Question

Represent the galvanic cell in which the reaction
Zn(s) + Cu^+2(aq) -------> Zn^+2(aq) + Cu(s)
takes
place. Explain the electrochemical theory of rusting of iron and write reactions
involved in the rusting of iron.

Answer 1

By the end of this section, you will be able to:

Use cell notation to describe galvanic cells

Describe the basic components of galvanic cells

Galvanic cells, also known as voltaic cells, are electrochemical cells in which spontaneous oxidation-reduction reactions produce electrical energy. In writing the equations, it is often convenient to separate the oxidation-reduction reactions into half-reactions to facilitate balancing the overall equation and to emphasize the actual chemical transformations.

Explanation:

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Answer 2

Answer:

 galvanic cells

Galvanic cells, also known as voltaic cells, are electrochemical cells in which spontaneous oxidation-reduction reactions produce electrical energy. In writing the equations, it is often convenient to separate the oxidation-reduction reactions into half-reactions to facilitate balancing the overall equation and to emphasize the actual chemical transformations.

Consider what happens when a clean piece of copper metal is placed in a solution of silver nitrate . As soon as the copper metal is added, silver metal begins to form and copper ions pass into the solution. The blue color of the solution on the far right indicates the presence of copper ions. The reaction may be split into its two half-reactions. Half-reactions separate the oxidation from the reduction, so each can be considered individually.

oxidation:Cu(s)Cu2+(aq)+2e−reduction:2×(Ag+(aq)+e−Ag(s))or2Ag+(aq)+2e−⟶Ag(s)overall:2Ag+(aq)+Cu(s)2Ag(s)+Cu2+(aq)" role="presentation" style="box-sizing: border-box; color: rgb(0, 0, 0); display: table-cell !important; line-height: 0; text-indent: 0px; text-align: center; text-transform: none; font-style: normal; font-weight: normal; font-size: 18.18px; letter-spacing: normal; word-wrap: normal; word-spacing: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 31.958em; min-height: 0px; border: 0px; margin: 0px; padding: 1px 0px; width: 10000em; position: relative;">oxidation:Cu(s)Cu2+(aq)+2e−reduction:2×(Ag+(aq)+e−Ag(s))or2Ag+(aq)+2e−⟶Ag(s)overall:2Ag+(aq)+Cu(s)2Ag(s)+Cu2+(aq)oxidation:Cu(s)Cu2+(aq)+2e−reduction:2×(Ag+(aq)+e−Ag(s))or2Ag+(aq)+2e−⟶Ag(s)overall:2Ag+(aq)+Cu(s)2Ag(s)+Cu2+(aq)

The equation for the reduction half-reaction had to be doubled so the number electrons “gained” in the reduction half-reaction equaled the number of electrons “lost” in the oxidation half-reaction.

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