Question

a. Describe the direction of electron flow between the electrodes when switch S is closed.b. Write the balanced half-reaction equation that will occur in half-cell 1 when switch S is closed.c. In which half-cell will oxidation occur when switch S is closed?

Answer 1

The direction of electron flow is from the Zn electrode to the Pb electrode.

b. The balanced half-reaction equation for the oxidation of Zn in half-cell 1 is: Zn(s) → Zn2+(aq) + 2e−

c. Oxidation will occur in half-cell 1.

When switch S is closed, electrons will flow from the more active metal (Zn) to the less active metal (Pb). This is because the Zn atoms have a lower ionization energy and are therefore more likely to lose electrons.

The electrons flow through the external circuit to the Pb electrode, where they reduce Pb2+ ions to Pb atoms. The overall reaction is:

Zn(s) + Pb2+(aq) → Zn2+(aq) + Pb(s)

So, the direction of electron flow is from the Zn electrode to the Pb electrode.

c. Oxidation is the loss of electrons, so it will occur in the half-cell where electrons are flowing away. so, electrons are flowing away from the Zn electrode, so oxidation will occur in half-cell 1.

Answer 2

answer and explanation

a. the direction of electron flow when switch S is closed is from the Zn to Pb.

the Zn electrode will lose mass while the Pb electrode will gain mass

b. in half cell 1 the following reaction will occur:

Pb²⁺(aq) + 2e -> Pb(s)

c. oxidation occurs on the Zn half cell according to the equation

Zn(s) => Zn²⁺ (aq) + 2e