Question

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

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

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
$A{g}^{+}+F{e}^{2+}\to F{e}^{3+}+Ag$
The value of EMF will be,
$\Delta E^{\circ }=E_{\text{oxidation }}^{\circ }-E_{\text{reduction }}^{\circ }$
$=-0.3995-(-0.7120)$
$=+0.3125V$