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Consider a galvanic (voltaic) cell that has the generic metals X and Y as electrodes. If X is more reactive than Y (that is, X more readily reacts to form a cation than Y does), classify the following descriptions by whether they apply to the X or Y electrode.

i. anode
ii. cathode
iii. electrons in the wire flow toward
iv. electrons in the wire flow away
v. cations from salt bridge flow toward
vi. anions from salt bridge flow toward
vii. gains mass
viii. loses mass

User Dan Dye
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1 Answer

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Answer:

X

anode

electrons in the wire flow away

anions from salt bridge flow toward

loses mass

Y

cathode

electrons in the wire flow toward

cations from salt bridge flow toward

gains mass

Step-by-step explanation:

In a galvanic cell, oxidation occurs at the anode while reduction occurs at the cathode. The metal that is more reactive functions as the anode while the less reactive metal functions as the cathode.

Electrons leave the anode and travel via a wire to the cathode. At the anode cations give up electrons and enter into the solution.

At the cathode, cations pick up electrons and are deposited on the cathode leading to a gain in mass at the cathode.

Positive ions from the salt bridge flow towards the cathode while negative ions from the salt bridge flow towards the anode.

User Brett Rigby
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