The standard electrode potential of the half cells is given below:
$Z n^{2 +} + 2 e^{-} \to Z n; E = - 0.76 V$
$F e^{2 +} + 2 e^{-} \to F e; E = - 0.44 V$
The emf of the cell $F e^{2 +} = Z n^{-} \to Z n^{2 +} + F e$

Question

The standard electrode potential of the half cells is given below-Zn-2-2e-rightarrow Zn- E-0-76 VFe-2-2e-rightarrow Fe- E-0-44 VThe emf of the cell Fe-2-Zn-rightarrow Zn-2-Fe 1-54 V-1-54 V0-32V-0-19 V

Toppr Admin · Accepted Answer

The correct option is C 0-32VE0cell-E0Fe2-Fe-E0Zn-Zn2-xA0- -xA0- -xA0- -xA0- -xA0-x2212-0-44V-0-76-xA0- -xA0-E0-0-32Volt

The standard electrode potential of the half cells is given below:
$Z n^{2 +} + 2 e^{-} \to Z n; E = - 0.76 V$
$F e^{2 +} + 2 e^{-} \to F e; E = - 0.44 V$
The emf of the cell $F e^{2 +} = Z n^{-} \to Z n^{2 +} + F e$
$1.54 V$
$- 1.54 V$
$0.32 V$
$+ 0.19 V$

The correct option is C $0.32 V$
$E_{c e l l}^{0} = (E_{{F e}^{2 +} | F e}^{0} + E_{Z n | {Z n}^{2 +}}^{0})$
$= (- 0.44 V + 0.76)$
$E^{0} = + 0.32 V o l t$

The standard electrode potential of the half cells is given below:Zn2++2e−→Zn;E=−0.76VFe2++2e−→Fe;E=−0.44VThe emf of the cell Fe2+=Zn−→Zn2++Fe1.54V−1.54V0.32V+0.19V

The correct option is C 0.32VE0cell=(E0Fe2+|Fe+E0Zn|Zn2+) =(−0.44V+0.76) E0=+0.32Volt

The standard electrode potential of the half cells is given below:
$Z n^{2 +} + 2 e^{-} \to Z n; E = - 0.76 V$
$F e^{2 +} + 2 e^{-} \to F e; E = - 0.44 V$
The emf of the cell $F e^{2 +} = Z n^{-} \to Z n^{2 +} + F e$
$1.54 V$
$- 1.54 V$
$0.32 V$
$+ 0.19 V$

The correct option is C $0.32 V$
$E_{c e l l}^{0} = (E_{{F e}^{2 +} | F e}^{0} + E_{Z n | {Z n}^{2 +}}^{0})$
$= (- 0.44 V + 0.76)$
$E^{0} = + 0.32 V o l t$