CHM 105 & 106 MO1 UNIT SIX, LECTURE THREE 1 THE LAST TIME WE MET WE WERE TALKING ABOUT BALANCING OXIDATION-REDUCTION

CHM 105 & 106 MO1 UNIT SIX, LECTURE THREE 1 CHM 105/106 Program 50: Unit 6 Lecture 3 THE LAST TIME WE MET WE WERE TALKING ABOUT BALANCING OXIDATION-REDUCTION EQUATIONS AND WE HAD JUST BEGAN TO TALK ABOUT THE DIFFERENCE BETWEEN DIRECT OXIDATION-REDUCTION PROCESSES WHICH CHEMICAL EQUATIONS TYPICALLY SHOW, AND INDIRECT OXIDATION-REDUCTION PROCESSES. BUT BEFORE I RETURN TO LOOKING AT COMPARISON BETWEEN DIRECT AND INDIRECT OXIDATION-REDUCTION PROCESSES I WANT TO LOOK AT ONE MORE BALANCING AN OXIDATION-REDUCTION EQUATION. ALL THE ONES THAT WE HAD DONE UP TO THIS POINT WE ENDED UP IN WHAT WE WOULD CALL AN ACID CONDITION. SO THEY WERE BALANCED AS THOUGH THERE WAS ACID WHICH WOULD BE HYDRONIUM IONS OR HYDROGEN IONS IN THE SYSTEM, BUT OCCASIONALLY OXIDATION REDUCTION REACTIONS ARE ONLY GOING TO OCCUR IF IN FACT THE SYSTEM IS BASIC. AND IF IT S BASIC IT MEANS THAT THE SYSTEM SHOULD HAVE HYDROXIDE IONS IN IT RATHER THAN HYDRONIUM IONS IN IT WHEN WE GET DONE. AND SO WE LL TAKE ONE MORE LOOK HERE AT BALANCING THEN THIS TIME NOW WE RE GOING TO BALANCE AND END UP WITH A SYSTEM THAT LOOKS BASIC INSTEAD OF ACIDIC. NOW WE RE GOING TO USE THE SAME PROCESS THAT WE DID BEFORE. WE RE GOING TO TAKE EACH HALF -REACTION AND BALANCE IT. WE LL ADD WATERS AS NEEDED TO BALANCE THE OXYGEN. WE LL MULTIPLY THROUGH TO GET THE ELECTRONS LOST AND GAINED THE SAME AND THEN WE LL MAKE ONE FINAL ADDITIONAL CHANGE AND WE LL POINT THAT OUT THEN RIGHT AT THE END AS WE CHANGE IT FROM THEN AN ACIDIC TO FINALLY A BASIC CONDITION. WELL WE CAN SEE HERE THAT OUR TWO CHANGES, OUR TWO HALF-REA CTIONS ARE GOING TO INVOLVE THE PHOSPHORUS COMPOUND AND THE OTHER HALF-REACTION IS GOING TO INVOLVE THE IODINES. SO WE LL START FIRST OF ALL THEN WITH THE PHOSPHITE AS IT WOULD BE KNOWN OVER HERE. SO WE START WITH PO 3. THREE MINUS, AND ON THE PRODUCT SIDE WE HAVE PO 4, THREE MINUS. NOW WE HAVE ONE PHOSPHORUS ON EACH SIDE SO THAT S ALREADY BALANCED ATOMIC-WISE. WELL WE SEE THAT WE HAVE THREE OXYGENS ON THIS SIDE AND FOUR ON THIS SIDE AND THE RULE THAT WE SAID WE WOULD USE IS WE WOULD ADD ONE MOLECULE OF WATER FOR EACH OXYGEN WE WERE SHORT. WE RE

CHM 105 & 106 MO1 UNIT SIX, LECTURE THREE 2 SHORT ONE ON THIS SIDE SO WE ADDED ONE MOLECULE OF WATER AND WHEN WE DID THIS OF COURSE WE ADDED TWO HYDROGENS TO THAT SIDE SO WE MUST LIKEWISE SHOW TWO HYDROGENS TO THE OTHER SIDE. AND AGAIN I STRESS THAT WE NEED TO SHOW THESE HYDROGENS AS H + IONS, THAT S THE FORM THAT THEY ARE IN THE WATER, H +. ALRIGHT, ATOMS ARE BALANCED. WE NOW LOOK AT THE CHARGE, AND ON THIS SIDE WE SEE THAT WE HAVE A NET CHARGE OF MINUS THREE. WATER IS ZERO. ON THIS SIDE WE HAVE A NET CHARGE OF A MINUS THREE PLUS TWO SO WE HAVE A MINUS ONE ON THIS SIDE. SO I HAVE A MINUS THREE OVER HERE AND A MINUS ONE OVER THERE. NOW THE ONLY THING THAT WE CAN ADD ARE ELECTRONS WHICH ARE NEGATIVE PARTICLES. SO THEREFORE TO GET THE CHARGE AS THE SAME WE WOULD NEED TO ADD THEN TWO ELECTRONS TO THE RIGHT-HAND SIDE BECAUSE NOW WE WOULD HAVE FIVE MINUSES AND TWO PLUSES WHICH WOULD GIVE US A MINUS THREE THERE AND A MINUS THREE OVER THERE. AND NOW WE HAVE A BALANCED HALF- REACTION, AND THIS IS HALF -REACTION REPRESENTS WHICH PROCESS? OXIDATION OR REDUCTION? OXIDATION, I HEARD SOMEBODY SAY, THAT S CORRECT BECAUSE THE PO 3 HAS LOST ELECTRONS. REMEMBER IF SOMETHING LOSES ELECTRONS THAT S AN OXIDATION - PROCESS. OKAY, SO THE OXIDATION. NOW, THE OTHER HALF REACTION THEN AND WE HAVE I 3 GOING TO I -. ALRIGHT, WE START OF COURSE BY BALANCING ATOMS. WE SEE THAT WE HAVE THREE IODINE ATOMS ON THIS SIDE. WE HAVE ONLY ONE OVER THERE. SO Y FIRST STEP WILL BE TO BALANCE THE ATOMS, PUT A THREE THERE. ATOMS ARE BALANCED. NO OXYGEN INVOLVED, NO WATER, NO HYDROGENS NEEDED IN THIS CASE, SO THE LAST THING IS NOW GOING TO BE TO CHECK THE CHARGE. ON THIS SIDE WE HAVE A MINUS ONE. ON THIS SIDE WE HAVE A MINUS THREE. SO AGAIN IN ORDER TO BALANCE THE CHARGES OUT WE WOULD NEED TO ADD TWO ELECTRONS TO THE LEFT-HAND SIDE. SO NOW WE HAVE A MINUS ONE PLUS TWO MINUSES SO IT LL GIVE US A NEGATIVE THREE CHARGE HERE AND A NEGATIVE THREE CHARGE THERE. AND NOW WE VE BALANCED OUR REDUCTION HALF -REACTION. GAINED ELECTRONS. SO I 3 IS GAINING ELECTRONS FROM SOMEWHERE. IT S BEING REDUCED. SO THIS IS THE REDUCTION HALF -REACTION. WE RE READY NOW TO ADD THE TWO HALF -REACTIONS TOGETHER. AGAIN WE MUST CHECK TO MAKE SURE ELECTRONS LOST AND GAINED ARE THE

