For the given electrochemical cells -Cell 1-E-Fe2-aq-Fes-0-44 VE-H-aq-H2g-0 VCell 2-E-Mg2-aq-Mgs-2-37 VE-H-aq-H2g-0 VChoose the correct option-

For cell 1: Fe gets oxidised to Fe^2+ because it has a lower SRP than H_2 At anode: Fe (s)⇌ nbsp;Fe^2+(aq)+2e^ nbsp; At cathode: 2H^+(aq)+2e^ ⇌ H_2 (g) E ...

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Standard XII

Chemistry

Electrolytic Cell

For the given...

Question

For the given electrochemical cells :

Cell 1:
$E_{F e^{2 +} (a q) / F e (s)}^{\circ} = - 0.44 V$
$E_{H^{+} (a q) / H_{2} (g)}^{\circ} = 0 V$

Cell 2:
$E_{M g^{2 +} (a q) / M g (s)}^{\circ} = - 2.37 V$
$E_{H^{+} (a q) / H_{2} (g)}^{\circ} = 0 V$

Choose the correct option:

Cell 1 reaction is more spontaneous than cell 2 reaction.

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Cell 2 reaction is more spontaneous than cell 1 reaction.

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E_{c e l l 1}^{\circ} = - 0.44 V

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Both (b) and (c).

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Solution

The correct option is B Cell 2 reaction is more spontaneous than cell 1 reaction.
For cell 1:
$F e$ gets oxidised to $F e^{2 +}$ because it has a lower SRP than $H_{2}$
At anode:
$F e (s) ⇌ F e^{2 +} (a q) + 2 e^{-}$
At cathode:
$2 H^{+} (a q) + 2 e^{-} ⇌ H_{2} (g)$

$E_{c e l l 1}^{\circ} = SRP of cathode - SRP of anode$
$E_{c e l l 1}^{\circ} = E_{H^{+} (a q) / H_{2} (g)}^{\circ} - E_{F e^{2 +} (a q) / F e (s)}^{\circ}$
$E_{c e l l 1}^{\circ} = 0 - (- 0.44) = + 0.44 V$

Similarly for cell 2:
$M g$ gets oxidised to $M g^{2 +}$ because it has a lower SRP than $H_{2}$
At anode:
$M g (s) ⇌ M g^{2 +} (a q) + 2 e^{-}$
At cathode:
$2 H^{+} (a q) + 2 e^{-} ⇌ H_{2} (g)$

$E_{c e l l 2}^{\circ} = SRP of cathode - SRP of anode$
$E_{c e l l 2}^{\circ} = E_{H^{+} (a q) / H_{2} (g)}^{\circ} - E_{M g^{2 +} (a q) / M g (s)}^{\circ}$
$E_{c e l l 2}^{\circ} = 0 - (- 2.37) = + 2.37 V$

Higher the $E_{c e l l}^{\circ}$ value, more is the spontaneity.
$E_{c e l l 2}^{\circ} > E_{c e l l 1}^{\circ}$
Cell 2 reaction is more spontaneous than cell 1 reaction.
Theory :
The magnitude of the cell potential is the measure of the driving force behind a reaction. The larger the value of the cell potential,the farther is the reaction
from equilibrium.

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Similar questions

Q. For a given electrochemical cell:

Given that:

A (s) ⇌ A^{n +} (a q) + n e^{-}

E^{\circ} = - 1.5 V

B (s) ⇌ B^{n +} (a q) + n e^{-}

E^{\circ} = - 3.1 V

Choose the correct statement.

Q. What will be the

E^{0}

value for the given half cell reaction?

F e^{3 +} (a q) + 3 e^{-} \to F e (s)

Given:

E^{0}

value for the half cell reaction,

F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q); E^{0} = 0.77 V

F e^{2 +} (a q) + 2 e^{-} \to F e (s); E^{0} = - 0.44 V

Q. The standard reduction potential at 298 K for the following half all reactions are given below:

Z n^{2 +} (a q) + 2 e^{-} \to Z n (s), E^{0} = - 0.762 V C r^{3 +} (a q) + 3 e^{-} \to C r (s), E^{0} = - 0.740 V 2 H^{+} (a q) + 2 e^{-} \to H_{2} (g), E^{0} = 0.0 V F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q), E^{0} = 0.77 V

Which is the strongest reducing agent?

Q. A galvanic cell is constructed by coupling of an iron and a nickel electrode. If at

298 K

E_{N i^{2 +} (a q) / N i (s)}^{\circ} = - 0.24 V

E_{F e^{2 +} (a q) / F e (s)}^{\circ} = - 0.44 V

then calculate the EMF of the cell

Q. Two electrochemical cells are assembled in which the following reactions occur.

V^{2 +} (a q) + V O^{2 +} (a q) + 2 H^{+} (a q) \to 2 V^{3 +} (a q) + H_{2} O (l); E_{c e l l}^{0} = 0.616 V

V^{3 +} (a q) + A g^{+} (a q) + H_{2} O (l) \to V O^{2 +} (a q) + 2 H^{+} (a q) + A g (s); E_{c e l l}^{0} = 0.439 V

Calculate the

E^{0}

for the half-cell reaction,

V^{3 +} (a q) + e \to V^{2 +} (a q)

Given :

(E_{A g^{+} (a q) / A g (s)}^{0} = 0.799 V)

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Electrolytic Cell

Standard XII Chemistry

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For the given electrochemical cells : Cell 1: E∘Fe2+(aq)/Fe(s)=−0.44 V E∘H+(aq)/H2(g)=0 V Cell 2: E∘Mg2+(aq)/Mg(s)=−2.37 V E∘H+(aq)/H2(g)=0 V Choose the correct option:

For the given electrochemical cells :

Cell 1:
$E_{F e^{2 +} (a q) / F e (s)}^{\circ} = - 0.44 V$
$E_{H^{+} (a q) / H_{2} (g)}^{\circ} = 0 V$

Cell 2:
$E_{M g^{2 +} (a q) / M g (s)}^{\circ} = - 2.37 V$
$E_{H^{+} (a q) / H_{2} (g)}^{\circ} = 0 V$

Choose the correct option: