What is powering this clock?. How much Voltage You can see the battery is missing and the clips are attached to the terminals. What is the voltage required.

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Presentation transcript:

What is powering this clock?

How much Voltage You can see the battery is missing and the clips are attached to the terminals. What is the voltage required to run the clock?

Make a Battery Place a drop of Copper nitrate on one end of the paper. Place a piece of copper in the center of the wet spot

Make a Battery On the other end place a drop of zinc nitrate and place a piece of zinc in the wet solution.

Make a Battery Add a couple of drops of KNO 3 in the middle of the two solutions to make a salt bridge.

Make a Battery V Touch the probe leads to the two metals as pictured here. Record the voltage.

What’s the sign? If the reading is negative, switch the leads to the other metals. You want to get a positive voltage reading. Record the metal that is at the red lead and the metal at the black lead.

REDUCE RED CATS This is the way I remember that reduction occurs at the cathode and it is at the red lead. Reduction ? Oxidation ?

Look at the Standard Reduction Potential Table Cu 2+ Cu Zn 2+ Zn

Find the voltage for each pair of metals you have. ?

Cu e -  Cu 0.34 volts Zn  Zn e volts 1.10 volts 1.1 volts Cu(NO 3 ) 2 Cu Zn Zn(NO 3 ) 2

What is the purpose of the salt bridge? 1.1 volts Cu(NO 3 ) 2 Cu Zn Zn(NO 3 ) 2

What is powering this clock?

How much Voltage You can see the battery is missing and the clips are attached to the terminals. What is the voltage required to run the clock?

After adding the phenolphthalein around the strip of magnesium a pink color is observed. Also there are tiny bubbles all along the sides of the magnesium

Lead Battery Anode : Pb (s) + HSO 4 -  PbSO 4(s) +H + (aq) 2 e - cathode: PbO 2(s) + 3 H + +HSO 4 + 2e -  PbSO 4 + H 2 O V V V

Mercury Battery STEEL cathode HgO in KOH Zn(OH) 2 Zn container anode Watches, pacemakers, calculators

Rechargeable Nickel-cadmium anode Cd + OH -  Cd(OH) e - cathode NiO(OH) S + H 2 O  Ni(OH) 2 + OH - Recharge many times because the solid products adhere to the surface of the electrode renewing the battery.

Corrosion Corrosion is the oxidative deterioration of a metal such at rust. Drop of water Fe --> Fe e - anode O 2 from the air O 2 + 4H e -  2 H 2 O cathode Rust

How can you prevent corrosion? Look at the equation and prevent the reaction from happening. What can you do?

Electrochemical Cells There are 2 types of cells –Galvanic also called voltaic is a spontaneous reaction that produces an electric current - Electrolytic requires an outside source to supply the current such as a battery or electrical outlet

Electroplating Example of an Electrolytic cell –Silverplated dinnerware - Silver is a soft metal what would happen if you used a solid silver fork?

Electrolysis Battery Red lead + anode Black lead cathode - Graphite electrodes Na 2 SO 4(aq)

What is happening?? Reduction: 2 H 2 O (l)  H 2(g) + 2 H - (aq) V 2 H + (aq) 3 e -  H 2(g) 0.00 V Na + (aq) + e -  Na (s) V Oxidation: 2 H 2 O  O 2(g) + 4H + (aq) + 4e V 2 SO 4 2-  S 2 O e V

See bubbles? What is the clue? Look at the data table again and see which reactions you think took place

What is happening?? Reduction: 2 H 2 O (l)  H 2(g) + 2 H - (aq) V 2 H + (aq) 3 e -  H 2(g) 0.00 V Na + (aq) + e -  Na (s) V Oxidation: 2 H 2 O  O 2(g) + 4H + (aq) + 4e V 2 SO 4 2-  S 2 O e V

Answer Reduction was water or Na + We know is must be water for 3 reasons - –1. a gas was produced –2. sodium reacts with water violently –3. It became more basic 2H 2 O + 2 e - --> H 2(g) + 2 OH V

Oxidation was either water of sulfate ion Oxidation of water produces H + and a gas. Do we have evidence of that? H 2 O --> O 2(g) + 4 H e V

Energy Involved 2H 2 O + 2 e - --> H 2(g) + 2 OH V H 2 O --> O 2(g) + 4 H e V V What does the negative sign mean?

Change the electrodes to Copper Battery Red lead + anode Black lead cathode - Copper electrodes Na 2 SO 4(aq)

What is happening?? Reduction: 2 H 2 O (l)  H 2(g) + 2 H - (aq) V 2 H + (aq) 3 e -  H 2(g) 0.00 V Na + (aq) + e -  Na (s) V Cu e - -  Cu V Oxidation: 2 H 2 O  O 2(g) + 4H + (aq) + 4e V 2 SO 4 2-  S 2 O e V Cu (s) -  Cu e V