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arsen [322]
3 years ago
15

A galvanic (voltaic) cell has the generic metals X and Y as electrodes. X is more reactive than Y , that is, X more readily reac

ts to form a cation than Y does. Classify the descriptions by whether they apply to the X or Y electrode.
Chemistry
1 answer:
Margaret [11]3 years ago
7 0

Answer:

(1) cathode: Y

(2) anode X

(3) electrons in the wire flow toward: Y

(4) electrons in the wire flow away from: X

(5) anions from the salt bridge flow toward X

(6) cations from the salt bridge flow toward Y

(7) gains mass: Y

(8) looses mass X

Explanation:

The voltaic cell uses two different metal electrodes, each in an electrolyte solution. The anode will undergo oxidation and the cathode will undergo reduction. The metal of the anode will oxidize, going from an oxidation state of 0 (in the solid form) to a positive oxidation state, and it will become an ion. At the cathode, the metal ion in the solution will accept one or more electrons from the cathode, and the ion’s oxidation state will reduce to 0. This forms a solid metal that deposits on the cathode. The two electrodes must be electrically connected to each other, allowing for a flow of electrons that leave the metal of the anode and flow through this connection to the ions at the surface of the cathode. This flow of electrons is an electrical current that can be used to do work, such as turn a motor or power a light.

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Answer:

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Explanation:

Based on the balanced equation:

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<em>2 moles of HCl produce 1 mole of sulfuric acid</em>

<em />

To solve the problem we need to find the moles of sulfuric acid produced based on the chemical equation. Then, using its molar mass -<em>Molar mass H2SO4 = 98g/mol- </em>we can find the mass of sulfuric acid produced:

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<em>Mass sulfuric acid:</em>

4mol H2SO4 * (98g / mol) =

392g sulfuric acid are produced

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