CHM 105 & 106 MO1 UNIT SIX, LECTURE THREE 3 SAME. TWO WERE LOST AND TWO WERE GAINED SO THIS TIME WE DON T HAVE TO DO ANY MULTIPLYING OR ANYTHING LIKE THAT. ELECTRONS ALREADY BALANCE. REMEMBER MOST OF THE EXAMPLES WE DID PREVIOUSLY THEY WEREN T THE SAME SO WE HAD TO MULTIPLY EITHER ONE OR BOTH HALF-REACTIONS THROUGH BY SOME NUMBER. YES, A QUESTION? (STUDENT RESPONSE NOT AUDIBLE) THAT IS CORRECT. OKAY. BUT UNTIL WE DO IT WE DON T KNOW WHICH WAS OXIDATION AND WHICH WAS THE REDUCTION SO IT S KIND OF THE CHICKEN AND THE EGG THING BUT WE HAVE TO BALANCE IT FIRST TO FIND OUT WHERE THEY ARE. THEN OF COURSE YES IF THEY RE ON THE RIGHT THAT S OXIDATION, IF THEY RE HERE ON THE LEFT THAT S REDUCTION. OKAY, ALRIGHT, SO ANYWAY WE CAN THEN GO AHEAD AND CANCEL OUT THE ELECTRONS. AND WE CAN NOW COMBINE THE TWO HALF -REACTIONS. SO WE LL WRITE EVERYTHING DOWN THEN. WE HAVE PO 3-3 PLUS AN H 2 O PLUS AN I - 3-3 YIELDS PO 4 PLUS 3I - PLUS 2H +. THAT S THE BALANCED OXIDATION-REDUCTION EQUATION. OKAY. BUT WE WANTED TO BALANCE IT SHOWING A BASIC CONDITION NOT AN ACID CONDITION. IF WE HAVE HYDROGEN IONS IN EXCESS IN THE SYSTEM WE KNOW THAT THAT S AN ACID. SO THE QUESTION IS HOW CAN WE GET RID OF THE HYDROGEN IONS? WELL WE RECALL FROM OUR CHAPTER FIVE DISCUSSION ON ACIDS AND BASES IF WE HAVE A HYDROGEN IONS REACTS WITH A HYDROXIDE ION WE END UP FORMING A MOLECULE OF WATER. SO WHAT WE RE GOING TO DO HERE NOW IS AS LONG AS WE DO THE SAME THING TO BOTH SIDES OF THE EQUATION, WE DON T CHANGE ANYTHING. SO I M GOING TO ADD TWO HYDROXIDE IONS TO THIS SIDE AND I M GOING TO ADD TWO HYDROXIDE IONS TO THIS SIDE. NOW THE REASON I ADDED TWO HYDROXIDE IONS IS BECAUSE I WANTED TO GET RID OF TWO HYDROGEN IONS, TWO HYDRONIUM IONS. NOW WHEN I DO THAT, WHEN I ADD THESE TWO TOGETHER I M NOT GOING TO HAVE THE IONS LEFT, I M GOING TO HAVE ACTUALLY NOW TWO MOLECULES OF WATER THEY HYDROGEN ION AND THE HYDROXIDES ARE GOING TO REACT AND FORM TWO MOLECULES OF WATER. NOW WE SEE THAT WE HAVE HYDROXIDE ION THAT S GOING TO BE OVER HERE AND WE RE NOW SHOWING A BASIC SOLUTION. SO WE HAVE COMPLETED IT. THE ONLY THING IS THAT WE NOW HAVE WATER SHOWING UP ON THE REACTANT SIDE AND WATER SHOWING UP ON THE PRODUCT SIDE AND NORMALLY WE WOULDN T WANT THAT. SHOULD BE

CHM 105 & 106 MO1 UNIT SIX, LECTURE THREE 4 ON ONLY ONE SIDE OR THE OTHER. WE SEE THAT WE HAVE TWO WATERS OVER HERE, BUT ONLY ONE WATER OVER HERE. SO THEREFORE WE COULD SUBTRACT ONE WATER FROM EACH SIDE AND IF I DO THAT OF COURSE IT S JUST GONNA CANCEL THAT ONE OUT MINUS A WATER, AND IF I COME OVER HERE AND SUBTRACT ONE WATER FROM THIS I M GOING TO BE LEFT NOW WITH JUST ONE H 2 O. SO NOW WE SHOULD HAVE EVERYTHING FINISHED UP AND WE SHOULD NOW BE SHOWING A BASIC EQUATION. SO LET S GO AHEAD AND REWRITE IT ONE MORE TIME THEN, NOW HOPEFULLY WITH EVERYTHING THERE. SO WE HAVE PO 3-3, THAT PEN ISN T DOING TOO WELL, LET S TRY A DIFFERENT COLOR HERE. OKAY WE LL START A GAIN. PO 3-3 PLUS, WE VE GOT THESE HERE 2OH - PLUS I - 3 YIELDS PO 3-4 PLUS 3I - PLUS H 2 O. NOW WE HAVE A TOTAL BALANCED OXIDATION-REDUCTION EQUATION AND IT IS SHOWING IT TO BE BASIC NOW. BUT NOTICE THAT THE FIRST STEPS WE DID WE DID THE SAME AS WE VE DONE ALL OF THE OXIDATION-REDUCTION EQUATIONS. THE ONLY FINAL STEP IS IF WE WANTED TO SHOW IT AS A BASIC CONDITION IS WE MUST ADD ENOUGH HYDROXIDES TO NEUTRALIZE TO CANCEL OUT THE HYDROGEN IONS THAT ARE PRESENT. IF WE D HAVE HAD 12 HYDROGEN IONS IN THERE WE WOULD VE ADDED 12 HYDROXIDES TO BOTH SIDES AND FORMED 12 WATERS ON THE PRODUCT SIDE. OKAY, ALRIGHT NOW JUST A QUICK CHECK IN TERMS OF CHARGE HERE WE HAVE MINUS THREE MINUS TWO MINUS ONE. SO WE HAVE A TOTAL CHARGE ON THIS SIDE OF MINUS SIX. WE HAVE MINUS THREE AND A NOTHER MINUS THREE AND WATER ZERO, MINUS SIX. SO THE TOTAL EQUATION DOES BALANCE IN TERMS OF CHARGE AND ATOMS AND SO WE RE COMFORTABLE THAT WE DO HAVE A TOTAL BALANCED EQUATION. ANY QUESTION ON CONVERTING FORM AN ACID TO A BASE-TYPE BALANCING? ALRIGHT, LET S TURN THEN ONCE AGAIN TO LOOKING AT DIRECT VERSUS INDIRECT OXIDATION-REDUCTION PROCESSES. AGAIN IF WE WERE TO TAKE A PENNY MADE OF COPPER METAL AND PLACE IT IN A BEAKER AND WE WERE TO ADD SOME NITRIC ACID, AND ACTUALLY THE NITRIC ACID IS COLORLESS, TO THAT PENNY, WE GET A REACTION OCCURRING. IMMEDIATELY. AS SOON AS WE POUR THE NITRIC ACID IN THERE WE START SEEING SOME THINGS HAPPENING. WE NOTICE THE SOLUTION BEGINNING TO TURN BLUE OR BLUE-GREEN. YOU SEE THE EVOLUTION OF A REDDISH-BROWN GAS. SO WE KNOW THINGS ARE OCCURRING, THIS JUST SHOWS IT IN A LATER STATE, STILL BUBBLING AND MORE REDDISH-

CHM 105 & 106 MO1 UNIT SIX, LECTURE THREE 5 BROWN GAS COMING OFF. I USED TO DO THIS AS A DEMONSTRATION BUT SEEING THAT THEY NOW CLAIM THAT NITROGEN DIOXIDE IS A TOXIC POISON THEY DISCOURAGE ME FROM DOING THAT IN THE CLASSROOM. SO IF I HAD A VENT OR A HOOD HERE IN THE SINK, YEAH I COULD STILL DO THAT. BUT IT S A FAIRLY NICE DRAMATIC REACTION. IN THE PROCESS THE COPPER HAS CHANGED FROM COPPER METAL TO COPPER PLUS TWO. IT AHS UNDERGONE THEN OXIDATION. IT HAS LOST IT S TWO ELECTRONS. THE NITRIC ACID, THE NITRATE PART OF THE NITRIC ACID NOW HAS UNDERGONE A CHANGE TO NO 2 GAS IN WHICH THE NITROGEN WHICH WAS A PLUS FIVE HAS GONE TO A PLUS FOUR, ITS GAINED ONE ELECTRON. OBVIOUSLY IF WE WERE GOING TO BALANCE THIS WE WOULD HAVE TO HAVE TWO OF THESE THROUGH TO GET OUR TWO ELECTRONS. WE D HAVE TO BALANCE SOME OXYGENS AND HYDROGENS IN THERE SO WE RE NOT TRYING TO BALANCE EVERYTHING THERE. BUT WE SEE THAT WE DO HAVE AN OXIDATION AND A REDUCTION PROCESS OCCURRING. THIS IS A DIRECT OXIDATION- REDUCTION.. BY THAT I MEAN THAT THE ELECTRONS FROM THE COPPER ATOMS ARE GOING DIRECTLY TO THE NITRATE IONS THAT ARE COLLIDING WITH THE COPPER SURFACE. SO THIS IS A DIRECT OXIDATION-REDUCTION PROCESS. BUT THERE ARE MANY CASES, MANY IMPORTANT APPLICATIONS OF OXIDATION-REDUCTION REACTIONS WHICH ARE INDIRECT, AND INDIRECT OXIDATION-REDUCTION PROCESSES LEAD US TO WHAT WE CALL ELECTROCHEMICAL CELLS. OKAY. LET S SUPPOSE THAT WE WERE TO SET UP THE FOLLOWING SYSTEM. THIS SIDE WE HA VE A SOLUTION CONTAINING SILVER NITRATE. SILVER NITRATE S A COLORLESS SOLUTION. ON THIS SIDE OVER HERE WE HAVE A SOLUTION OF COPPER NITRATE A LIGHT BLUE SOLUTION. WE PUT A PIECE OF SILVER METAL IN THIS SIDE AND A PIECE OF COPPER METAL IN THAT SIDE, AND THEN WE HOOK THE TWO PIECES OF METAL TOGETHER WITH A WIRE. NOW, IF I WERE TO TAKE A PIECE OF SILVER METAL AND STICK IT DIRECTLY INTO COPPER NITRATE SOLUTION I WOULD FIND THAT IN FACT I DID GET A DIRECT OXIDATION-REDUCTION PROCESS. SO THE QUESTION IS WILL IT DO IT INDIRECTLY? SO LET ME JUST AGAIN, WE LL DESIGN THIS HERE REAL QUICKLY. HERE S THE BEAKER AND IF I WERE TO PUT SOME COPPER NITRATE SOLUTION IN HERE, OR EXCUSE ME, I SAID THAT WRONG I WANT TO DO IT THE OTHER WAY. PUT SOME SILVER NITRATE SOLUTION IN HERE AND I M GOING TO PUT A PIECE OF COPPER WIRE OR

CHM 105 & 106 MO1 UNIT SIX, LECTURE THREE 6 SOMETHING DOWN IN HERE. I WOULD FIND THAT IMMEDIATELY A CHEMICAL REACTION WOULD BEGIN TO OCCUR AND WHAT I WOULD FIND HAPPENING IS THAT THIS COIL OF WIRE WOULD BECOME COVERED WITH SILVER METAL. SO APPARENTLY THE SILVER ION IS GAINING AN ELECTRON AND BECOMING SILVER METAL AND I WOULD BE ABLE TO TELL THAT BECAUSE THE WATER WOULD BECOME COATED WITH A SILVER-LOOKING SOLID WHICH OBVIOUSLY IS NOT THE COPPER. BUT IN ORDER FOR THAT TO HAPPEN OF COURSE IF THE SILVER IS REDUCED THEN SOMETHING HAS TO BE OXIDIZED AND WE FIND IN FACT THAT THE COPPER IS OXIDIZED TO COPPER TWO PLUS, GIVING OFF TWO ELECTRONS, AND IF WE WERE TO BALANCE THESE AND ADD THEM TOGETHER IT WOULD LOOK LIKE THAT. AND WHAT WE FIND S THAT THE COLORLESS SOLUTION THAT WE HAD BEGINS TO TURN BLUE, AND THE PRESENCE OF THE BLUE COLOR IN THE BEAKER IS INDICATIVE OF THE FACT THAT COPPER IONS ARE BEGINNING TO BE FORMED. SO WE KNOW IN FACT THAT COPPER METAL DIRECTLY IN CONTACT WITH SILVER IONS IN SOLUTION WILL SPONTANEOUSLY UNDERGO A DIRECT OXIDATION-REDUCTION PROCESS. I DON T HAVE TO DO ANYTHING. I DON T HAVE TO HEAT IT, I DON T HAVE TO IGNITE IT I DON T HAVE TO DO ANYTHING. I JUST PUT THE COPPER WIRE IN THERE AND THE REACTION BEGINS TO OCCUR. SPONTANEOUS DIRECT OXIDATION-REDUCTION. WELL THE QUESTION THEN WAS IF THAT IN FACT DOES HAPPEN DIRECTLY WOULD IT HAPPEN INDIRECTLY, AND THAT S WHAT WE RE TRYING TO SHOW HERE. NOTICE THAT THIS TIME INSTEAD OF ACTUALLY PLACING THE COPPER STRIP DIRECTLY INTO THE SILVER NITRATE THE COPPER STRIP IS IN CONTACT WITH THE SILVER NITRATE INDIRECTLY THROUGH THE SILVER STRIP, BUT THE COPPER ATOMS OVER HERE STILL EXPERIENCE THE PRESENCE OF THE SILVER IONS OVER HERE. HOOKED TOGETHER, AND WE KNOW FROM THE DISCUSSION OF SOLIDS METALS HAVE FAIRLY MOBILE ELECTRONS THAT S ONE OF THEIR CHARACTERISTICS. METALS ARE GOOD CONDUCTORS AND SILVER AND COPPER ARE BOTH GOOD CONDUCTORS. OKAY. NOW BEFORE ANYTHING COULD HAPPEN WE RE GOING TO HAVE TO COMPLETE THIS CIRCUIT. IN OTHER WORDS WE CAN T JUST LET ELECTRONS MOVE ONE WAY OR THE OTHER IN HERE BECAUSE WE CAN T HAVE A CHARGED BEAKER. WE CAN T HAVE A BEAKER OVER HERE CHARGED MINUS OR A BEAKER OVER THERE CHARGED PLUS. WE HAVE TO ALWAYS MAINTAIN ELECTRIC NEUTRALITY IN A SOLUTION, AND

CHM 105 & 106 MO1 UNIT SIX, LECTURE THREE 7 SO IN ORDER FOR THIS TO OPERATE, SORT OF LIKE WE HAVE IF I WALK OVER HERE, IF I DO THIS THE REASON THE LIGHTS GO OUT IS BECAUSE I HAVE NOT COMPLETED A CIRCUIT, ELECTRONS CAN T FLOW, THE LIGHT CAN CONDUCT ELECTRICITY. THERE S NO PROBLEM THERE. WE KNOW ELECTRICITY IS AVAILABLE IN THE BUILDING THERE S NO PROBLEM THERE, BUT UNTIL I COMPLETE THAT CIRCUIT OF COURSE THE ELECTRONS CAN T FLOW. THAT S WHAT THE SWITCH DOES AND THAT S WHAT THIS LITTLE THING IN HERE DOES IS COMPLETES. IT S JUST LIKE THE SWITCH. IT COMPLETES THE ELECTRICAL CONNECTION. THIS IS REFERRED TO AS A SALT- BRIDGE, AND THE SALT -BRIDGE CONTAINS SOME SORT OF IONIC COMPOUND THAT WILL NOT REACT WITH EITHER OF THE SOLUTIONS IN THE BEAKERS. IN THIS PARTICULAR CASE WE RE SHOWING A SALT-BRIDGE CONTAINING POTASSIUM IONS AND NITRATE IONS. NOW WE LL TALK HERE IN JUST A SECOND ABOUT WHAT THE SALT -BRIDGE ACTUALLY DOES. NOW, LET S SUPPOSE THAT WE DO HOOK THESE TOGETHER. WHAT EVIDENCE WOULD WE HAVE THAT A REACTION WAS OCCURRING? WELL IF WE HOOK THIS TOGETHER AND LET IT GO LONG ENOUGH WHAT WE WOULD FIND IS OVER HERE NOTICE THAT THE COPPER NOW MUST UNDERGO OXIDATION ION THE PROCESS AND SO WE WOULD HAVE COPPER IONS COMING INTO SOLUTION WHICH MEANS THAT THIS SOLUTION SHOULD INCREASE IN COLOR, IF IN FACT THE PROCESS IS OCCURRING. ANOTHER FACTOR WOULD BE IF WE WERE TO WEIGH THE COPPER WIRE BEFORE WE STARTED THE COPPER STRIP I SHOULD SAY. IF WE WEIGHED THE COPPER STRIP BEFORE WE STARTED AND THEN HOOKED IT TOGETHER AND CAME BACK A DAY LATER AND WEIGHED IT WE SHOULD HAVE THEN A MASS LOSS. IN OTHER WORDS THE COPPER STRIP SHOULD WEIGHT LESS THAN IT DID BEFORE WE HOOKED THE THING TOGETHER. THAT S ASSUMING THAT THIS THING OPERATES AT ALL. WE DON T KNOW THAT UNTIL WE MAKE SOME TESTS. OVER HERE THEN NOW THE SILVER WE WOULDN T BE ABLE TO TELL BECAUSE SILVER SOLUTION IS COLORLESS SO THERE WOULDN T BE ANY CHANGE IN COLOR ANYWAY, BUT IF IN FACT THIS PROCESS DID OCCUR WE KNOW THAT WE SEE THE SILVER ION IS MOVING OVER AND WE VE ALREADY SHOWN THE REACTION THAT THE SILVER ION WOULD GAIN AN ELECTRON AND BECOME A SILVER ATOM AND THAT WOULD MEAN THEN THAT THIS ELECTRODE SHOULD HAVE A MASS GAIN. IN THEO WORDS, AS THESE SILVER ATOMS ARE FORMED THEY RE GOING TO BE

