Calculate e.m.f. of the following cell:
`Zn(s)|Zn^(2+)(0.1M)||(0.01M)Ag^(+)|Ag(s)`
Given: `E^(@)` for `Zn^(2+)|Zn=-0.76V`, `E^(@)` for `Ag^(+)|Ag=+0.80V` [Given: `log10=1`]

Write the nearest equation and calculate the e.m.f. of the following cell at 298 K Cu(s)|Cu^(2+)(0.130M)||Ag^(+)(1.00xx10^(-4)M)|Ag(s) Given :E_(Cu^(2+)//Cu)^(@)=0.34V and E_(Ag^(+)//Ag)^(@)=+0.80V

The value of the reaction quotient Q for the cell Zn(s)|Zn^(2+)(0.01M)||Ag^(+)(0.05M)|Ag(s) is

The value of reaction quotient Q for the cell Zn(s)|Zn^(2+)(0.01M)||Ag^(o+)(1.25M)|Ag(s) is

Find the emf of the cell Zn(s)|Zn^(+2)(0.01 M)|KCI "saturated" |Zn^(+2)(1.0 M)|Zn(s)

Find the emf of the cell Zn(s)|Zn^(+2)(0.01 M)|KCI "saturated" |Zn^(+2)(1.0 M)|Zn(s)

Calculate the e.m.f. of the cell Zn//Zn^(2+) (0.1M) "||" Cu^(2+) (0.01M) "|"Cu E_(Zn^(2+)//Zn)^(Theta)= -0.76 " V and "E_(Cu^(2+)//Cu)^(Theta)=0.34V

(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.

The value of the reaction quotient Q for the cell Zn(s)∣Zn^(2+)(0.01M)∣∣Ag^(+)(0.25M)∣Ag(s) is :

The emf of the cell, Ni|Ni^(2+)(1.0M)||Ag^(+)(1.0M)|Ag [E^(@) for Ni^(2+)//Ni =- 0.25 volt, E^(@) for Ag^(+)//Ag = 0.80 volt] is given by : [E^(@) for Ag^(+)//Ag = 0.80 volt]

For the galvanic cell Zn(s)[Zn^2+ (aq)||Ag^+(aq) Ag(s)] , G° of the reaction in joule will be E°_ (Zn^(2+) / Zn )= - 0.76V , E°_( Ag_+/Ag )= 0.80 V