The electrode potentials for Cu2-aq-e- Cu-aq and Cu2-aq-e- Cu s are-0-15 V and-0-50 V respectively- The value of E0Cu2-Cu will be-

Cu^2+(aq) + e^ → Cu^+(aq) nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp;(i) Cu^+(aq) + e^ → Cu(s) nbsp; nbsp; nbsp; nbsp; nbsp; ...

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Byju's Answer

Standard VII

Chemistry

Gibb's Energy and Nernst Equation

The electrode...

Question

The electrode potentials for $C u^{2 +} (a q) + e^{-} \to C u^{+} (a q) and C u^{2 +} (a q) + e^{-} \to C u (s) are + 0.15 V and + 0.50 V$ respectively. The value of $E_{C u^{2 +} / C u}^{0}$ will be:

0.150 V

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0.500 V

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0.325 V

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0.650 V

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Solution

The correct option is C 0.325 V
$C u^{2 +} (a q) + e^{-} \to C u^{+} (a q)$ (i)
$C u^{+} (a q) + e^{-} \to C u (s)$ (ii)
Adding (i) and (ii)
$C u^{2 +} (a q) + 2 e^{-} \to C u (s)$ (iii)
$Δ G_{3} = Δ G_{1} + Δ G_{2}$
$- n F E_{3}^{\circ} = - n F E_{1}^{\circ} - n F E_{2}^{\circ}$
$- 2 F E_{3}^{\circ} = - 1 F E_{1}^{\circ} - 1 F E_{2}^{\circ}$
$2 E_{3}^{\circ} = E_{1}^{\circ} + E_{2}^{\circ}$
$2 E_{3}^{\circ} = 0.15 + 0.5$
$2 E_{3}^{\circ} = 0.65$
$\Rightarrow E_{3}^{\circ} = \frac{0.65}{2} = 0.325 V$

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Similar questions

Q. The electrode potentials for

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and

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The value of

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E_{C u^{2 +} / C u}^{0} = 0.34 V, E_{Z n^{2 +} / Z n}^{0} = - 0.76 V

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The electrode potentials for Cu2+(aq)+e−→Cu+(aq) and Cu2+(aq)+e−→Cu(s) are+0.15 V and+0.50V respectively. The value of E0Cu2+/Cu will be:

The correct option is C 0.325 VCu2+(aq)+e−→Cu+(aq) (i) Cu+(aq)+e−→Cu(s) (ii) Adding (i) and (ii) Cu2+(aq)+2e−→Cu(s) (iii) ΔG3=ΔG1+ΔG2 −nFE∘3=−nFE∘1−nFE∘2 −2FE∘3=−1FE∘1−1FE∘2 2E∘3=E∘1+E∘2 2E∘3=0.15+0.5 2E∘3=0.65 ⇒E∘3=0.652=0.325 V

The electrode potentials for $C u^{2 +} (a q) + e^{-} \to C u^{+} (a q) and C u^{2 +} (a q) + e^{-} \to C u (s) are + 0.15 V and + 0.50 V$ respectively. The value of $E_{C u^{2 +} / C u}^{0}$ will be: