Respuesta :
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.
Further explanation
- Anode: X
- Cathode : Y
- Electrons in the wire flow toward : Y
- Electrons in the wire flow away : X
- Cations from salt bridge flow toward : y
- Anions from salt bridge flow toward : X
- Gains mass : Y
- Loses mass: X
Because X is more reactive than Y, So X is oxidized to X²⁺ and Y²⁺ is reduced to Y. Therefore the overall cell reaction is:
X(s) + Y²⁺(aq) → X²⁺(aq) + Y(s)
X electrode: Electrons in the wire flow away from, Anions from the salt bridge flow toward, Anode, Loses mass
Y electrode: Electrons in the wire flow toward, Cations from the salt bridge flow toward, Cathode, Gains mass
Learn more
- Learn more about galvanic (voltaic) cell https://brainly.com/question/10161182
- Learn more about the generic metals https://brainly.com/question/2146012
- Learn more about cathode https://brainly.com/question/3218763
Answer details
Grade: 9
Subject: chemistry
Chapter: a galvanic (voltaic) cell
Keywords: galvanic (voltaic) cell, the generic metals, cathode, anode, salt bridge
Electrons flow from anode to cathode in a voltaic cell.
A voltaic cell is one in which electrical energy is produced by a redox reaction. Electrons flow from anode to cathode in a voltaic cell and the cell reaction is spontaneous.
Given a generic voltaic cell in which metal X is more reactive than metal Y, it follows that;
- X will be the anode and Y will be the cathode
- Electrons in the wire will flow away from X and flow towards Y
- Anions in the salt bridge flow towards X while cations in the salt bridge flow towards Y.
- There will be a gain in mass at Y and a loss in mass at X.
Learn more: https://brainly.com/question/1340582