Question

The two half cell reaction of an electrochemical cell is given as 

\[\ce{Ag+ + e- -> Ag}\], `"E"_("Ag"^+//"Ag")^circ` = - 0.3995 V

\[\ce{Fe^{2+} -> Fe^{3+} + e-}\], `"E"_("Fe"^{3+}//"Fe")^{2+}` = - 0.7120 V

The value of EMF will be ______.

Options

• - 0.3125 V

• 0.3125 V

• 1.114 V

• - 1.114 V

Answer 1

0.3125 V

Explanation:

Species that are more harmful when E° (standard reduction potential) has a lower negative value, it works as an oxidising agent instead of a reducing one. As a result, the general response is

\[\ce{Ag+ + Fe^{2+} -> Fe^{3+} + Ag}\]

The value of EMF will be,

`Delta"E"^circ = "E"_"oxidation"^circ - "E"_"reduction"^circ`

= - 0.3995 - (- 0.7120)

= + 0.3125 V