The voltaic cell made up of cobalt, copper and their M²⁺ ions, has E $\degree$_cell = 0.62 V. If E $\degree$ of the cathode half-cell is 0.34 V, what is E $\degree$ of the anode half-cell? Cu²⁺(aq) + Co(s) $\rightarrow$ Cu(s) + Co²⁺(aq)

Question

Quizplus · Accepted Answer

The given voltaic cell can be represented as follows:

Cu(s) | Cu2+(aq) || Co2+(aq) | Co(s)

where the anode half-cell is Co2+(aq) | Co(s) and the cathode half-cell is Cu2+(aq) | Cu(s).

Using the given EF1F1F1S1F1F1F10 value of the cell, we can write:

EF1F1F1S1F1F1F10 = EF1F1F1S1F1F1F10(cathode) - EF1F1F1S1F1F1F10(anode)

0.62 V = 0.34 V - EF1F1F1S1F1F1F10(anode)

EF1F1F1S1F1F1F10(anode) = -0.28 V

Therefore, the answer is A) -0.28 V.

The Voltaic Cell Made Up of Cobalt, Copper and Their °\degree°

The Voltaic Cell Made Up of Cobalt, Copper and Their $\degree$