Question

IQ2.

Some electrode potential data are shown.
Zn²+ (aq) + 2 e → Zn(s)
Pb²+ (aq) + 2 e → Pb(s)

E = -0.76 V
E = -0.13 V

Which is a correct statement about this cell?

Zn(s)| Zn²+ (aq)||Pb²+ (aq)|Pb(s)

A Electrons travel in the external circuit from zinc to lead.

B The concentration of lead(II) ions increases.

C The maximum EMF of the cell is 0.89 V

D Zinc is deposited.

Answer 1

Step-by-step explanation:

The correct statement about this cell is:

A) Electrons travel in the external circuit from zinc to lead.

This is because the electrode potential of zinc is lower than that of lead. In a galvanic cell, the electron flow goes from the negative electrode (anode), which is where oxidation occurs, to the positive electrode (cathode), where reduction occurs. In this case, the anode is the zinc electrode and the cathode is the lead electrode. Since the zinc electrode has a more negative electrode potential, it is where oxidation occurs and electrons are released, and these electrons travel in the external circuit to the lead electrode, where reduction occurs.