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Cells and Voltage.

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Presentation on theme: "Cells and Voltage."— Presentation transcript:

1 Cells and Voltage

2 Volta ( ) Electrical current made with spontaneous redox. Separated half – reactions - called half-cells. Transfer of electrons occurs through wire. Load in circuit is run with high enough voltage. Voltaic cell converts chemical energy into electrical energy from a spontaneous redox reaction.

3 2 Ag+(aq) + Cu(s) → 2 Ag(s) + Cu2+(aq)

4 Electric current (I) – a flow of electrons.
Electrons flow through conductors – metals – called electrodes. Voltage (V) – ability of the electrons to do work. – also called electrical potential (E°).

5 The following applies to ALL electrochemical cells:
node – electrode where xidation occurs. – e- produced at anode. – negative electrode. A o athode – where eduction occurs. – e- consumed at cathode. – positive electrode. C r

6 Electrons move from anode to cathode ("A to C").
Cathode gains in mass: - e- are accepted by the cations in solution. - Cations are reduced into solid. - “Plate out” onto the cathode. Anode loses mass: - Metal atoms are ionized as they lose e-. - Ions dissolve in solution. - Eventually, anode is completely oxidize to ions.


8 Ions build-up in both half-cells as the cell operates.
Salt bridge - maintains charge balance. Joins the two half-cells. Filled with an electrolyte solution (salt or acid). Allows movement of ions, without mixing solutions.

9 Salt bridge ions neutralize the products at the electrodes allowing the redox reaction to continue.


11 2 Ag+(aq) + Cu(s) → 2 Ag(s) + Cu2+(aq)
Short hand notation of a voltaic cell is often called the line notation. 2 Ag+(aq) + Cu(s) → 2 Ag(s) + Cu2+(aq)

12 A electrochemical cell with Cu (II) oxidizing Zn is constructed.
Identify the anode and cathode. Write the net equation for the reaction. What direction do the electrons move? What is the line notation?

13 Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)
1. Copper is the cathode – reduction. Zinc is the anode – oxidation. 2. Oxidation: Zn(s) → Zn2+(aq) + 2 e– Reduction: Cu2+(aq) + 2 e– → Cu(s) Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) 3. e- move from zinc half-cell to copper half-cell. 4. Zn(s) / Zn2+(aq) // Cu2+(aq) / Cu(s)

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