Question

Define emf. Calculat the emf of the following galvanic cell :
`Zn_((s))+Cu_((aq))^(+2)to Zn_((aq))^(+2)+Cu_((s))`
`E_(zn^(+2//Zn))^(0) =0.76V(anode) ,E_(cu^(+3//Cu))^(0)=+0.34` (Cathode)

Answer 1

EMF : The difference in electrode potentials between radiction potential of cathode and reduction potential of anode is called emf.
emf = reduction potential of cathode - reduction potential of anode
`emf = + 0.34 -(-0.76) implies+ 0.34 +0.76 =1.1V`