Question

Calculate \[\ce{E^0_{cell}}\] for the following cell.

\[\ce{Cr_{(s)} | Cr^{3+}_{( aq)} || Fe^{2+}){( aq)} | Fe_{(s)}}\]

Given: `"E"_("Cr"^(3+)//"Cr")^0` = −0.74 V,

`"E"_("Fe"^(2+)//"Fe")^0` = −0.44 V

विकल्प

• −1.18 V

• +1.18 V

• +0.3 V

• −0.3 V

Answer 1

+0.3 V

Explanation:

Cr is anode and Fe is a cathode.

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

= −0.44 − (−0.74)

= +0.3 V