Electrode potential for the following half-cell reactions areZn→Zn2++2e-,Eo=+0 76VFe→Fe2++2e-,Eo=+0 44VThe EMF for the cell reaction Fe2++Zn→2++Fe will be(1)-0 32 V(2)+1 - Chemistry - Redox Reactions

Question

E l e c t r o d e   p o t e n t i a l   f o r   t h
e   f o l l o w i n g   h a l f - c e l l   r e a c
t i o n s   a r e Z n → Z n 2 + + 2 e - , E o = + 0 76 V
F e → F e 2 + + 2 e - , E o = + 0 44 V T h e   E M F
  f o r   t h e   c e l l   r e a c t i o n
  F e 2 + + Z n → 2 + + F e   w i l l   b e (
1 ) - 0 32   V ( 2 ) + 1 20   V ( 3 ) - 1 20   V ( 4
) + 0 32   V

Raveena Sharma · Accepted Answer

The reaction is:

Fe2+ + Zn → Fe +
Znâ€‹2+

E0cell = Ecathode -
Eanode
= (-0 44) - (-0 76)
= -0 44+0 76 = 0 32 V

Correct option is (d)