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

II, IV

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II, III

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III, IV

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I, II

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Solution

The correct option is B II, III
As we know,
$E_{c e l l} = E_{c a t h o d e} - E_{a n o d e}$
$I I$ and $I I I$ $\Rightarrow E_{c e l l}^{⊖}$
$= 1.25 (- 1.25)$
$= 2.5$

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Similar questions

Deducted from the following

E^{o}

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^{o} = 1.5 V

(ii)

B^{2 +} + e \to B^{+}; E^{o} = 2.1 V

(iii)

C^{2 +} + e \to C^{+}; E^{o} = + 0.5 V

(iv)

D \to D^{2 +} + 2 e; E^{o} = - 1.5 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) $D \to D^{2 +} + 2 e; E^{\circ} = - 1.5 V$

Q. 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	$E + 4 e^{-} ⟶ E^{4 -}$	$- 0.38 V$