Given-i Cu2- - 2e- Cu- E- - 0-337 Vii Cu2- - e- Cu- E- - 0-153 VElectrode potential- E- for the reaction- Cu- - e- Cu- will be

The correct option is D 0.52 V Cu ^ 2+ + e ^ ⟶ Cu can be written fromAt anode: #160; #160; Cu ^ + ⟶ Cu ^ 2+ + e ^ ; E _ ox ^ o = 0.1 ...

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Given:i Cu2...

Question

Given:
(i) $C u^{2 +} + 2 e^{-} \to C u, E^{\circ} = 0.337 V$
(ii) $C u^{2 +} + e^{-} \to C u^{+}, E^{\circ} = 0.153 V$
Electrode potential, $E^{\circ}$ for the reaction, $C u^{+} + e^{-} \to C u$ , will be

0.38 V

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0.52 V

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0.90 V

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0.30 V

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Solution

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C u^{2 +} + 2 e^{-} \to C u; E^{o} = 0.337

C u^{2 +} + e^{-} \to C u^{+}; E^{o} = - 0.153

E^{o}

C u^{+} + e \to C u

E_{C u^{2 +} / C u}^{\circ} = 0.337 V and E_{C u^{+} / C u^{2 +}}^{\circ} = - 0.153 V

E_{C u^{+} / C u}^{\circ}

{C u}^{2 +} + {2 e}^{-} \to C u, E^{\circ} = 0.337 V

{C u}^{2 +} + e^{-} \to C u^{+}, E^{\circ} = 0.153 V

E^{\circ}

{C u}^{+} + e^{-} \to C u,

{C u}^{2 +} + {2 e}^{-} \to C u, E^{o} = + 0.337 V

{C u}^{2 +} + e^{-} \to C u^{+} E^{o} = + 0.153 V

E^{0}

{C u}^{+} + e^{-} \to C u

C u^{2 +} + 2 e^{-} ⟶ C u

E^{0} = 0.337 V

C u^{2 +} + e^{-} ⟶ C u^{+}

E^{0} = 0.153 V

E^{0}

C u^{+} + e^{-} ⟶ C u

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