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 Pb 2- - Pb 0-0-13 V and E Fe 2- - Fe 0-0-44 V at -298 K A- More of P b and F e 2- ionsB- More of Fe and P b 2- ionsC- More of Pb and FeD- More of P b 2- and F e 2- ions

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Coordination and Oxidation Number

Adding powder...

Question

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)

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
As the value of standard electrode potential for $P b^{2 +}$ is more than that of $F e^{2 +}$ i.e.
$E_{P b^{2 +} / P b}^{\circ} > E_{F e^{2 +} / F e}^{\circ}$
So, $F e$ will get oxidised and $P b^{2 +}$ will get reduced. Hence, we will have 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 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)

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 :

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

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..............

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Adding powdered Pb and Fe to a solution containing 1.0 M each of Pb2+ and Fe2+ ions would result in the formation of: (Given : EoPb2+/Pb= - 0.13 V and EoFe2+/Fe = -0.44 V at 298 K)

The correct option is A More of Pb and Fe2+ ionsAs the value of standard electrode potential for Pb2+ is more than that of Fe2+ i.e. E∘Pb2+/Pb>E∘Fe2+/Fe So, Fe will get oxidised and Pb2+ will get reduced. Hence, we will have more of Pb and Fe2+ ions.

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)