Home
Class 12
CHEMISTRY
E(Ag^(+),Ag)^(0) = 0.8 V . Calculate the...

`E_(Ag^(+),Ag)^(0) = 0.8 V` . Calculate the single electrode potential in `0.01 M Ag^(+)`

Promotional Banner

Topper's Solved these Questions

  • ELECTROCHEMISTRY

    AAKASH SERIES|Exercise Subjective Exercise -4 (Long Answer Questions)|3 Videos

  • ELECTROCHEMISTRY

    AAKASH SERIES|Exercise Subjective Exercise -4 (Short Answer Questions)|5 Videos

  • ELECTROCHEMISTRY

    AAKASH SERIES|Exercise Subjective Exercise -3 (Short Answer Questions)|24 Videos

  • DILUTE SOLUTIONS

    AAKASH SERIES|Exercise EXERCISE - 1.2|55 Videos

  • ELEMENTS OF D - BLOCK

    AAKASH SERIES|Exercise EXERCISE - 5.2|33 Videos

Similar Questions

Explore conceptually related problems

Calculate single electrode potential of Pt, Cl_(2)//Cl^(-) (0.01)(E^(@)= +1.36V)

E^@ of zinc electrode is -0.762V . Calculate the single electrode potential of Zn electrode in decimolar ZnSO_4 solution.

E_(1//2Cl_(2)|Cl^(-))^(@)= + 1.36V . The single electrode potential of Pt, Cl_(2)(1 "atm") |Cl^(-)(0.1M) is

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^(+)

E^(@)Zn^(2+)//Zn = -0.77V . What is the 'E' value of the electrode containing 0.01M Zn^(2+) ions

If the standard electrode protential of Cu^(2+)// Cu electrode is 0.34 V , what is the electrode potential at 0.01 M concentration of Cu^(2+) ? (T = 298^(@) K)

Calculate the Electrode potential of single electrode. Cu^(2+)(0.01M)//Cu" "(E^(@)= +0.337V)

E^(0) of silver electrode is 0.8 V and solubility product of Agl is 1 xx 10^(-16) . Calculate the potential of silver electrode at 25^@C in a saturated Ago solution in water.

Calculate the concentration of silver ions in the cell constructed by using 0.1 M concentration of Cu^(2+) and Ag^(+) ions. Cu and Ag metals are used as electrodes. The cell potential is 0.422 V. [E _(Ag^(2+)//Ag)=0.80V,E_(Cu^(2+)//Cu)=+0.34V]

E_0 for the reaction , Ag^(+) + e^(-) to Ag , is 0.8 V . Calculate E^@ for the reaction. Ag(NH_3)_(2)^(+) + e^(-) to Ag + 2NH_3, if the dissociation constant for. Ag(NH_3)_(2)^(+) into NH_3 and Ag^(+) is 6 xx 10^(-14).