Home
Class 12
CHEMISTRY
A zinc electrode is placed in 0.1M solut...

A zinc electrode is placed in `0.1M` solution of `ZnSO_(4)` at `25^(@)C`. Assuming salt is dissociated to the extent of `20%` at this dilution. The potential. The potential of this electrode at this temperature is `:`
`(E^(c-)._(Zn^(2+)|Zn)=-0.76V)`
`a. 0.79V" ".b. -0.79V" "c. -0.81V." "d. 0.81V`

Promotional Banner

Similar Questions

Explore conceptually related problems

A zinc electrode is placed in 0.1M solution of ZnSO_(4) at 25^(@)C . Assuming salt is dissociated to the extent of 20% at this dilution. The potential of this electrode at this temperature is : (E_(Zn^(2+)|Zn)^(@)=-0.76V) a. 0.79V" ".b. -0.79V" "c. -0.81V." "d. 0.81V

A zinc electrode is placed in 0.1M solution of ZnSO_(4) at 25^(@)C . Assuming salt is dissociated to the extent of 20% at this dilution. The potential of this electrode at this temperature is : (E_(Zn^(2+)|Zn)^(@)=-0.76V) a. 0.79V" ".b. -0.79V" "c. -0.81V." "d. 0.81V

A zinc rod is dipped in 0.1M sol^ of ZnSO_4 . The sald is 95% dissociated at this dilution at 298K. Calculate the electrode potential. (E_(Zn^(2+)//Zn)^@=-0.76V)

Zinc rod is dipped in 0.1 M solution of ZnSO4. The salt is 95% dissociated at this dilution at 298 K. Calculate the electrode potential. [Given EZn2+/Zn = +(-0.76 V]

A zinc rod is dipped in 0.1 M ZnSO_(4) solution. The salt is 95% dissociated of this dilution at 298 K. Calculate electrode potential. (E_(Zn^(2+)//Zn)=-0.76 V) .