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Standard XII

Chemistry

Electrolysis and Electrolytes

Given the fol...

Question

Given the following half-cell reactions and corresponding reduction potentials:

S NO Reaction $E^{0}$ in volts

I $A + e^{-} ⟶ A^{-}$ $- 0.24 V$

II $B + 2 e^{-} ⟶ B^{2 -}$ $+ 1.25 V$

III $C + 3 e^{-} ⟶ C^{3 -}$ $- 0.25 V$

IV $D + 2 e^{-} ⟶ D^{2 -}$ $+ 0.68 V$

V $E + 4 e^{-} ⟶ E^{4 -}$ $- 0.38 V$

Which combination of two half-cell would result in a cell with largest positive $E_{c e l l}^{0}$ value for a galvanic cell?

II and V

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I and II

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I and III

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II and IV

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Solution

The correct option is A II and V
Electrochemical series helps in the selection of electrode assemblies to construct the galvanic cells of desired EMFs.

For a galvanic cell to provide high cell voltage, the half cell having higher reduction potential and lower reduction potential should be connected.

In the given half cell reactions,
$B + 2 e^{-} \to B^{2 -}; (Highest value)$
$E + 4 e^{-} \to E^{4 -}; (Lowest value)$

Hence, the cell reaction of cell with largest positive $E^{0}$ value will be,
$E^{4 -} + 2 B \to E + 2 B^{2 -}$
$E_{c e l l}^{0} = 1.25 - (- 0.38)$
$E_{c e l l}^{0} = 1.63 V$

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Similar questions

Deduce from the following $E^{⊖}$ values of half cells, what combination of two half cells would results in a cell with the largest potential ?

I. $A + e^{-} \to A^{⊖}, E^{⊖} = - 0.24 V$

II. $B^{⊖} + e^{-} \to B^{2 -}, E^{⊖} = + 1.25 V$

III. $C^{⊖} + 2 e^{-} \to C^{3 -}; E^{⊖} = - 1.25 V$

IV.

A + 2 e^{-} \to D^{2 -}; E^{⊖} = + 0.68 V

Deduced from the following $E^{\circ}$ values of half cells, what combination of two half cells would result in a cell with the largest potential?
(i) $A^{3 -} \to A^{2 -} + e; E^{\circ} = 1.5 V$ (ii) $B^{2 +} + e \to B^{+}; E^{\circ} = 2.1 V$
(iii) $C^{2 +} + e \to C^{+}; E^{\circ} = + 0.5 V$ (iv) $D \to D^{2 +} + 2 e; E^{\circ} = - 1.5 V$

Half cell reactions for some electrodes are given below:

A + e^{-} ⟶ A^{-}; E^{0} = 0.96 V

II.

B^{-} + e^{-} ⟶ B^{2 -}; E^{o} = - 0.12 V

III.

C^{+} + e^{-} ⟶ C; E^{o} = + 0.18 V

IV.

D^{2 +} + {2 e}^{-} ⟶ D; E^{o} = - 1.12 V

The largest potential will be generated in which cell?

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. What will be the

E^{0}

value for the given half cell reaction?

C r^{2 +} (a q) + 2 e^{-} \to C r (s)

Given:

E^{0}

value for the half cell reaction,

C r^{3 +} (a q) + e^{-} \to C r^{2 +} (a q); E^{0} = - 0.41 V

C r^{3 +} (a q) + 3 e^{-} \to C r (s); E^{0} = - 0.74 V

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Electrolysis and Electrolytes

Standard XII Chemistry

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S NO	Reaction	$E^{0}$ in volts
I	$A + e^{-} ⟶ A^{-}$	$- 0.24 V$
II	$B + 2 e^{-} ⟶ B^{2 -}$	$+ 1.25 V$
III	$C + 3 e^{-} ⟶ C^{3 -}$	$- 0.25 V$
IV	$D + 2 e^{-} ⟶ D^{2 -}$	$+ 0.68 V$
V	$E + 4 e^{-} ⟶ E^{4 -}$	$- 0.38 V$