CHM 105 & 106 MO1 UNIT SIX, LECTURE THREE 8 STICKING TO THIS ELECTRODE, AND AS THEY STICK TO THAT ELECTRODE THEY RE GOING TO INCREASE THE MASS OF THE ELECTRODE. ALRIGHT, SO THE ELECTROCHEMICAL CELL WAS FORMED AND SURE ENOUGH THEY FOUND THAT IN FACT IT WAS SPONTANEOUS. THEY DIDN T HAVE TO DO ANYTHING BUT HOOK IT TOGETHER AND IN FACT THE COPPER LOST WEIGHT, THE SOLUTION GOT MORE BLUE, THE SILVER GAINED WEIGHT. AND SO OXIDATION-REDUCTION WAS OCCURRING BY AN INDIRECT PROCESS. IN OTHER WORDS THE COPPER METAL WAS NEVER IN DIRECT CONTACT WITH THE SILVER SOLUTION BUT YET THE SAME OXIDATION-REDUCTION PROCESS OCCURRED. BY THIS THEN WE HAD PRODUCED WHAT WE CALL AS I SAID AN ELECTROCHEMICAL CELL. NOW I MENTIONED THAT WE WOULD LOOK A LITTLE BIT AT THE SALT -BRIDGE. WHEN THIS HAPPENS, WHEN COPPER IONS MOVE OUT INTO SOLUTION THAT WOULD MEAN THAT THIS SOLUTION OVER HERE WOULD NOW BECOME POSITIVE, WHICH OBVIOUSLY IT CAN T BE THAT WAY. SO WHAT HAPPENS IS THAT AS THIS HERE IS FORMED, POTASSIUM IONS OVER HERE WILL MOVE OUT INTO SOLUTION ON THIS SIDE SO THAT EVERY TIME WE PUT A COPPER PLUS TWO IN SOME OF THE POTASSIUMS, TWO OF THE POTASSIUMS WOULD MOVE OUT AND SO WE LL MAINTAIN A CHARGE THROUGH THIS TUBE AND WE LL MAINTAIN ZERO CHARGE IN THE BEAKER AND ZERO CHARGE OVER HERE. BECAUSE YOU SEE ON THIS SIDE NOW WHEN THE TWO POTASSIUMS MOVE IN TWO SILVERS HAVE JUST MOVED OUT BECAUSE AGAIN WE HAVE TO HAVE TWO OF THESE FOR EACH COPPER THAT S UNDERGOING A CHANGE. SO TWO SILVER IONS HAVE LEFT THE SOLUTION BUT TWO POTASSIUM IONS HAVE MOVED INTO THE SOLUTION SO IT HAS REMAINED ELECTRICALLY NEUTRAL. ALRIGHT, NOW LET S EXPAND ON THIS JUST A LITTLE BIT. LET S SUPPOSE WE DO EXACTLY THE SAME THING AGAIN EXCEPT THIS TIME WE RE GOING TO CUT THIS WIRE RIGHT HERE. AND WE RE GOING TO UT IN A MEASURING DEVICE, AND THE MEASURING DEVICE THAT WE PUT IN IS CALLED A VOLTMETER. AND A VOLTMETER IS GOING TO MEASURE THE ELECTRON FLOW OR THE ELECTROCHEMICAL POTENTIAL THEN DIFFERENCE BETWEEN THE TWO BEAKERS, BETWEEN THE TWO WHAT WE CALL HALF -CELLS. ONE CELL HAVE REDUCTION OCCURRING IN IT, ONE CELL HAS OXIDATION OCCURRING IN IT. WE REFER TO THESE THEN AS HALF -CELLS. SO A HALF -CELL IS LIKE A HALF-REACTION THAT WE LOOKED AT. NOW WE NOTICE THAT IN THIS PARTICULAR

CHM 105 & 106 MO1 UNIT SIX, LECTURE THREE 9 CASE IT NOT ONLY READS BUT IT READS A NUMERIC VALUE, AND THAT NUMERIC VALUE IS REPRESENTATIVE OF THE POTENTIAL DIFFERENCES BETWEEN THESE TWO HALF-CELLS. IN OTHER WORDS, BOTH OF THEM HAVE SOME TENDENCY TO WANT TO UNDERGO REDUCTION. BOTH HAVE SOME TENDENCY TO WANT TO UNDERGO OXIDATION. BUT APPARENTLY IN THIS CASE THE COPPER HAS A GREATER POTENTIAL TO BE OXIDIZED THAN THE SILVER. OR WE CAN SAY IT THE OTHER WAY; SILVER HAS A GREATER POTENTIAL TO BE REDUCED THAN DOES THE COPPER ION. BECAUSE THE COPPER ION DOESN T GO OUT OF SOLUTION, SILVER ION GOES OUT OF SOLUTION. ALRIGHT, AND WE VE SHOWN THE ELECTRON FLOW HERE THEN TO BE OUT OF THIS HALF-CELL INTO THIS HALF-CELL. NOW WE THEN IDENTIFY THE TWO HALF-CELLS BY NAMES. A TOTAL ELECTROCHEMICAL CELL IS COMPOSED OF TWO HALF-CELLS, ONE IN WHICH OXIDATION OCCURS THAT IS CALLED THE ANODE. AND THE ANODE THEN IS WHERE OXIDATION OCCURS. AND THAT S WHAT S HAPPENING OVER HERE. COPPER IS BEING OXIDIZED FROM COPPER ZERO TO COPPER TWO PLUS. SO THAT S WHAT S HAPPENING IN THIS CELL. THAT S THE ANODE. THE OTHER ELECTRODE THEN WHERE REDUCTION IS OCCURRING IS CALLED THE CATHODE. SO THE CATHODE IS THE ELECTROCHEMICAL CELL AT WHICH REDUCTION IS OCCURRING, AND OVER HERE WE SHOW THAT REACTION TO BE Ag + PLUS THE ELECTRON TO BECOME Ag 0. ALRIGHT, SO EVERY ELECTROCHEMICAL CELL IS COMPOSED OF TWO ELECTRODES THEN AN ANODE IN WHICH OXIDATION OCCURS AND A CATHODE AT WHICH REDUCTION OCCURS. NOW LOOKING AT THIS RIGHT HERE, WHICH IS WHAT WE CALL THE CELL POTENTIAL, THE CELL POTENTIAL AND WE HAVE A LITTLE UPPER CASE ZERO. THAT MEANS AT STANDARD CONDITIONS, STANDARD CONDITIONS FOR ELECTROCHEMISTRY MEANS 25 CELSIUS DEGREES, ONE ATMOSPHERE PRESSURE, AND ONE MOLAR CONCENTRATION FOR ALL IONS. DON T NEED TO REMEMBER THAT BUT THAT S WHAT THIS LITTLE NOT MEANS. E-NOT (PHONETIC SPELLING) OF THE CELL IS DEFINED AS THE REDUCTION E-NOT FOR THE CATHODE MINUS THE REDUCTION POTENTIAL, E-NOT, FOR THE ANODE. AND I SAID REDUCTION POTENTIAL. THE POTENTIAL FOR THE THING TO BE REDUCED. EVEN THOUGH THIS ONE S BEING OXIDIZED WE RE STILL LOOKING AT IT S POTENTIAL TO BE REDUCED. NOW THE QUESTION IS HOW CAN WE NUMERICALLY MEASURE THESE REDUCTION POTENTIALS? ONCE AGAIN IF I GO

