Question

The standard reduction potential $E^o$ for half reations are $Zn = Zn^{2+} + 2e^- ;\,\,E^o = +0.76 V$ $Fe = Fe^{2+} + 2e^-;\,\, E^o = + 0.41 V$ The EMF of hte cell reaction $Fe^{2+} + Zn = Zn^{2+} + Fe$ is

• A. $-0.35\, V$
• B. $+ 0.35\, V$
• C. $+1.17 V$
• D. $-1.17 V$

Answer 1

The Correct Option is B

$Zn \longrightarrow Zn ^{2+}+2 e^{-} ; E_{ oxi }^{\circ}=+0.76\, V $
or $ E_{\text {red }}^{\circ}=-0.76 \,V $
$ Fe \longrightarrow Fe ^{2+}+2 e^{-} ; E_{ oxi }^{\circ}=+0.41\, V $
or $E_{ red }^{\circ}=-0.41 \,V $
$\therefore E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}(R P)-E_{\text {anode }}^{\circ}(R P) $
$ =-0.41-(0.76) $
$=+0.35 \,V $