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Cells and Voltage. Volta (1745-1827). 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.
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Volta (1745-1827) • 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.
Electriccurrent (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°).
The following applies to ALL electrochemical cells: node – electrode where xidation occurs. – e- produced at anode. – negative electrode. A o C athode – where eduction occurs. – e- consumed at cathode. – positive electrode. r
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.
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, • withoutmixing solutions.
Salt bridge ions neutralize the products at the electrodes allowing the redox reaction to continue.
Short hand notation of a voltaic cell is often called the line notation. 2Ag+(aq)+Cu(s)→ 2 Ag(s)+Cu2+(aq)
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?
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)