CHM 105 & 106 MO1 UNIT SIX, LECTURE THREE 10 BACK TO THE DRAWING THAT WE JUST HAD HERE WE DO SEE A NUMERIC VALUE. THAT S E-NOT IN THE CASE, THAT S E-NOT OF THIS CELL. WHAT THAT REPRESENTS DOESN T REPRESENT THE REDUCTION POTENTIAL OF THIS SIDE AND IT DOESN T REPRESENT THE REDUCTION POTENTIAL OF THIS SIDE. WHAT E-NOT REPRESENTS IS THE REDUCTION POTENTIAL DIFFERENCE OF THE TWO SIDES. WELL SEEING THAT WE CAN T HAVE OXIDATION OCCUR WITHOUT REDUCTION, CAN T HAVE REDUCTION OCCUR WITHOUT OXIDATION, HOW CAN WE EVER GO ABOUT FIGURING OUT THE REDUCTION POTENTIAL OF A HALF -CELL? BECAUSE WE ALWAYS HAVE TO HAVE BOTH OF THEM, AND ANYTIME WE HOOK TWO TOGETHER WE RE NOT MEASURING ONE OR THE OTHER, WE RE MEASURING THE DIFFERENCE. OKAY. WELL TO GET AROUND THAT PROBLEM THEN WE HAD TO HA VE SOME SYSTEM DESIGNED TO DEVELOP REDUCTION POTENTIALS. FIRST OF ALL WE LL JUST TAKE A LITTLE LOOK AT THIS TABLE HERE, REDUCTION POTENTIAL, AND THE REASON THAT I VE PUT THIS UP IS BECAUSE I WANT TO POINT OUT THIS ONE RIGHT HERE. NOW WHAT THESE REDUCTION POTENTIALS SHOW US FIRST OF ALL IS THAT IF SOMETHING HAS A REDUCTION POTENTIAL POSITIVE IT MEANS THAT IT IS MORE EASILY REDUCED THAN HYDROGEN IONS. FOR INSTANCE, IF I GO DOWN HERE AND LOOK AT PERMANGANATE I SEE THAT PERMANGANATE AS A REDUCTION POTENTIAL OF PLUS 1.52 VOLTS. IT HAS A MUCH GREATER POTENTIAL TO GAIN ELECTRONS THAN DOES THE HYDROGEN ION TO GAIN ELECTRONS. IT HAS MORE POTENTIAL TO BE REDUCED THAN DOES HYDROGEN TO BE REDUCED. THAT S WHAT THIS NUMBER SHOWS. BUT IF I LOOK OVER HERE AT POTASSIUM NOW I NOTICE THAT POTASSIUM S REDUCTION POTENTIAL IS A NEGATIVE 2.92 VOLTS. IT AS MUCH LESS DESIRE TO BE REDUCED THAN DOES HYDROGEN ION AND SO THAT S WHY IT HAS THE NEGATIVE SIGN. IF YOU SEE A NEGATIVE E-NOT THAT MEANS IT HAS LESS POTENTIAL TO BE REDUCED THAN HYDROGEN. A POSITIVE E-NOT MEANS THAT IT HAS GREATER POTENTIAL TO BE REDUCED. ALRIGHT, HYDROGEN ZERO. DOES THAT MEAN THAT THE HYDROGEN ION HAS NO POTENTIAL TO BE REDUCED? NO. IT MEANS THAT WE HAD TO HAVE A STARTING POINT AND SO THEY ARBITRARILY SAID LET S SET THE REDUCTION OF HYDROGEN ION AS ZERO. BECAUSE THEN WE CAN USE THAT, WE CAN USE THAT HALF -CELL, HOOK ALL OTHER HALF -CELLS TO IT, READ ON THE METER,. AND WHATEVER THE METER READS WE WILL ASSIGN TO THESE OTHER

CHM 105 & 106 MO1 UNIT SIX, LECTURE THREE 11 HALF-CELLS. BECAUSE WE ARBITRARILY ARE CALLING HYDROGEN ZERO. NOW, THE STANDARD HYDROGEN ELECTRODE AS WE REFER TO IT, STANDARD HYDROGEN ELECTRODE, THAT MEANS THAT WE HAVE AN ELECTRODE AT ONE ATMOSPHERE OF PRESSURE, 25 CELSIUS DEGREES, ONE MOLAR CONCENTRATION. THE STANDARD HYDROGEN ELECTRODE IS THE REFERENCE ELECTRODE FOR DETERMINING THEN ALL OF THESE REDUCTION POTENTIALS. THIS IS WHAT THE STANDARD HYDROGEN ELECTRODE LOOKS LIKE. OKAY. WE HAVE A TUBE BRINGING HYDROGEN GAS IN, GASEOUS STATE. SO THIS IS H 2 GAS UNDER ONE ATMOSPHERE OF PRESSURE. THE BUBBLES OF HYDROGEN THEN PASS OVER THIS SURFACE OF A PIECE OF PLATINUM. THE PLATINUM IS THERE ONLY TO PROVIDE A SURFACE FOR OXIDATION- REDUCTION. IT IS NOT INVOLVED IN THE OXIDATION-REDUCTION PROCESS ITSELF. DOWN HERE WE HAVE HYDROGEN IONS, ONE MOLAR CONCENTRATION AND WE HAVE A WIRE COMING OUT THAT CAN BE HOOKED TO OF COURSE THE OTHER HALF-CELL. SO WE TAKE THIS HALF-CELL, HOOK IT TO ONE OF THOSE WITH A PIECE OF COPPER METAL, COPPER ION IN IT, HOOK THE TWO TOGETHER WITH A METER BETWEEN THEM AND SEE WHAT THE VOLTAGE IS. WHATEVER THAT VOLTAGE IS WOULD BE ASSIGNED TO THE COPPER BECAUSE WE RE CALLING THIS ONE ZERO. IN OTHER WORDS THIS ONE IS 2H + PLUS TWO ELECTRONS TO PRODUCE H 2 AND THIS ONE WE RE SAYING BY ARBITRATION HAS A VALUE OF ZERO. OKAY. SO THE STANDARD HYDROGEN ELECTRODE IS OUR REFERENCE POINT. WE COULD HAVE CALLED THIS TWO VOLTS AND COMPARED EVERYTHING. IT WOULDN T MAKE ONE BIT OF DIFFERENCE IN OUR CALCULATION OF CELLS, BUT IF YOU RE GOING TO TAKE SOMETHING AS A REFERENCE POINT IT PROBABLY MAKES MORE SENSE TO GO AHEAD AND PICK YOUR SIMPLEST NUMBER WHICH IS ZERO. OKAY. NOW BACK TO THIS THEN. SO E-NOT OF ANY ELECTROCHEMICAL CELL WHICH IS MADE UP OF A CATHODE AND AN ANODE, THAT CELL POTENTIAL CAN BE DETERMINED BY TAKING THOSE VALUES FROM THE CHART THAT WE JUST HAD AND UTILIZING THEM THEN IN WHATEVER FASHION WE NEED FORT EH CATHODE AND ANODE POSITION. WE LL TAKE A LOOK AT A COUPLE OF EXAMPLES OF THIS IN JUST A SECOND BUT LET ME SAY ONE ADDITIONAL THING BEFORE WE GO ON TO MAKE A QUANTITATIVE CALCULATION. OF E-NOT OF THE CELL, NOT THE HALF-CELL, NO, THE E-NOT OF THE CELL IS GREATER THAN ZERO, OR IN OTHER WORDS WHEN

