is in equilibrium stage, value of logarithms of equilibrium constant is

The metal-insoluble salt electrode consists of a metal M coated with a porous insoluble salt MX In a solution of X^(-), A good example is Ihe silver, silver-chloride electrode for which the half-cell reaction is AgCI(s)+e Leftrightarrow Ag(s) + CI (aq), where, the reduction of sold silver chloride produces solid silver and releases chloride ion into solute for the cell M(s)|M^(2+) (aq)||Ag^(+) (aq)|Ag(s)" "E_(M//M^(2+))^(@)=+2.37V and E_(Ag//Ag^(+))^(@)=+0.80V Maximum work done by cell under standard conditions is:

The dissociation constant for [Ag(NH 3 ​ ) 2 ​ ] + into Ag + is 10 −13 at 298K. If E 0 Ag + /Ag ​ =0.8V then E 0 for the half cell [Ag(NH 3 ​ ) 2 ​ ] + +e − →Ag+2NH 3 ​ will be:

RESONANCE ENGLISH-ELECTROCHEMISRY-Advanced Level Problems
  1. Which is/are correct among the following ? Given the half cell emf's E...

    Text Solution

    |

  2. When a solution of conductanes 1.342 mho m^(-1) was placed in a conduc...

    Text Solution

    |

  3. The pK(sp) of AgI is 16.07 if the E^(@) value for Ag^(+)//Ag is 0.7991...

    Text Solution

    |

  4. Voltage of the cell Pt, H(2) (1atm)|HOCN(1.3 xx 10^(-3)M)||Ag^(+) (0....

    Text Solution

    |

  5. K(d) for dissociation of [Ag(NH(3))(2)]^(+) into Ag^(+) and NH(3) is 6...

    Text Solution

    |

  6. The resistance of an aqueous solution containing 0.624g of CuSO(4).5H(...

    Text Solution

    |

  7. Calculate the e.m.f. of the cell Pt|H(2)(1.0atm)|CH(3)COOH (0.1M)||N...

    Text Solution

    |

  8. Calculate the equilibrium concentration of all ions in an ideal soluti...

    Text Solution

    |

  9. The expermental setup for a typical Zn-Ni galvanic cell as shown below...

    Text Solution

    |

  10. At 18^(@)C the mobilities of NH(4)^(+) and CIO(4)^(-) ions are 6.6 xx ...

    Text Solution

    |

  11. 10g fairly concentrated solution of CuSO(4) is electrolyzed using 0.01...

    Text Solution

    |

  12. An electric current is passed through electrolytic cells in series one...

    Text Solution

    |

  13. After electrlysis of NaCI solution with inert electrodes for a certain...

    Text Solution

    |

  14. Same quantity of electricity is being used to liberate iodide (at anod...

    Text Solution

    |

  15. The resistance of two electrolytes X and Y ere found to be 45 and 100 ...

    Text Solution

    |

  16. For 0.0128 N solution fo acetic at 25^(@)C equivalent conductance of t...

    Text Solution

    |

  17. Specific conductance of pure water at 25^(@)C is 0.58 xx 10^(-7) mho c...

    Text Solution

    |

  18. The reduction potential diagram for Cu in acid solution is : Calc...

    Text Solution

    |

  19. For the cellls in opposition, Zn(s)|ZnCl(2)(sol.)|AgCl(s)|Ag|AgCl(s)...

    Text Solution

    |

  20. Prove that Moment of couple= Force times couple arm

    Text Solution

    |