The standard electrode potential E - for Daniel cell is -1-1 V- Calculate the - G - for the reaction Zn s - Cu 2- aq - Zn 2- aq - Cu s 1 F -96500 C mol -1 -

212.3 kJ/Mol Δ G^∘ = nFE^∘_cell = 2 × 96500 × 1.1 = 212300 J mol^ 1 = 212.300 KJ mol^ 1 ...

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Gibb's Free Energy

The standard ...

Question

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

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(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)

1.1

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

= 96487

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 +}]}

2 V

300 K

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

300 K

(R = 8 J K^{- 1} m o l^{- 1}, F = 96000 C m o l^{- 1})

1.1 V

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

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