Question

The absolute value of standard reduction potential for $CuS|Cu$ electrode in volts is
Given: $K_{sp}$ of $CuS=8.0\times {10}^{-36};E_{Cu|C{u}^{2+}}^o=-0.34V$.
(take $log2=0.3$)

Answer 1

Half cell reactions are
at anode: $Cu\leftrightharpoons C{u}^{2+}+2e^{-}$
at cathode: $CuS+2e^{-}\leftrightharpoons Cu+S^{2-}$
so, overall cell reaction is: $CuS\leftrightharpoons C{u}^{2+}+S^{2-}$
now, as $E_{cell}^o=(E_{reduction}^o{)}_C-(E_{reduction}^o{)}_A$
$=E_{CuS|Cu}^o-E_{C{u}^{2+}|Cu}^o$
$=x-0.34V$
Cell is in equilibrium, therefore, $E_{cell}=0$
So, $E_{cell}=E_{cell}^o-\frac{0.059}{2}log\left[C{u}^{2+}\right]\left[S^{2-}\right]$
$0=(x-0.34)-\frac{0.059}{2}log\left[C{u}^{2+}\right]\left[S^{2-}\right]$
$E_{cell}^o=\frac{0.059}{2}log{K}_{sp}$
$=0.03log(8\times {10}^{-36})$
$=0.03[3log2-36]$
$=0.03[3\times 0.3-36]$
$=0.03(-35.1)=-1.05V$
$=x-0.34=-1.05V$
$=-0.71V$
hence, $E_{CuS|Cu}^o=-0.71V$