Question

The two half-cell reactions of an electrochemical cell is given as
$Ag ^{+}+e^{-} \rightarrow Ag ; E_{ Ag ^{+} / Ag }^{\circ}=-0.3995\, V$
$ Fe ^{2+} \rightarrow Fe ^{3+}+e^{-} ; E_{ Fe ^{3+}}^{\circ} / Fe ^{2+}=-0.7120\, V$
The value of EMF will be

• A. $-0.3125 \,V$
• B. $0.3125\, V$
• C. $1.114 \,V$
• D. $-1.114\, V$

Answer 1

Answer: (B)

Species with more negative $E^{\circ}$ (standard reduction potential) generally acts as reducing agent, while with less negative value, $E^{\circ}$ act as oxidising agent. Thus, the overall reaction is
$Ag ^{+}+ Fe ^{2+} \rightarrow Fe ^{3+}+ Ag$
The value of EMF will be,
$ \Delta E^{\circ}=E_{\text {oxidation }}^{\circ}-E_{\text {reduction }}^{\circ} $
$ =-0.3995-(-0.7120)$
$ =+0.3125\, V$