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Gibb's Free Energy

The electrode...

Question

The electrode potential for the Daniell cell given is $1.1$ V.
$Z n (s) | Z n^{2 +} (a q) | | C u^{2 +} (a q) | C u (s)$
Write overall cell reaction and calculate the standard Gibb's energy for the reaction. [F $= 96487$ c/mol].

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Solution

$Z n + C u^{2 +} \to Z n^{2 +} + C u$
$Δ G^{o} = - n F E_{c e l l}^{o}$
$= - 2 \times 96487 \times 1.1$
$= - 212271.4$ Joules.

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Similar questions

1.1 V

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

Z n_{(s)} + C u_{(a q)}^{2 +} ⇌ Z n_{(a q)}^{2 +} + C u_{(s)}

= 1.1 v o l t

F = 96500

(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}) .

(E^{\circ})

+ 1.1 V

Δ G^{\circ}

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

(1 F = 96500 C / m o l)

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

M_{(a q)}^{n +} + n e^{-} \to M_{(s)}

E_{M^{n +} / M} = E^{o}

M^{n +} / M - \frac{2.303 R T}{n F} l o g \frac{[M]}{[M^{n +}]}

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