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Electrochemical Series

Arrange Z n ,...

Question

Arrange $Z n$ , $P b$ and $A l$ in the decreasing order of their reducing powers under standard conditions.
Given: $E_{Z n^{2 +} | Z n}^{0} = - 0.76 V, E_{P b^{2 +} | P b}^{0} = - 0.13 V$
$E_{A l^{3 +} | A l}^{0} = - 1.66 V$

Z n > A l > P b

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A 1 > P b > Z n

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Z n > A l > P b

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A l > Z n > P b

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Solution

The correct option is D $A l > Z n > P b$
Reducing power $\propto \frac{1}{Reduction potential}$
So going by the given numerical values of reduction potential at standard conditions, $A l > Z n > P b$ is the order of reducing power.

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Similar questions

Z n

P b

A l

E_{Z n^{2 +} | Z n}^{0} = - 0.76 V, E_{P b^{2 +} | P b}^{0} = - 0.13 V

E_{A l^{3 +} | A l}^{0} = - 1.66 V

Z n

P b

A l

E_{Z n^{2 +} | Z n}^{0} = - 0.76 V, E_{P b^{2 +} | P b}^{0} = - 0.13 V

E_{A l^{3 +} | A l}^{0} = - 1.66 V

A l, C u, F e, M g

Z n

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

L i

K

C a

N a

M g

A l

Z n

F e

P b

C u

H g

A g

I

Z n (s) + C u {({N O}_{3})}_{2} (a q) \to C u (s) + Z n {({N O}_{3})}_{2} (a q)

I I

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