The cell reaction in Daniel cell is $Z n (s) + C u_{(a q)}^{2 +} \to Z n_{(a q)}^{2 +} + C u_{(s)}$ and Nernst equation for single electrode potential for general electrode reaction $M_{(a q)}^{n +} + n e^{-} \to M_{(s)}$ is $E_{M^{n +} / M} = E^{o}$ $M^{n +} / M - \frac{2.303 R T}{n F} l o g \frac{[M]}{[M^{n +}]}$
Derive Nernst equation for Daniel Cell.

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Solution

Daniell’s cell consists of zinc and copper electrodes. The electrode reactions in Daniell’s cell are :
$A n o d e : {Z n}_{(s)} \to {Z n}^{2 +} + 2 e^{-} C a t h o d e : {C u}^{2 +} + 2 e^{-} \to {C u}_{(s)} N e t : {Z n}_{(s)} + {C u}^{2 +} \to {Z n}^{2 +} + {C u}_{(s)}$
$E_{c e l l} = E_{c e l l}^{0} - \frac{2.303 R T}{2 F} l o g \frac{[{C u}_{(s)}] [{Z n}^{2 +}]}{[{Z n}_{(s)}] [{C u}^{2 +}]}$
Since, the activities of pure copper and zinc metals are taken as unity, hence the Nernst equation for the Daniell’s cell is,
$E_{c e l l} = E_{c e l l}^{0} - \frac{2.303 R T}{2 F} l o g \frac{[{Z n}^{2 +}]}{[{C u}^{2 +}]}$

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1.1

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

= 96487

(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}) .

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

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

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