Question

If e fe2+/fe =-0.4410 and e fe3+/fe=0.7710 the standard
e.M.F of the reaction fe+2fe3+=3fe2+ will be

Answer 1

Answer:

$Fe^{0} +2Fe^{3+} ----->3Fe^{2+}$    or

$Fe+2Fe^{3+} ----->Fe^{2+} +2Fe^{2+}$ oxidation(Anode)

So the formula in red potentials

$EMF_{cell} =e_{cathode} -e_{Anode}$

                         =0.771V-(-0.441V)

                         =1.212V

Here red potential values are $E^{0} _{Fe^{2+}/Fe } =-0.441V$

                                                 $e^{0} _{Fe^{3+}/Fe^{21}  } =0.771V$