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Nernst Equation

Adding powder...

Question

Adding powdered $P b$ and $F e$ to a solution containing 1 M each of $P b^{2 +}$ and $F e^{2 +}$ ions would result in the formation of: $(E_{P b^{2 +} / P b}^{0} = - 0.126 V, E_{F e^{2 +} / F e}^{0} = - 0.44 V)$

more of

P b

and

F e^{2 +}

ions

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more of

F e

and

P b^{2 +}

ions

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more of

P b

and

F e

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more of

P b^{2 +}

and

F e^{2 +}

ions

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Solution

The correct option is A more of $P b$ and $F e^{2 +}$ ions
The standard reduction potential of $P b^{+ 2} | P b$ electrode is $- 0.126$ V which less negative than the standard reduction potential of $F e^{2 +} | F e$ electrode which is $- 0.44$ V
Hence, $P b^{2 +}$ ions are reduced to Pb and Fe is oxidized to $F e^{2 +}$ .
Adding powdered $P b$ and $F e$ to a solution containing 1 M each of $P b^{2 +}$ and $F e^{2 +}$ ions would result in the formation of more of $P b$ and $F e^{2 +}$ ions.

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Similar questions

Q. Adding powdered

P b

and

F e

to a solution containing

1.0 M

each of

P b^{2 +}

and

F e^{2 +}

ions would result in the formation of:
(Given :

E_{P b^{2 +} / P b}^{o}

= - 0.13 V and

E_{F e^{2 +} / F e}^{o}

= -0.44 V at 298 K)

Adding powdered Pb and Fe to a solution containing 1 M in each of $P b^{2 +}$ and $F e^{2 +}$ ions would result into the formation of :

Q. Adding powdered

P b

and

F e

to a solution containing

1.0 M

each of

P b^{2 +}

and

F e^{2 +}

ions would result in the formation of:
(Given :

E_{P b^{2 +} / P b}^{o}

= - 0.13 V and

E_{F e^{2 +} / F e}^{o}

= -0.44 V at 298 K)

Q. If

E_{(P b^{2 +} | P b)}^{⊖} = - 0.126 V, E_{(Z n^{2 +} | Z n)}^{⊖} = - 0.763 V

25^{\circ} C

, calculate EMF of the cell:

Z n | Z n^{2 +} (1 M) | | P b^{2 +}

(1 M) | Pb. If an excess of metallic zn added to 1 M concentration of

P b^{2 +}

ions, The concentration of

P b^{2 +}

ions at equilibrium is

2.5 \times 10^{- x} M

value of x is..............

Given: $F e_{(s)} \to F e^{2 +} + 2 e^{-}; E^{o} = + 0.44 V$

$P b_{(s)} \to P b^{2 +} + 2 e^{-}; E^{o} = + 0.13 V$

$A g^{+} e^{-} \to A g; E^{o} = + 0.8 V$

$C u^{2 +} 2 e^{-} \to C u; E^{o} = + 0.34 V$

Which of the following metals will oxidise iron?

a) $A g$

b) $C u$

c) $P b$

d) None of these

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Adding powdered Pb and Fe to a solution containing 1 M each of Pb2+ and Fe2+ ions would result in the formation of: (E0Pb2+/Pb=−0.126V,E0Fe2+/Fe=−0.44V)

Adding powdered $P b$ and $F e$ to a solution containing 1 M each of $P b^{2 +}$ and $F e^{2 +}$ ions would result in the formation of: $(E_{P b^{2 +} / P b}^{0} = - 0.126 V, E_{F e^{2 +} / F e}^{0} = - 0.44 V)$