CHM 105 & 106 MO1 UNIT SIX, LECTURE THREE 12 IT S POSITIVE WHEN WE PUT THESE TWO TOGETHER, IF E-NOT IS GREATER THAN ZERO WE SAY THAT THE CELL IS SPONTANEOUS AS IT IS HOOKED TOGETHER, AND WE ALSO REFER TO IT THEN AS A GALVANIC OR VOLTAIC ELECTROCHEMICAL CELL. THESE TWO NAMES GALVANIC AND VOLTAIC, GALVANIC CAME FROM GALVANI, WHO DID A LOT OF EARLY WORK WITH ELECTRICITY AND MUSCLES IN FROGS. A FROG LEG, TOUCH IT WITH A LITTLE BIT OF ELECTRICITY AND THE MUSCLE REFLEXES, AND GALVANI THEN BEGAN EXPLORING ELECTRICAL BEHAVIOR. AND VOLTA, ANOTHER ITALIAN CHEMIST WAS DOING A LOT OF WORK WITH ELECTROCHEMICAL CELLS HIMSELF IN MAKING SOME OF THESE MEASUREMENTS, SO THE TWO TERMS YOU MAY SEE EITHER ONE GIVEN TO A SITUATION FOR A SPONTANEOUS ELECTROCHEMICAL CELL. IF E-NOT OF THE CELL IS LESS THAN ZERO, OR IN OTHER WORDS IF IT S NEGATIVE THEN IT IS A NON-SPONTANEOUS PROCESS AND IF IT S NON-SPONTANEOUS THAT MEANS THAT IN ORDER TO GET IT TO WORK YOU RE GOING TO HAVE TO APPLY SOME ADDITIONAL FORCE, SOME EXTERNAL FORCE TO THE SYSTEM. IT S NOT JUST GOING TO OCCUR. YOU RE GOING TO HAVE TO DO SOMETHING TO IT, LIKE HEATING UP AN ENDO THERMIC REACTION. A NON-SPONTANEOUS ELECTROCHEMICAL CELL WE RE GOING TO HAVE TO DO SOMETHING TO IT TO GET IT TO OPERATE, AND THESE ARE REFERRED TO AS ELECTROLYTIC ELECTROCHEMICAL CELLS, THE NON-SPONTANEOUS. AND WE RE GOING TO TAKE A LOOK HER A LITTLE BIT LATER IN THE CHAPTER OF SOME PRACTICAL SPONTANEOUS ELECTROCHEMICAL CELLS, AND WE RE GOING TO ALSO LOOK AT SOME PRACTICAL APPLICATIONS OF NON- SPONTANEOUS ELECTROCHEMICAL CELLS, VERY LARGE INDUSTRY, THE ELECTROCHEMICAL INDUSTRY. WELL LET S GO BACK FOR A MOMENT THEN AND TAKE A LOOK AT HOW WE CAN QUANTITATIVELY CALCULATE THE CELL POTENTIAL. OUR FIRST ONE HERE, NOW NOTICE THAT I DON T HAVE DRAWINGS OF TWO BEAKERS WITH THINGS IN THEM. I M MERELY SHOWING TWO CHEMICALS IN THIS PROCESS. EVERY ELECTROCHEMICAL CELL CAN BE DESCRIBED BY A BALANCED OXIDATION-REDUCTION EQUATION. SO WE DON T HAVE TO MAKE DRAWINGS, WE CAN WRITE DOWN THE CHEMICAL EQUATION FOR THE ELECTROCHEMICAL CELL, AND THAT S WHAT THIS REPRESENTS. THIS REPRESENTS AN ELECTROCHEMICAL CELL COMPOSED OF SOME AQUEOUS IODINE WITH IODIDE ION AND SOME SOLID NICKEL WITH SOME NICKEL ION. THE

CHM 105 & 106 MO1 UNIT SIX, LECTURE THREE 13 FIRST THING WE NEED TO DO IS OF COURSE WE NEED TO IDENTIFY WHICH IS UNDERGOING OXIDATION AND WHICH IS UNDERGOING REDUCTION BECAUSE THAT S WHAT S GOING TO DEFINE WHICH IS THE CATHODE AND WHICH IS THE ANODE. LET S LOOK AT NICKEL BECAUSE I THINK THAT S GOING TO BE THE MOST OBVIOUS. WHAT S HAPPENING TO THE NICKEL? IS IT BEING OXIDIZED OR BEING REDUCED? GOES FROM ZERO TO A PLUS TWO. IT S OXIDIZED. OKAY, ABSOLUTELY CORRECT. REMEMBER OXIDATION IS AN INCREASE IN THE CHARGE NUMBER. WE VE INCREASED FROM ZERO TO TA PLUS TWO. WELL IF WE LOOK AT THE IODINE WE CAN ALSO SEE THAT THE IODINE HAS GONE FROM A ZERO TO A MINUS ONE. SO THAT S A DECREASE IN CHARGE NUMBER SO THIS IS THE REDUCTION AND THIS IS THE OXIDATION. SO NOW WE CAN SAY E-NOT OF THE CELL IS EQUAL TO E-NOT OF THE CATHODE AND THE CATHODE WAS THE HALF-CELL THAT REPRESENTS REDUCTION. OKAY, SO THIS IS GOING TO BE AN I 2 -SLASH-I IS WHAT WE RE GOING TO BE LOOKING FOR THERE, MINUS E-NOT OF THE ANODE, AND THE ANODE IS THE OXIDATION PROCESS, AND THAT IS THE Ni SOLID-SLASH-Ni + 2. ALRIGHT NOW WE TAKE OUR CHART HERE AND WE RE GOING TO BE LOOKING FOR TWO THINGS IODINE GOING TO IODIDE, AND NICKEL ION GOING TO NICKEL. ALRIGHT SO FIRST OF ALL COMING DOWN HERE THIS LOOKS LIKE THE FIRST ONE THAT WE WANT TO FIND RIGHT HERE, I 2 TO I -, AND WE SEE THAT THE REDUCTION POTENTIAL IS A PLUS.53 VOLTS. SO UP HERE WE RE GOING TO PUT PLUS.53 VOLTS. VERY IMPORTANT TO CARE FOR YOUR SIGNS SO THAT YOU DON T FORGET WHETHER IT S POSITIVE OR NEGATIVE. THE OTHER ONE WE WANNA FIND IS THE NICKEL, AND WE FIND THE NICKEL RIGHT HERE. AND THE NICKEL HAS A REDUCTION POTENTIAL, NOW YOU MIGHT SAY BUT IT S NOT UNDERGOING REDUCTION IT S UNDERGOING OXIDATION, THAT S FINE, THAT DOESN T MAKE ANY DIFFERENCE BECAUSE NOTICE THAT WE RE NOT ADDING THE TWO THINGS TOGETHER, WE RE SUBTRACTING. THE NEGATIVE SIGN IN THAT SELL EQUATION CALCULATION ALREADY COMPENSATES FOR THE FACT THAT THIS IS NOT GOING TO BE GOING THIS DIRECTION THAT IT S GOING TO BE GOING TO OTHER DIRECTION WHICH WOULD BE OPPOSITE SIGN. SO WE VE ALREADY CORRECTED FOR IT SO WE USE IT JUST THE WAY IT SHOWS IN THE CHART. MINUS.25. ALRIGHT, SO WE HAVE IDENTIFIED THAT. SO WE HAVE MINUS A MINUS.25 VOLTS. OF COURSE MINUS A MINUS IS GOING TO GIVE USA PLUS AND WE RE GOING TO END UP WITH A

