Question

Standard electrode potential of half-cell reactions are given below $C u^{2+}+2 e^{-} \rightarrow C u ; E^{\circ}=0.34\, V $ $Z n^{2+}+2 e^{-} \rightarrow Z n ; E^{\circ}=-0.76 \,V$ What is the emf of the cell?

• A. + 1.10 V
• B. -1.10 V
• C. -0.42 V
• D. +0.42 V

Answer 1

The Correct Option is A

$Z n \rightarrow Z n^{2+}+2 e^{-}$(oxidation half-reaction)
$C u^{2+}+2 e^{-} \rightarrow C u$ (reduction half-reaction)
Emf of the cell $=$ standard oxidation potential of zinc
electrode $+$ standard reduction potential of Cu electrode
$=0.76 \,V +0.34\, V =1.10 \,V$

Answer 2

Ans: Formula for calculating the EMF of the cell 

EMF= E°(cathode) - E°(anode)

Where, 

E°(cathode)= where reduction occur 

E°(anode) = where oxidation occur 

According to the question, 

Standard electrode potentials are: 

E°(Cu2+ + 2e- → Cu) = 0.34 V (cathode)

E°(Zn2+ + 2e- → Zn) = -0.76 V (anode)

Now substitute it in the formula: 

EMF(cell) = E°(cathode) - E°(anode)

 EMF(cell) = 0.34 V - (-0.76 V)

 EMF(cell) = 0.34 V + 0.76 V

 EMF(cell) = 1.10 V

The positive EMF suggest that the reaction is spontaneous.