Question

To find the standard potential of $M^{3+}|M$ electrode, the following cell is constituted:
$Pt|M|M^{3+}\left(0.0018mo{l}^{-1}L\right)\parallel A{g}^{+}\left(0.01mo{l}^{-1}L\right)|Ag$
The emf of this cell is found to be $0.42volt$. Calculate the standard potential of the half reaction $M^{3+}+3e^{-}\to M$.
$E_{A{g}^{+}/Ag}^{\circ }=0.80volt$.

Answer 1

The cell reaction is
$Mg+3Ag+\to 3Ag+M^{3+}$
Applying Nernst equations,
$E_{cell}=E_{cell}^{\circ }-\frac{0.0591}{3}\log \frac{\left[M^{3+}\right]}{[A{g}^{+}{]}^3}$
$0.42=E_{cell}^{\circ }-\frac{0.0591}{3}\log \frac{\left(0.0018\right)}{(0.01{)}^3}=E_{cell}^{\circ }-0.064$
$E_{cell}^{\circ }=(0.42+0.-064)=0.484volt$
$E_{Cell}^{\circ }=E_{Cathode}^{\circ }-E_{Anode}^{\circ }$
or $E_{Anode}^{\circ }=E_{Cathode}^{\circ }-E_{Celll}^{\circ }$
$=(0,80-0.484)=0.32volt$.