What is the standard electrode potential of the half-reaction:
\[
\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \, (\text{solid})?
\]
Given that the standard electrode potential for the half-reaction:
\[
\text{Ag}^+ + e^- \rightarrow \text{Ag} \, (\text{solid}) \quad \text{is} \, +0.80 \, \text{V}.
\]
Also, the cell potential for the following reaction is:
\[
\text{Cu}^{2+} + 2\text{Ag} \rightarrow \text{Cu} + 2\text{Ag}^+
\]
is \( 0.46 \, \text{V} \).
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The cell potential can be determined by subtracting the electrode potential of the anode from that of the cathode. For a reduction half-reaction, the standard electrode potential is positive.
The cell potential \( E_{\text{cell}} \) for the reaction is given by:
\[
E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}}
\]
Where:
- \( E_{\text{cathode}} \) is the standard electrode potential for the cathode (where reduction occurs),
- \( E_{\text{anode}} \) is the standard electrode potential for the anode (where oxidation occurs).
In the given reaction:
\[
\text{Cu}^{2+} + 2\text{Ag} \rightarrow \text{Cu} + 2\text{Ag}^+
\]
The cathode is where \( \text{Cu}^{2+} \) is reduced to \( \text{Cu} \), and the anode is where \( \text{Ag} \) is oxidized to \( \text{Ag}^+ \).
We are given:
- The cell potential \( E_{\text{cell}} = 0.46 \, \text{V} \),
- The standard electrode potential of the silver half-reaction is \( E_{\text{Ag}^+/Ag} = +0.80 \, \text{V} \).
Using the equation for cell potential:
\[
0.46 = E_{\text{Cu}^{2+}/Cu} - 0.80
\]
Now solve for \( E_{\text{Cu}^{2+}/Cu} \):
\[
E_{\text{Cu}^{2+}/Cu} = 0.46 + 0.80 = 1.26 \, \text{V}
\]
Thus, the standard electrode potential of the half-reaction is \( 1.26 \, \text{V} \).
