POSTLAB 1. Using a table of E° values, place sodium,...

POSTLAB 1. Using a table of E° values, place sodium, magnesium and silver in the appropriate places in your activity series. Na+ (aq) + e- ? Na(s) E° Na = -2.71V Mg2+ (aq) + e- ? Mg(s) E° Mg = -2.37V Ag+ (aq) + e- ? Ag(s) E° Ag = 0.80V 2. What is the highest cell voltage that could be constructed from the half cells used in this experiment? the highest cell voltage that could be constructed from the half cells used in this experiment is zinc anode + Izli- cathode 3. Supposed that you had used Fe/Fe2+ as the reference reaction instead of the one that included the most active metal. How would your answer to question 2 have changed. 4. Determine the value of ?G° for the following systems using your value for E°. a) Zn + Fe2+ ? Zn2+ + Fe b) Cu + Zn2+ ? Cu2+ + Zn

Transcript

00:01 So when the sodium is been reacted sodium with the aqueous sodium this is nothing but the aqueous with the one electron is going to get it this is a sodium solid we are going to get it and the a node of n a will be having minus two or we can say seven a two point seven one volt and the magnesium is m g uh 2 plus here it is used to come up to plus aqueous plus equius plus electron we are going to get it mg solid a knot of mg will be minus 7 .27 volt and the ag plus here it will be aqueous and electron one electron we are going to get it a g plus plus yeah this is what we are going to get it and the e knot of a cell that of a g 0 .80 volt so this is all about the first part solution moving ahead with the second part let us try to understand what is been given in the second part of the solution the highest cell volt could be construct with a half cell use the experiment of a zinc iod plus ion iodide cathode so it has been representing as an e node of cell is equal to e knot of reduction, e not of reduction plus e knot of oxidation.

01:34 So it's going to become like this way.

01:37 So from the standard value we are going to get it zinc anode plus iodide as in cathode.

01:43 So it will be the value will be putting here e knot of cell is equal to plus 0 .54 plus 0 .76.

01:53 Fold so it's become 1 .30 volt from the another cell experiment when we are using the enode cell value which is a less than this so from the other cell when we are using the experimental value so a node of cell we are going to say that the value is less than this so this is the solution part for the b next is what when we are going to see the part of a third this is a second and this is the third one now suppose that f e f e plus two is been included the most active part most active metal face then the zinc is going to be act as what anode and the f e this is going to be act as in what it is going to be act as in cathode and the zinc is going to be act as in uh like we are knowing that most active much is anode so let us try to write it a node of cell is equal to it's going to become minus 0 .45 plus 0 .76 the answer will be 0 .31 volt so this is a required solution for the 3 third part next we are moving ahead with the 4th part like here we can say that the reaction is being given zn plus f e plus 2 we are going to get it over here that z and plus 2 plus fv then the delta and g value we are knowing that the formula n minus n f e not so then n is nothing but the number of electron transfer in the reactions and f is ferrida constant that is a 965 0 value we are knowing so we are using the formula of a not of reduction as we are used the above one same plus e not this is nothing but the e not is in order of oxidation and reduction.

03:59 So oxidation and reductions, putting the values we are getting minus 0 .45 plus 0 .76 volt.

04:08 So the value will be plus 0 .31 volt.

04:12 Substituting this value in this delta g formula, what we are going to get it, that we are going to get it that the delta g is equal to minus 59 .83 kilojoules divided by mole.

04:26 The reaction is a feasible as the change in the grip energy will be negative so when we are going to be considering as in what negative so here the reaction will be z in c u plus z in plus two we are getting c u plus two plus z n so number of electron again we are going to be writing with that e knot of cell so e not is equal to reduction oxidation so putting the value 0 .76 minus plus 0 .15 volt so the value will be here it is going to become minus 0 .91 volt...

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