Given that the standard potential E- of Cu2-Cu and Cu-Cu are 0-340 V and 0-522 V respectively- The E- of Cu2-Cu- is-

Cu^2++2e^ → Cu E^∘=0.340 V Cu→ Cu^++e^ E^∘= 0.522 V Cu^2++e^ → Cu^+ E^∘=? Applying Δ G= nFE^∘ We get, ( 1× F× E^∘)=( 2× F× 0.340)+( 1× F× 0.522) Solving ...

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Question

Given that the standard potential ( $E^{\circ}$ ) of $C u^{2 +} / C u$ and $C u^{+} / C u$ are $0.340 V$ and $0.522 V$ respectively. The $E^{\circ}$ of $C u^{2 +} / C u^{+}$ is:

- 0.182 V

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- 0.158 V

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+ 0.158 V

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0.182 V

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Solution

The correct option is C $+ 0.158 V$
$C u^{2 +} + 2 e^{-} \to C u E^{\circ} = 0.340 V$
$C u \to C u^{+} + e^{-} E^{\circ} = - 0.522 V$
$C u^{2 +} + e^{-} \to C u^{+} E^{\circ} = ?$
Applying $Δ G = - n F E^{\circ}$
We get,
$(- 1 \times F \times E^{\circ}) = (- 2 \times F \times 0.340) + (- 1 \times F \times - 0.522)$
Solving, we get, $E^{\circ} = 0.158 V$

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Similar questions

Q. Given the standard potentials

E^{o} (C u^{2 +} / C u)

and

E^{o} (C u^{+} / C u)

0.340

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0.522

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E^{o} (C u^{2 +} / C u^{+})

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Q. Given that the standard potential (

E^{\circ}

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C u^{2 +} / C u

and

C u^{+} / C u

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respectively. The

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Q. The standard reduction potentials of

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and

C u^{2 +} / C u^{+}

are

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respectively. The standard electrode potential of

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Q. Given that

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

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{C u}^{2 +} / C u

and

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Given that the standard potential (E∘) of Cu2+/Cu and Cu+/Cu are 0.340 V and 0.522 V respectively. The E∘ of Cu2+/Cu+ is:

The correct option is C +0.158 VCu2++2e−→Cu E∘=0.340 V Cu→Cu++e− E∘=−0.522 V Cu2++e−→Cu+ E∘=? Applying ΔG=−nFE∘ We get, (−1×F×E∘)=(−2×F×0.340)+(−1×F×−0.522) Solving, we get, E∘=0.158 V

Given that the standard potential ( $E^{\circ}$ ) of $C u^{2 +} / C u$ and $C u^{+} / C u$ are $0.340 V$ and $0.522 V$ respectively. The $E^{\circ}$ of $C u^{2 +} / C u^{+}$ is: