Question

The standard oxidation potentials, \[\ce{E{^{\circ}_{oxi}}}\] for the half cell reactions are as follows:

\[\ce{Zn -> Zn^2+ + 2e-}\]; \[\ce{E{^{\circ}_{oxi}}}\] = +0.76 V

\[\ce{Fe -> Fe^2+ + 2e-}\]; \[\ce{E{^{\circ}_{oxi}}}\] = +0.41 V

The emf of the cell, \[\ce{Fe^2+ + Zn -> Zn^2 + Fe}\], is ______.

पर्याय

• +0.35 V

• −0.35 V

• +1.17 V

• −1.17 V

Answer 1

The emf of the cell, \[\ce{Fe^2+ + Zn -> Zn^2 + Fe}\], is +0.35 V.

Explanation:

By using formula, we can find out the EMF of a galvanic cell:

EMF of cell = EMF of cathode − EMF of anode

At the cathode, reduction of iron takes place.

Reaction at cathode: \[\ce{Fe^2+ + 2e- -> Fe}\]

At anode, oxidation of zinc takes place.

Reaction at anode: \[\ce{Zn -> Zn^2+ + 2e-}\]

\[\ce{E{^{\circ}_{oxi}}}\] = +0.76 V for Zn

\[\ce{E{^{\circ}_{oxi}}}\] = +0.41 V for Fe

∴ \[\ce{E{^{\circ}_{red}}}\] = −0.76 V for Zn

\[\ce{E{^{\circ}_{red}}}\] = −0.41 V for Fe

\[\ce{E{^{\circ}_{cell}} = E{^{\circ}_{cathode}} - E{^{\circ}_{anode}}}\]

= −0.41 − (−0.76)

= +0.35 V