CHM 105 & 106 MO1 UNIT SIX, LECTURE THREE 14 POSITIVE.78 VOLTS. SO, WILL IODINE, WILL AQUEOUS IODINE SPONTANEOUSLY REACT WITH NICKEL SOLID? YES. REMEMBER IF E-NOT IS GREATER THAN ZERO IT IS A SPONTANEOUS PROCESS AS IT IS WRITTEN. IF WE HAD WRITTEN THE REACTION THE OTHER DIRECTION, NICKEL ION REACTING WITH IODINE TO GIVE US THAT WE WOULD HAVE ENDED UP WITH A NEGATIVE.78 WHICH WOULD HAVE TOLD US THAT IN FACT IT WASN T SPONTANEOUS. ALRIGHT, WELL LET S TRY ONE MORE HERE QUICKLY. WE HAVE SOLID COPPER REACTING WITH TIN ION TO FORM, NOT COPPER, EXCUSE ME COBALT, TO FORM COBALT TWO AND TIN METAL. AGAIN WE CAN SHOW THIS IF WE WANT TO THIS IS WHAT WE RE LOOKING AT HERE SO WE HAVE OVER HERE A PIECE OF COBALT, COBALT TWO PLUS IONS OVER HERE WE HAVE A STRIP OF TIN AND Sn 2+ IONS. WE RE GOING TO PUT A SALT BRIDGE IN THERE, WE RE GONNA HOOK THIS TOGETHER THROUGH A METER AND THAT S WHAT WE RE ASKING. WHAT S THE POTENTIAL WHEN WE DO THAT? NOW THAT S ONE WAY TO SHOW IT, DRAW IT OUT OR THE OTHER WAY TO SHOW IT IS MERELY WRITE THE OXIDATION-REDUCTION EQUATION FOR THE PROCESS. SO, WE GO UP HERE AND WE SAY COPPER, EH, COBALT, IS GOING FROM ZERO TO PLUS TWO. IS IT BEING OXIDIZED OR REDUCED? OXIDIZED, THANK YOU. SO THIS IS THE OXIDATION PROCESS, AND TIN IS GOING FROM PLUS TWO TO A ZERO SO OBVIOUSLY IT IS BEING REDUCED. SO THE COBALT IS GOING TO BE THE ANODE AND THE TIN, TIN ION IS GOING TO BE THE CATHODE. SO E-NOT OF THE CELL IS EQUAL TO E-NOT OF THE CATHODE WHICH IS THE TIN. SO WE GO Sn 2+ SLASH Sn. I M JUST WRITING THAT DOWN SO THAT BOOKKEEPING WISE WE KNOW WHAT TO LOOK FOR IN THE CHART, MINUS E-NOT OF THE Co 2+ SLASH Co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

CHM 105 & 106 MO1 UNIT SIX, LECTURE THREE 15 SPONTANEOUS AND SO IT WOULD BE REFERRED TO AS A GALVANIC OR A VOLTAIC CELL. NOW IT S SPONTANEOUS IN THIS DIRECTION SO THAT WOULD MEAN THAT THE COBALT METAL HERE IS GIVING UP ELECTRONS TO BECOME COBALT PLUS TWO. SO THE ELECTRONS WOULD BE MOVING IN THIS DIRECTION, COMING BACK OVER HERE WHERE THE TIN WOULD BE GAINING THE ELECTRONS AND BECOMING TIN METAL. SO WE D HAVE TIN IONS MOVING OUT OF SOLUTION, BEING THEN REDUCED AND OVER HERE WE D HAVE COBALT IONS MOVING INTO SOLUTION AS THEY W ERE FORMED FORM THE COBALT METAL. AND IF I WERE TO HOOK THESE WIRES TOGETHER IT WOULD SPONTANEOUSLY OCCUR WITH A POTENTIAL OF PLUS.14 VOLTS SHOWING. A QUESTION? (STUDENT NOT AUDIBLE) DOESN T YES. THIS IS ALWAYS E-NOT OF THE CATHODE MINUS E-NOT OF THE ANODE. NEVER CHANGES. THESE PLUSSES UP HERE ARE OF CURSE MERELY IN STOICHIOMETRY SHOWING PLUS MEANS THAT THEY ARE IN CONTACT WITH EACH OTHER. ALRIGHT, ANY OTHER QUESTIONS ON THIS IDEA OF CALCULATING CELL POTENTIALS. ALRIGHT, IN OUR NEXT LOOK AT ELECTROCHEMISTRY WE RE GOING TO TAKE LOOK AT SOME PRACTICAL APPLICATIONS OF BOTH AS I SAID SPONTANEOUS AND NON- SPONTANEOUS ELECTROCHEMICAL CELLS, THERE S A LOT OF INDUSTRY, A LOT OF APPLICATION TO THIS USE OF ELECTROCHEMISTRY.