[" 6.A zinc rod is dipped in "0.1M" solution of "],[ZnSO_(4)" .The salt is "95%" dissociated at this "],[" dilution at "298K" .Calculate the electrode "],[" potential."[E_(Zn^(2+)/Zn)^(@)=-0.76V]" Delhi "2012C]

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

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) .

A Zn - electrode is dipped in a 0.1 (M) ZnSO_(4) solution. The salt in 95% dissociated in solution. Find the reduction potential of the electrode. Given : Temperature =298K, E_(Zn^(2+)|Zn)^(@)=-0.76V

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