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Byju's Answer

Standard IX

Chemistry

Electrode Potential

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

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.
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. Some standard electrode potentials at

298 K

are given below:

P b^{2 +} / P b - 0.13 V N i^{2 +} / N i - 0.24 V C d^{2 +} / C d - 0.40 V F e^{2 +} / F e - 0.44 V

To a solution containing

0.001 M

X^{2 +}

and

0.1 M

Y^{2 +}

, the metal rods

X

and

Y

are inserted (at

298 K

) and connected by a conducting wire. This resulted in dissolution of

X

. The correct combination(s) of

X

and

Y

, respectively, is(are)
(Given: Gas constant,

R = 8.314 J K^{- 1} m o l^{- 1}

,
Faraday constant,

F = 96500 C m o l^{- 1}

)

Δ G^{o}

for the reaction,

C u^{2 +} + F e \to F e^{2 +} + C u

is:

(Given:

E_{C u^{2 +} | C u}^{o} = + 0.34

E_{F e^{2 +} | F e}^{o} = - 0.44

V).

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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)