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Cu^(2+)(aq.) is unstable in solution and...

`Cu^(2+)(aq.)` is unstable in solution and under goes simultaneous oxidation and reduction according to the reaction
`2Cu^(+)(aq.)hArr Cu^(2+)(aq.)+Cu(s)`
Choose the correct `E^(@)` for the above reaction if
`E_(Cu^(2+))^(@)//Cu = 0.34 V and E_(Cu^(2+))^(@)//Cu^(+) = 0.15 V`

A

`-0.38 V`

B

`+0.49 V`

C

`+0.38 V`

D

`+0.19 V`

Text Solution

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The correct Answer is:
To solve the problem, we need to determine the standard electrode potential (E°) for the reaction: \[ 2Cu^+(aq) \rightleftharpoons Cu^{2+}(aq) + Cu(s) \] We are given the following standard electrode potentials: - \( E^\circ_{Cu^{2+}/Cu} = 0.34 \, V \) - \( E^\circ_{Cu^{2+}/Cu^+} = 0.15 \, V \) ### Step 1: Write the half-reactions 1. The reduction half-reaction for \( Cu^{2+} \) to \( Cu^+ \): \[ Cu^{2+} + 2e^- \rightarrow Cu^+ \quad (E^\circ = 0.15 \, V) \] 2. The oxidation half-reaction for \( Cu^+ \) to \( Cu \): \[ Cu^+ + e^- \rightarrow Cu \quad (E^\circ = 0.34 \, V) \] ### Step 2: Reverse the oxidation half-reaction To find the standard potential for the oxidation of \( Cu^+ \) to \( Cu^{2+} \), we reverse the first half-reaction: \[ Cu^+ \rightarrow Cu^{2+} + e^- \quad (E^\circ = -0.15 \, V) \] ### Step 3: Combine the half-reactions Now, we can add the two half-reactions together. The overall reaction will be: \[ 2Cu^+ \rightarrow Cu^{2+} + Cu \] ### Step 4: Calculate the standard electrode potential for the overall reaction To find the standard electrode potential for the overall reaction, we use the formula: \[ E^\circ_{cell} = E^\circ_{reduction} - E^\circ_{oxidation} \] Substituting the values we have: \[ E^\circ_{cell} = E^\circ_{Cu^{2+}/Cu^+} - E^\circ_{Cu^+/Cu} \] \[ E^\circ_{cell} = 0.15 \, V - 0.34 \, V \] \[ E^\circ_{cell} = -0.19 \, V \] ### Step 5: Finalize the answer The standard electrode potential for the reaction \( 2Cu^+ \rightleftharpoons Cu^{2+} + Cu \) is: \[ E^\circ = -0.19 \, V \] ### Summary of the steps: 1. Write the half-reactions for the reduction and oxidation processes. 2. Reverse the oxidation half-reaction to find its potential. 3. Combine the half-reactions to form the overall reaction. 4. Calculate the standard electrode potential for the overall reaction using the potentials of the half-reactions.

To solve the problem, we need to determine the standard electrode potential (E°) for the reaction: \[ 2Cu^+(aq) \rightleftharpoons Cu^{2+}(aq) + Cu(s) \] We are given the following standard electrode potentials: - \( E^\circ_{Cu^{2+}/Cu} = 0.34 \, V \) - \( E^\circ_{Cu^{2+}/Cu^+} = 0.15 \, V \) ...
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