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

To determine the standard electrode potential for \(Cu^{2+}/Cu\), we use the Nernst equation for the cell: \(E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.059}{n}\log \frac{[Cu^{2+}]}{[Ag^+]^2}\). Here, \(E_{\text{cell}} = 0.43\) V, \(E^\circ_{Ag^+/Ag} = 0.80\) V, and at \(298 K\), \(\frac{2.303RT}{F} = 0.06\) V. The overall cell reaction is \(Cu + 2Ag^+ \rightarrow Cu^{2+} + 2Ag\), hence \(n=2\).

Since \(E^\circ_{\text{cell}} = E^\circ_{Ag^+/Ag} - E^\circ_{Cu^{2+}/Cu}\), we write:
\(E^\circ_{\text{cell}} = 0.80\) V - \(E^\circ_{Cu^{2+}/Cu}\).
Now, calculating the reaction quotient \(Q = \frac{0.001}{(0.01)^2} = 10\).

Plugging into the Nernst equation:
\(0.43 = E^\circ_{Ag^+/Ag} - E^\circ_{Cu^{2+}/Cu} - \frac{0.06}{2}\log 10\).
Simplifying gives:
\(0.43 = 0.80 - E^\circ_{Cu^{2+}/Cu} - 0.03 \).
Solve for \(E^\circ_{Cu^{2+}/Cu}\):
\(E^\circ_{Cu^{2+}/Cu} = 0.80 - 0.43 - 0.03 = 0.34\) V.

Converting to the required format:
\(E^\circ_{Cu^{2+}/Cu} = 34 \times 10^{-2}\) V, which falls within the specified range of 34,34. Therefore, the magnitude of standard electrode potential for \(Cu^{2+}/Cu\) is \(34 \times 10^{-2}\) V.