Question

The $$\mathrm{E}{\phantom{A}}_{\mathrm{cell}}^0$$ for the following cell is:

$$\mathrm{Fe}{\phantom{A}}_{(\mathrm{s})}|\mathrm{Fe}{\phantom{A}}_{(\mathrm{aq})}^{2+}||\mathrm{Zn}{\phantom{A}}_{(\mathrm{aq})}^{2+}|\mathrm{Zn}{\phantom{A}}_{(\mathrm{s})}$$

$$\mathrm{E}{\phantom{A}}_{\mathrm{Fe}}^0$$ = −0.41 V,

$$\mathrm{E}{\phantom{A}}_{\mathrm{Zn}}^0$$ = −0.76 V

Options

• +1.17 V

• −1.17 V

• +0.35 V

• −0.35 V

Answer 1

−0.35 V

Explanation:

Fe is anode and Zn is cathode.

$$\mathrm{E}{\phantom{A}}_{\mathrm{cell}}^0=\mathrm{E}{\phantom{A}}_{\mathrm{cathode}}^0-\mathrm{E}{\phantom{A}}_{\mathrm{anode}}^0$$

= −0.76 − (−0.41)

= −0.35 V