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Electrochemical Series

The standard ...

Question

The standard electrode potentials of $Z n, A g$ and $C u$ are $- 0.76, 0.80$ and $0.34$ volt respectively then_________.

A g

can oxidies

Z n

and

C u

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A g

can reduce

Z n^{2 +}

and

C u^{2 +}

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Z n

can reduce

A g^{+}

and

C u^{2 +}

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C u

can oxidise

Z n

and

A g

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Solution

The correct option is A $A g$ can oxidies $Z n$ and $C u$
Given
$E z {Z n}^{+ 2} / Z n = - 0.76$ or $E Z n / {Z n}^{+ 2} = 0.76$
$E {A g}^{+} / A g = 0.80$ or $E A g / {A g}^{+} = - 0.80$
$E {C u}^{+ 2} / C u = 0.34$ or $E C u / {C u}^{+ 2} = - 0.34$
As oxidation potential order is
$A g < C u < Z n$
$\Rightarrow$ More the oxidation potential more the tendency to oxidise itself.
As $A g$ has less oxidation potential among all, hence $A g$ can oxidise $Z n$ and $C u$

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Similar questions

The standard reducation potentials of

$Z n^{2 +} | Z n, C u^{2 +} | C u$ and $A g^{+} | A g$

are respectively -0.76, 0.34 and 0.8 V. The following cells were constructed.

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

b) $Z n | Z n^{2 +} | | A g^{+} | A g$

c) $C u | C u^{2 +} | | A g^{+} | A g$

What is the correct order $E {^{0}}_{c e l l}$ of these cells?

C u^{2 +} / C u; Z n^{2 +} / Z n; L i^{+} / L i; A g^{+} / A g

H^{+} / H_{2}

+ 0.34 V, - 0.762 V, - 3.05 V, + 0.80 V

0.00 V

C u | C u^{2 +} (1 M) | | A g^{+} (1 M) | A g, E^{o} = 0.46 V

Z n | Z n^{2 +} (1 M) | | C u^{2 +} (1 M) | C u, E^{o} = 1.10 V

Z n | Z n^{2 +} | | A g^{+} (1 M) | A g

(E^{\circ})

A g / A g^{+} (1 M) ∥ C u^{2 +} (1 M) | C u; E^{\circ} = - 0.46 v o l t

Z n / Z n^{2 +} (1 M) ∥ C u^{2 +} (1 M) | C u; E^{\circ} = + 1.10 v o l t

Z n | Z n^{2 +} (1 M) ∥ A g^{+} (1 M) | A g

C u | C u^{2 +} (1 M) | | A g^{+} (1 M) | A g, E^{0} = 0.46 V

Z n | Z n^{2 +} (1 M) | | C u^{2 +} (1 M) | C u, E^{0} = 1.10 V

Z n | Z n^{2 +} (1 M) | | A g^{+} (1 M) | A g

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