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

Correct Answer: 34

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

So, the answer is \(34\).