NEET2018ChemistryElectrochemistryActual
If E ^ ( Zn ²⁺, Zn )=-0.763 ~V and E ^ ( Fe ²⁺, Fe )=-0.44 ~V , then the emf of the cell Zn | Zn ²⁺( a =0.00 l ) | Fe ²⁺( a =0.005) | Fe is
Options
- Aequal to 0.323 V
- Bless than 0.323 V
- Cgreater than 0.323 V
- Dequal to 1.103 V
Correct answer
C. greater than 0.323 V
Step-by-step solution
The cell reaction is Zn + Fe ²⁺ Zn ²⁺+ Fe From Nernst equation aligned & E _ cell = E _ cell ^ - 0.0591 n a _ Zn ²⁺ a _ Fe ²⁺ & =(0.763-0.44)- 0.0591 1 0.001 0.005 & =0.364 ~V & aligned