Question

You decide to construct a zinc/aluminum galvanic cell in which the electrodes are connected by a wire and the solutions are connected with a salt bridge. One electrode consists of an aluminum bar in a 1.0 M solution of aluminum nitrate. The other electrode consists of zinc bar in a 1.0 M solution of zinc nitrate. a. Which electrode is the cathode and which is the anode? b. What is the direction of electron flow? c. What chemical reactions are occurring at each electrode? d. What is the overall balanced chemical reaction? e. After a period of time, will the bar of zinc become heavier, lighter or stay the same? What about the aluminum bar? f. What is the standard cell potential?

Answer 1

a). To determine cathode and anode :

To answer this question we look at the reduction potential of both metals.

The metal which has more negative electrode potential is considered as anode and other is cathode.

E(red) of Zn²⁺/Zn = -0.762 V

E(red) of Al³⁺/Al = -1.66 V

From these values , anode is Aluminum and cathode is Zn.

b).

Direction of electron flow

Oxidation occur on anode and reduction occur on cathode.

Direction of electron flow is always from anode to cathode since anode has high tendency to loose electrons and cathode has tendency to accept it.

c).

Chemical reactions :

Oxidation: It occurs on anode

Al(s) -- > Al³⁺ +3e-

Reduction:

It occurs at cathode

Zn²⁺ + 2e^- -- > Zn(s)

d)

Overall balanced chemical reaction

For this we need to balance an electrons. Lets do it and write overall reaction.

2Al(s) -- >2 Al³⁺ +6e-

3Zn²⁺ + 6e^- -- > 3Zn(s)

Overall reaction is addition of oxidation and reduction half

2Al(s) -- >2 Al³⁺ +6e-

3Zn²⁺ + 6e^- -- > 3Zn(s)

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2Al(s)+ 3Zn²⁺ -- >2 Al³⁺ + 3Zn(s)

e).

Reduction occurs at cathode and Zn is cathode. In this process metal cations (Zn²⁺) get electrons on cathode and deposit on cathode so Zn becomes heavier.

Whereas Al is anode and oxidation occurs at anode. In this process metal cation goes into solution and leave electrons on anode. It reduces size of anode.

So Al becomes lighter.

f).

To find cell potential we need to use standard reduction potential of anode and cathode the formula is.

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E⁰cell = Ecathode – Eanode

E⁰cell = -0.762 V – (-1.66V)

=0.898 V

E⁰cell = 0.898 V