The following reactions occurs in the cell $M g_{(} s) + 2 A g^{+} (0.0001 M) ⇌ M g^{2 +} (0.130 M) + 2 A g_{(s)}$ . Calculate $E_{(c e l l)} E_{(c e l l)}^{0} = 3.17 v o l t$

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Solution

$E_{c e l l} = E_{c e l l}^{0} - \frac{0.0591}{n} l o g Q$

$E_{c e l l} = E_{c e l l}^{0} - \frac{0.0591}{n} l o g \frac{[M g^{+ 2}]}{[A g^{+}]^{2}}$
$M g$ changes to $M g^{+ 2}$ by losing 2 electrons and $2 A g^{+}$ changes to $2 A g$ by gaining 2 electrons. So, exchange of 2 electrons takes place. Hence, $n = 2$ .
$E_{c e l l} = 3.17 - \frac{0.0591}{2} l o g \frac{[0.130]}{[0.0001]^{2}}$
$E_{c e l l} = 3.17 - (0.02955 \times 7.114) V$
$E_{c e l l} = 3.17 - 0.21 = 2.96 V$

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Similar questions

M g (s) + 2 A g^{+} (0.0001 M) \to M g^{2 +} (0.130 M) + 2 A g (s)

E_{(c e l l)} if E_{(c e l l)}^{o} = 3.17 V

M g_{(s)} + 2 A g^{+} (0.0001 M) \to M g^{+ 2} (0.120 M) + 2 A g (s) if E_{c e l l}^{\circ} = 3.17 v o l t

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M g_{(s)} + 2 A g_{(a q)}^{+} \to M g_{(a q)}^{2 +} + 2 A g_{(s)}

E_{M g^{2 +} / M g}^{0} = - 2.37 V

E_{A g^{+} / A g}^{0} = 0.80 V

[M g^{2 +}] = 0.001 M; [A g^{+}] = 0.0001 M

(M g_{(s)}) + 2 A g^{+} (0.001 M) \to M g^{2 +} (0.130 M) + 2 A g_{(s)}

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E_{c e l l}^{o} = 3.17

M g + 2 {A g}^{+} (0.0001 M) \to {M g}^{2 +} (0.01 M) + 2 A g

E_{c e l l}^{o} = 3.177 V

E_{c e l l}

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