The electrode potentials for Cu2-aq - e- Cuaq- and Cuaq- - e- CusAre -0-15 V and -0-50 V respectively-The value of ECu2-Cu- will be

The $E^{\circ}$ for $C u^{2 +} / C u^{+}$ ; $C u^{+} / C u$ , $C u^{2 +} / C u$ are 0.15V, 0.50 V and 0.325 V respectively. The redox cell showing redox reaction $2 C u^{+} \to C u^{2 +} + C u$ is made. The $E^{\circ}$ of this cell reaction and $Δ G^{\circ}$ may be:

Q. Give Nernst equation:
Calculate the electrode potential of the following single electrode.

C u_{(a q)}^{+ +} (C = 0.01 M) / C u; (E^{\circ} = + 0.337 V)

Q. The equilibrium constant for the reaction will be:

C u (s) + C u^{2 +} (a q .) ⇌ 2 C u^{+} (a q .)

E_{C u^{2 +} / C u}^{\circ} = 0.34 V E_{C u^{2 +} / C u^{+}}^{\circ} = 0.15 V

(Given:

log 3.72 = 0.571

)

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The electrode potentials for Cu2+(aq)+e−→Cu+(aq) and Cu+(aq)+e−→Cu(s)Are +0.15V and +0.50V respectively.The value of E∘Cu2+/Cu will be

The correct option is B 0.325VCu2++e−→Cu+ E01=0.15VΔG01=−n1E01FCu++e−→Cu E02=0.50VΔG02=−n2E02FCu2++2e−→Cu ΔG0=ΔG01+ΔG02−nE0F=(−n1E01F)+(−n2E02F)E0=n1E01+n2E02n=0.15×1+0.50×12⇒0.325V

The electrode potentials for $C u_{(a q)}^{2 +} + e^{-} \to C u_{(a q)}^{+}$ and $C u_{(a q)}^{+} + e^{-} \to C u_{(s)}$
Are $+ 0.15 V$ and $+ 0.50 V$ respectively.
The value of $E_{C u^{2 +} / C u}^{\circ}$ will be