Question

The two half-cell reactions of an electrochemical cell is given as
$Ag ^{+}+e^{-} \longrightarrow Ag ; \,\,\,E_{ Ag ^{+} / Ag }^{\circ}=-0.3995\, V$
$Fe ^{2+} \longrightarrow Fe ^{3+}+e^{-} ; E_{ Fe ^{3+} / Fe ^{2+}}^{\circ}=-0.7120\, V$
The value of cell 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 $E^{\circ}$ acts as oxidisinq aqent. Thus, the overall reaction is
$Ag ^{+}+ Fe ^{2+} \longrightarrow Fe ^{3+}+ Ag$
$\Delta E^{\circ}=E_{O A}^{\circ}-E^{\circ}{ }_{ RA }$
$=-0.3995-(-0.7120)\, V$
$=-0.3995+0.7120\, V$
$=+0.3125\, V$