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Electrolytic Cell

An electroche...

Question

An electrochemical cell has two half cell reactions as,

$A^{2 +} + {2 e}^{-} ⟶ A; E^{o} = 0.34 V$

$X ⟶ X^{2 +} + {2 e}^{-}; E^{o} = 2.37 V$

Caclulate the $E_{c e l l}$ .

2.71V

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2.03V

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-2.71V

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-2.03V

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Solution

The correct option is A 2.71V
Given reactions,
$A^{2 +} + 2 e^{-} \to A, E_{c e l l} = 0.34 V$
$X \to X^{2 +} + 2 e^{-}, E_{c e l l} = 2.37 V$
We know that,
$E_{c e l l} = E_{c a t h o d e} - E_{a n o d e}$
At anode oxidation takes place and at cathode reduction takes place.
So,
$E_{c e l l} = 2.37 - (- 0.34) = 2.71 V$

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Similar questions

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

E^{o} = 0.15 V

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

E^{o} = 0.34 V

E^{o}

C u^{+} + e^{-} \to C u

C u^{+}

P b^{2 +} + 2 e \to P b

(E^{o} = - 0.13 v)

A g^{+} + e \to A g

(E^{o} = + 0.80 v)

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

C u \to C u^{+ +} + 2 e^{-}, E^{0} = 0.34 V

E^{o}

Z n \to Z n^{2 +} + 2 e^{-}, E^{o} = 0.76 V F e \to F e^{2 +} + 2 e^{-}, E^{o} = 0.41 V

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

E^{o}

Z n ⟶ {Z n}^{2 +} + 2 e^{-}

E^{o} = + 0.76 V

F e ⟶ {F e}^{2 +} + 2 e^{-}

E^{o} = + 0.41 V

{F e}^{2 +} + Z n ⟶ {Z n}^{2 +} + F e

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