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Electrochemical Cells, Galvanic Cells

Reaction that...

Question

Reaction that takes place at graphite anode in dry cell is:

Z n^{2 +} + 2 e^{-} \to Z n (s)

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Z n (s) \to Z n^{2 +} + 2 e^{-}

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M n^{2 +} + 2 e^{-} \to M n (s)

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M n (s) \to M n^{+} + e^{-} + 1.5 V

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Solution

The correct option is B $Z n (s) \to Z n^{2 +} + 2 e^{-}$
In the dry cell, the oxidation of zinc occurs at graphite anode. Zinc loses two electrons to form ${Z n}^{2 +}$ ion. The electrode reaction is $Z n \to {Z n}^{2 +} + {2 e}^{-}$
Hence, correct option is B.

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Similar questions

M n^{2 +} + 2 e^{-} \to M n; E^{\circ} = - 1.20 V

Z n^{2 +} + 2 e^{-} \to Z n; E^{\circ} = - 0.80 V

E^{⊝}

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

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

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

E^{\circ}

Z n \to Z n^{2 +} + 2 e^{-}; E^{\circ} = + 0.76 v o l t

F e \to F e^{2 +} + 2 e^{-}; E^{\circ} = + 0.41 v o l t

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

E^{\circ}

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

C u \to C u^{2 +} + 2 e^{-}; E^{\circ} = 0.34 V

Z n (s) + C u^{2 +} \to Z n^{2 +} + C u (s)

Z n^{2 +} (a q) + 2 e^{-} \to Z n (s)

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