Question

The \[\ce{E^0_{cell}}\] for the following cell is:

\[\ce{Fe_{(s)} | Fe^{2+}_{( aq)} || Zn^{2+}_{( aq)} | Zn_{(s)}}\]

\[\ce{E^0_{Fe}}\] = −0.41 V,

\[\ce{E^0_{Zn}}\] = −0.76 V

पर्याय

• +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.

\[\ce{E^0_{cell} = E^0_{cathode} - E^0_{anode}}\]

= −0.76 − (−0.41)

= −0.35 V