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 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 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. 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

(a) A cell is prepared by dipping a zinc rod in 1M zinc sulphate solution and a silver electrode in 1M silver nitrate solution. The standard electrode potential given : E^(@)Zn_(2+1//Zn) = -0.76V, E^(@)A_(g+//)A_(g) = +0.80V What is the effect of increase in concentration of Zn^(2+) " on the " E_(cell) ? (b) Write the products of electrolysis of aqueous solution of NaCI with platinum electrodes. (c) Calculate e.m.f. of the following cell at 298 K: "Ni(s)"//"Ni"^(2+)(0.01M)////"Cu"^(2+)(0.1M)//"Cu(s)" ["Given"E_(Ni2+//Ni)^(@) = -0.025 V E_(Cu2+//Cu)^(@) = +0.34V] Write the overall cell reaction.