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)

Define emf. Calculat the emf of the following galvanic cell : `Zn_((s)

1.	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» 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`

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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)

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