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Correct order...

Question

Correct order of the tendency of metals to loss of electrons.

Z n > C u > A g

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Z n < C u < A g

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Z n > C u < A g

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C u > Z n > A g

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Solution

The correct option is A $Z n > C u > A g$
The tendency of metals to lose electrons can be determined by their position in electrochemical series.
Elements that lose electrons more easily have lower reduction potential. If we consider the reduction potentials of $Z n, C u$ & $A g$ , their order of tendency to lose electrons is given by $Z n > C u > A g$
$E^{o} = - 0.763, + 0.153, + 0.799$

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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 | 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

(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

P b + C u (N O_{3})_{2} \to P b (N O_{3})_{2} + C u

C u + A g (N O_{3})_{2} \to C u (N O_{3})_{2} + A g

Z n + P b (N O_{3})_{2} \to Z n (N O_{3})_{2} + P b

T = 298 K)

C u (s) | C u^{2 +} (1.0 M) | | A g^{+} (1.0 M) | A g (s) E_{c e l l}^{\circ} = 0.46 V

Z n (s) | Z n^{2 +} (1.0 M) | | C u^{2 +} (1.0 M) | C u (s) E_{c e l l}^{\circ} = 1.10 V

E_{c e l l}

Z n | Z n^{2 +} (0.1 M) | | A g^{+} (1.0 M | A g

298 K

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