Question

Calculate the electrode potentials of the following half cell at $298K$.
(a) $Ag\left(s\right)|AgN{O}_3\left(0.1M\right)$
(b) $Co\left(s\right)|C{o}^{2+}\left(0.01M\right)$
Given: $E_{{Ag}^{+}/Ag}=+0.80,E_{C{o}^{2+}/Co}=-0.28V$

Answer 1

As $(E_{red}^o{)}_{A{g}^{+}/Ag}$ is greater than $(E_{red}^o{)}_{C{o}^{+2}/Co}$, thus at cathode reduction of $A{g}^{+}$ will take place

Cathode:- $2A{g}^{+}+2e^{-}⟶2Ag$ $E_{A{g}^{+}/Ag}^o=0.80V$

Anode: $Co⟶C{o}^{2+}+2e^{-}$ $E_{Co/C{o}^{2+}}^o=0.28V$

Cell reaction:-$2A{g}^{+}+Co⟶C{o}^{2+}+2Ag$

$n=2e^{-}$

$E_{cell}^o=E_{A{g}^{+}/Ag}^o+E_{Co/C{o}^{2+}}^o$

$E_{cell}^o=0.80+0.28=1.08V$

Using Nernst Equation

$\Rightarrow E_{cell}=E_{cell}^o-\frac{0.0591}{2}\log \frac{\left[C{o}^{2+}\right]}{[A{g}^{+}{]}^2}$

$\Rightarrow E_{cell}=E_{cell}^o-\frac{0.0591}{2}\log \frac{\left[0.01\right]}{0.1{]}^2}$

$\Rightarrow E_{cell}=E_{cell}^o-O$

Thus, $E_{cell}=1.08V$