The standard electrode potential E- for Daniel cell is -1-1 V- Calculate - G- for the reaction- Zns - Cuaq2- Znaq2- - Cus 1 F - 96500 C-mol

Δ G = nFE = 2× 96500× 2.2 #160;= 212300 J = 212.3 kJ ...

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Gibb's Energy and Nernst Equation

The standard ...

Question

The standard electrode potential $(E^{\circ})$ for Daniel cell is $+ 1.1 V$ . Calculate $Δ G^{\circ}$ for the reaction.
$Z n (s) + C u_{(a q)}^{2 +} \to Z n_{(a q)}^{2 +} + C u (s)$
$(1 F = 96500 C / m o l)$

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(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}) .

The EMF of the Daniell cell is 1.1V . The external EMF to be applied for the following reaction to take place in it is:

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

1.1

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

= 96487

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 +} \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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