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Representation of Cell

Cu2+aq+Zns→ Z...

Question

$C u^{2 +} (a q) + Z n (s) \to Z n^{2 +} (a q) + C u (s)$
In the above reaction which of the following is reduced?

Z n (s)

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C u^{2 +} (a q)

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Z n^{2 +} (a q)

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C u (s)

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H_{2} O

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Solution

The correct option is C $C u^{2 +} (a q)$
$C u^{2 +} (a q) + Z n (s) \to Z n^{2 +} (a q) + C u (s), E^{0} = 0.34 - (- 0.76) = 1.10 V$

In the above reaction $C u^{2 +} (a q)$ is getting reduced and $Z n$ is getting oxidised to $Z n^{2 +}$ .

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Similar questions

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

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

E^{\circ}

{Zn(s)|Zn}^{2 +} (aq) | | {Cu}^{2 +} (aq)|Cu(s)

1.10

25^{\circ} C

Zn(s) + {Cu}^{2 +} (aq) \to Cu(s) + {Zn}^{2 +} (aq)

1.94 \times 10^{x}

x

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

E^{\circ}

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

25^{\circ} C

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

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