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Representation of Cell

If E Cu | Cu ...

Question

If $E_{C u | C u^{+}}^{0} = - 0.53 V$ and
$E_{C u^{+} | C u^{2 +}}^{0} = - 0.15 V$
For given cell,
$P t (s) | H_{2} (g, 101325 p a s c a l) | H_{2} S O_{4} (a q .) | | C u^{2 +} (a q ., 1 M) | C u (s)$
$E_{c e l l} = 0.34 V a t 298 K$
Then, the molarity of $H_{2} S O_{4} (a q .)$ solution is equal to:

1 M

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0.5 M

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2 M

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0.25 M

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Solution

The correct option is B $0.5 M$
From the given data,
$E_{C u^{2 +} | C u}^{0} = 0.34 V$

Nernst equation:
$E_{c e l l} = E_{c e l l}^{0} - \frac{0.0591}{2} l o g [H^{+}]^{2}$

$0.34 = 0.34 - \frac{0.0591}{2} l o g [H^{+}]^{2}$

$[H^{+}] = 1$ moles per litre

So, molarity of $H_{2} S O_{4} = 0.5 M$

Suggest Corrections

Similar questions

E_{C u | C u^{+}}^{0} = - 0.53 V

E_{C u^{+} | C u^{2 +}}^{0} = - 0.15 V

P t (s) | H_{2} (g, 101325 p a s c a l) | H_{2} S O_{4} (a q .) | | C u^{2 +} (a q ., 1 M) | C u (s)

E_{c e l l} = 0.34 V a t 298 K

H_{2} S O_{4} (a q .)

K_{c}

C u^{2 +} (a q) + S n^{2 +} (a q) ⟶ S n^{4 +} (a q) + C u (s)

25^{\circ} C

E_{C u^{2 +} | C u}^{\circ} = 0.34 V; E_{S n^{4 +} | S n^{2 +}}^{\circ} = 0.15 V

E_{C e l l}^{o}

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

E_{C u^{2 +} / C u}^{\circ} = 0.34 V, E_{Z n^{2 +} / Z n}^{\circ} = - 0.76 V

K_{c}

25^{o} C

Z n (s) + {C u}^{2 +} (a q) = {Z n}^{3 +} (a q) + C u (s)

E_{{Z n}^{2 +} / Z n}^{o} = - 0.76 V E_{C u / {C u}^{2 +}}^{o} = - 0.34 V

E^{\circ}

C u_{(s)} + Z n_{(a q)}^{2 +} \to Z n_{(s)} + C u_{(a q)}^{2 +}

E_{C u^{2 +} / C u}^{0} = 0.34 V, E_{Z n^{2 +} / Z n}^{0} = - 0.76 V

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