Question

( quad Z n rightarrow Z n^{2+} quad E^{circ}=+0.76 mathrm{V} )
( +F e^{2+} longrightarrow F e )
( Z n+F e^{2+} longrightarrow Z n^{2+}+F e )
( E^{prime}=0.76 )
So ( E^{circ}=+0.32 mathrm{v} quad=0.32 )

# (4) pH OTU. IM HUIU Electrode potential for the following half-cell reactions are Zn → Zn2+ + 2e-; E° = +0.76 V; Fe → Fe2+ + 2e-; E° = +0.44 V The EMF for the cell reaction Fe2+ + Zn → Zn4+ Fe will be (1) - 0.32 V (2) + 1.20 V (3) - 1.20 V (4) + 0.32 V

Solution