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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?

a. Describe the direction of electron flow between the electrodes when switch S is-example-1
User Alex Alekser
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2 Answers

7 votes
7 votes

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.

User Aleksandrs Ulme
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21 votes
21 votes

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

User Czchen
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