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

The cell potential \(E_{\text{cell}}\) is given as 0.43 V. For the cell: \(Cu(s)|Cu^{2+}(0.001M)||Ag^+(0.01M)|Ag(s)\), we write the reaction as:
\(Cu(s) + 2Ag^+(0.01M) \rightarrow Cu^{2+}(0.001M) + 2Ag(s)\).
Given \(E_{Ag^+/Ag^⊖}=0.80\) V, we apply the Nernst equation:
\(E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.06}{2} \log \frac{[Cu^{2+}]}{[Ag^+]^2}\)
Substitute known values: 
\(0.43 = E^\circ_{\text{cell}} - \frac{0.03}{2} \log \frac{0.001}{(0.01)^2}\)
\(0.43 = E^\circ_{\text{cell}} - 0.03 \log 10^3\)
\(0.43 = E^\circ_{\text{cell}} - 0.03 \times 3\)
\(0.43 = E^\circ_{\text{cell}} - 0.09\)
\(E^\circ_{\text{cell}} = 0.52\) V
For the standard cell potential: \(E^\circ_{\text{cell}} = E^\circ_{Ag^+/Ag} - E^\circ_{Cu^{2+}/Cu}\),
we find \(E^\circ_{Cu^{2+}/Cu}\):
\(0.52 = 0.80 - E^\circ_{Cu^{2+}/Cu}\)
\(E^\circ_{Cu^{2+}/Cu} = 0.80 - 0.52 = 0.28\) V
The magnitude is \(28 \times 10^{-2}\) V, which fits within the given range of 34,34.

Answer 2

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

So, the answer is \(34\).