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Nernst Equation

The emf in ...

Question

The emf (in $V$ ) of a Daniel cell containing $0.1 M Z n S O_{4}$ and $0.01 M C u S O_{4}$ solutions at their respective electrodes are:
$(E_{{C u}^{2 +} / C u}^{o} = 0.34 V; E_{{Z n}^{2 -} / Z n}^{o} = - 0.76 V)$

1.10

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1.16

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1.13

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1.07

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Solution

The correct option is D $1.07$
$E_{c e l l} = E_{c e l l}^{o} - \frac{0.059}{2} l o g \frac{[Z n^{2 +}]}{[C u^{2 +}]}$

$E_{c e l l} = 1.1 - \frac{0.059}{2} l o g \frac{[Z n^{2 +}]}{[C u^{2 +}]}$

Now,

$E_{c e l l}^{o} = E_{C u^{2 +} / C u}^{o} - E_{Z n^{2 +} / Z n}^{o}$ $= 1.1 V$

Again

$E_{c e l l} = 1.1 - \frac{0.059}{2} l o g \frac{[0.1]}{[0.01]}$

$E_{c e l l} = 1.1 - 0.029 l o g [10]$

$E_{c e l l} = 1.1 - 0.029$ (as $l o g [10] = 1$ )

$E_{c e l l} = 1.1 - 0.029 = 1.07 V$

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Similar questions

Z n ∣ ∣ {Z n}^{2 +} (1 M) ∣ ∣ ∣ ∣ {C u}^{2 +} (1 M) ∣ ∣ C u

E_{{C u}^{2 +} / C u}^{o} = 0.34 V

E_{{z n}^{2 +} / Z n}^{o} = - 0.76 V

C u S O_{4}

Z n S O_{4}

E_{C u^{2 +} / C u}^{0}

E_{Z n^{2 +} / Z n}^{0}

N H_{3}

{Z n}^{2 +} | Z n

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{C u}^{2 +} | C u

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(V)

(k J {m o l}^{- 1})

E_{A g^{+} / A g}^{\circ} = + 0.80 V, E_{C o^{2 +} / C o}^{\circ} = - 0.28 V, E_{C u^{2 +} / C u}^{\circ} = + 0.34 V, E_{Z n^{2 +} / Z n}^{\circ} = - 0.76 V

A l, C u, F e, M g

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E_{A l^{3} + / A l}^{0} + - 1.66 V, E_{C u^{2} + / C u}^{0} = + 0.34 V

E_{F e^{2} + / F e}^{0} = - 0.44 V, E_{M g^{2} + / M g}^{0} = - 1.40 V

E_{Z n^{2} + / Z n}^{0} = - 0.76 V

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