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Faraday's Second Law

E^o for ...

Question

$E^{o}$ for $M g^{2 +} / M g = - 2.37 V, Z n^{2 +} / Z n = - 0.76 V$ and $F e^{2 +} / F e = - 0.44 V$

Which of the following statement is correct?

Zn reduces

F e^{2 +}

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Zn reduces

M g^{2 +}

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M g

oxidises

F e

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Z n

oxidises

F e

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Solution

The correct option is A Zn reduces $F e^{2 +}$
Higher the negative value of reduction potential, the higher will be the reducing power.

Hence, $M g$ reduces both $Z n^{2 +}$ and $F e^{2 +} . Z n$ can reduce $F e^{2 +}$ but not $M g^{2 +}$ .

Similarly, $F e^{2 +}$ can oxidise $Z n$ and $M g$ both.

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Similar questions

$E^{\circ}$ value of $M g^{2 +} | M g, Z n^{2 +} | Z n$ and $F e^{2 +} | F e$ are – 2.37 V, - 0.76 V and 0.44 V respectively. Which of the following is correct?

Q. SRP values of

M g^{2 +} | M g

F e^{2 +} | F e

and

Z n^{2 +} | Z n

are

- 2.37 V, - 0.44 V

and

- 0.76 V

respectively. The correct statement is:

Q. Standard reduction potential values for the electrodes are given below

M g^{2 +} + 2 e^{-} \to M g; E^{0} = - 2.37 V

Z n^{2 +} + 2 e^{-} \to Z n; E^{0} = - 0.76 V

F e^{2 +} + 2 e^{-} \to F e; E^{0} = - 0.44 V

Which of the following statements is correct?

E^{⊖} o f M g^{2 +} | M g, Z n^{2 +} | Z n,

and

F e^{2 +} | F e

are

- 2.37 V

- 0.76 V

, and -

0.44 V

, respectively. Which of the following is correct ?

Q. The standard reduction potential

E^{o}

for half reactions are,

Z n \to Z n^{2 +} + 2 e^{-}; E^{o} = - 0.76 V

F e^{2 +} + 2 e^{-} \to F e; E^{o} = + 0.41 V

The emf of the cell reaction;

F e^{2 +} + Z n \to Z n^{2 +} + F e

is:

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Eo for Mg2+/Mg=−2.37V,Zn2+/Zn=−0.76V and Fe2+/Fe=−0.44V Which of the following statement is correct?

The correct option is A Zn reduces Fe2+Higher the negative value of reduction potential, the higher will be the reducing power. Hence, Mg reduces both Zn2+ and Fe2+.Zn can reduce Fe2+ but not Mg2+. Similarly, Fe2+ can oxidise Zn and Mg both.

$E^{o}$ for $M g^{2 +} / M g = - 2.37 V, Z n^{2 +} / Z n = - 0.76 V$ and $F e^{2 +} / F e = - 0.44 V$

Which of the following statement is correct?

The correct option is A Zn reduces $F e^{2 +}$
Higher the negative value of reduction potential, the higher will be the reducing power.

Hence, $M g$ reduces both $Z n^{2 +}$ and $F e^{2 +} . Z n$ can reduce $F e^{2 +}$ but not $M g^{2 +}$ .

Similarly, $F e^{2 +}$ can oxidise $Z n$ and $M g$ both.