Question

The two half-cell reactions of an electrochemical cell is given as
$A{g}^{+}+e^{-}⟶Ag;E_{A{g}^{+}/Ag}^{\circ }=-0.3995V$
$F{e}^{2+}⟶F{e}^{3+}+e^{-};E_{F{e}^{3+}/F{e}^{2+}}^{\circ }=-0.7120V$
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
$A{g}^{+}+F{e}^{2+}⟶F{e}^{3+}+Ag$
$\Delta E^{\circ }=E_{OA}^{\circ }-E^{\circ }{}_{RA}$
$=-0.3995-(-0.7120)V$
$=-0.3995+0.7120V$
$=+0.3125V$