Question

For the following two problems use the given half reactions to "construct" a galvanic cell. Ca2+ + 2 e- ? Ca Eocell=-2.76 V Cu2+ + 2 e- ? Cu Eocell=0.34 V Which of the following statements about this galvanic cell is FALSE? Electrons will flow from the Ca electrode toward the Cu electrode Cu will be the electrode at the cathode The potential of this cell (at standard conditions) will be 3.10 The mass of the Ca electrode will increase during the course of the reaction Negatively charged ions will flow from the salt bridge toward the anode

Answer 1

False statement is The mass of the Ca electrode will increase during the course of the reaction.

The two half cell reactions are

Reduction reaction occurs at cathode

Cu2+ + 2 e- = Cu

Ered = 0.34 V

Oxidation reaction occurs at anode

Ca = Ca2+ + 2e-

Eox = - (-2.76) = 2.76 V

Overall cell reaction

Cu2+ + Ca = Cu + Ca2+

E°cell = Eox + Ered

= 2.76 + 0.34 = 3.10 V

Electrons will flow from anode to cathode or Ca electrode to Cu electrode.

Cu is cathode

Ca is anode

Negatively charged ions will flow from the salt bridge toward the anode.

The mass of the Ca electrode will decrease during the course of the reaction.