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Building the Supply Curve

The standard ...

Question

The standard electronde potential for Daniell cell is $1.1 V .$ Calculate the standard Gibbs energy for the reaction:

$Z n (s) + C u^{2 +} (a q) \to Z n^{2 +} (a q) + C u (s)$

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Solution

Given:
$E_{(c e l l)}^{Θ} = 1.1 V$

Standard Gibbs energy

We know that,

$Δ_{r} G^{Θ} = - n F E_{(c e l l)}^{Θ}$

number of exchange electrons
$(n) = 2, F = 96487 C {mol}^{- 1}$

By putting the values, we get

$Δ_{r} G^{Θ} = - 2 \times 1.1 V \times 96487 C {mol}^{- 1}$

$Δ_{r} G^{Θ} = - 212271.4 J {mol}^{- 1}$

$Δ_{r} G^{Θ} = - 212.27 k J {mol}^{- 1}$

Final answer:

The standard Gibbs energy for the reaction is $- 212.27 k J {mol}^{- 1} .$

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Similar questions

Z n | Z n^{2 +} (a q) (1.0 M) | | C u^{2 +} (a q) (1.0 M) | | C u

0.350

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1.1 V

Z n (s) + C u^{2 +} (a q .) \to Z n^{2 +} (a q .) + C u (s)

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Z n (s) | Z n^{2 +} (a q) | | C u^{2 +} (a q) | C u (s)

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Z n (s) + C u^{2 +} (a q) \to C u (s) + Z n^{2 +} (a q)

E_{c e l l}^{\circ} = 1.10 V, △ G^{\circ}

(E^{\circ})

Δ G^{\circ}

Z n (s) + C u^{2 +} (a q) ⟶ Z n^{2 +} (a q) + C u (s)

(1 F = 96500 C m o l^{- 1}) .

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