Question

For a cell, $Cu(s)|Cu^{2+}(0.001M)||Ag^+(0.01M)|Ag(s)$
the cell potential is found to be $0.43 V$ at $298 K$. The magnitude of standard electrode potential for $Cu^{2+}/{Cu}$ is ____ $× 10^{–2}$ V.
[Given: $E_{Ag^+/Ag^⊖}=0.80$ V and $\frac {2.303RT}{F}=0.06 V$]

Answer 1

$E=E^∘–\frac {0.06}{2}log⁡\frac {[Cu^{2+}]}{[Ag^⊕]^2}$
$E=E^∘–\frac {0.06}{2}log⁡\frac {0.001}{(0.01)^2}$
$0.43=E^∘–0.03$
$E^∘=0.46 V$
$E^∘_{Ag^⊕/Ag}−E^∘_{Cu^{2+}{Cu}}=0.46$
$∴E^∘_{Cu^{2+}{Cu}}=0.8–0.46$
= $0.34 V$
$= 34 × 10^{-2}V$

So, the answer is $34$.