सेल `Ni//Ni^(2+)(1.0M)||Au^(3+)(1.0M)//Au` (यदि `Ni^(2+)//Ni` के लिये `E^(@)=-0.25V,Au^(3+)//Au` के लिये `E^(@)=1.50V` )का विo वाo बo है

Emf of the cell Ni|Ni^(2+) (0.1M)||Au^(3+) (1.0M)|Au will be (Given E_(Ni^(2+)//Ni)^(@)= 0.25V, E_(Au^(2+)//Au)^(@)= 1.5V )

Ni|Ni^(2+)(1.0M)||Au^(3+)(1.0M)| Au (where E^(@) for Ni^(2+)//Niis -0.25and V and E^(@) for Au^(3+)//Au is (0.150V). What is the emf of the cell ?

emf of cell Ni, Ni^(2+)(1.0M) || Au^(3+)(1.0M), Au is …………If E^@ for Ni^(2+)|Ni is 0.25V, E^@ for Au^(3+)| Au is 1.50 V.

Calculate emf of the following cell reaction at 2968 K : Ni(s)//Ni^(2+)(0.01 M) || Cu^(2+)(0.1 M)//Cu(s) [Given E_(Ni^(2+)//Ni)^(@)=-0.25" V ",E_(Cu^(2+)//Cu)^(@)=+0.34 " V " ] Write the overall cell reaction.

The emf of the cell: Ni//Ni^(2+) (1.0 M)//Au^(3+) (1.0M)//Au (E^(@) = -0.25V " for " Ni^(2+)//Ni and E^(@)= 1.5V "for " Au^(3+)//Au) is

Emf of the cell Ni| Ni^(2+) ( 0.1 M) | Au^(3+) (1.0M) Au will be E_(Ni//Ni(2+))^@ = 0.5=25. E_(Au//Au^(3+))^@ = 1.5 V .