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Calculate E"for [ CIOz (a4) + 3H,0() O3(g) +Hzo() Consider the following two the spontancaus 3 hatereaction 2 HOZ - redoxrejction 6OH 2 (aq) their standard 1,2...

Question

Calculate E"for [ CIOz (a4) + 3H,0() O3(g) +Hzo() Consider the following two the spontancaus 3 hatereaction 2 HOZ - redoxrejction 6OH 2 (aq) their standard 1,246 V 3 reduction potentials.arecoupled Ml See Perodic Table See Hint

Calculate E"for [ CIOz (a4) + 3H,0() O3(g) +Hzo() Consider the following two the spontancaus 3 hatereaction 2 HOZ - redoxrejction 6OH 2 (aq) their standard 1,246 V 3 reduction potentials. arecoupled Ml See Perodic Table See Hint



Answers

Use the table of standard reduction potentials (Appendix $\mathrm{M}$ ) to calculate $\Delta_{\mathrm{r}} \mathrm{G}^{\circ}$ for the following reactions at $298 \mathrm{K}$ (a) $3 \mathrm{Cu}(\mathrm{s})+2 \mathrm{NO}_{3}^{-}(\mathrm{aq})+8 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow$ $3 \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{NO}(\mathrm{g})+4 \mathrm{H}_{2} \mathrm{O}(\ell)$ (b) $\mathrm{H}_{2} \mathrm{O}_{2}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq})+2 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow$ $\mathrm{Cl}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\ell)$

To calculate the standard reduction potential in art reduction for the reduction off Berry um, two plus sign to gain electrons and from Perry. Um, so we have given and this there and barium react with Barry. Um, I react with the chloride ions to farm Very um chloride. The Delta Jeannot for this reaction is given, which is 16 point seven killer job. So to determine the standard reduction potential for this reaction, we will use the equation. The TelDta jean art is equal to minus. And if in art, rearranging this equation for inert, we will get in. Art is equal to minus delta g not divide by. And if putting the values so we will get in art is equal to minus where Delta Jean art is 16.7 kill a whole are 16 7 double zero chow. So we will put the value of Delta jean art and job, and this is equal to 16 7 double zero job where n is the number of moles of electron, which is to more off electrons. If it's Ferrari constant, which is 96,500 Cool, um, part moral of electrons so calculating the value off the energy we get in art is equal toe 2.906 Jowell, Pirkle, um are in art is equal toe minus two point 906 r 91 and straight up. Jowell Pirkle Um we will unit right the unit work because child per column is equal to one world. This is the required solution to this question.

In each part of this problem were given an overall cell reaction, and we want to determine what the balanced cell equation is, as well as value for the cell potential for these reactions as they're written, you also want to determine whether or not the reaction is spontaneous, starting with Port A. This is the reaction that were given. And so we know that we can look up from the table that these were the 2/2 reactions that are taking place. And now, in order to determine which one of these half reactions we need to flip which ones we need to keep. We have to look at the given reaction in the problem because we know that we need to rate the overall cell potential as it is written. What that means is that for Emma No. Four minus going to mn two plus the half reaction Emma, No. Four minus going to MN two plus has that same form, and so that half reaction for the standard reduction potential would stay the same value of positive 1.51 because it matches the way that this overall reaction is written. And then for I minus going to I to the half reaction for the reduction. Is I two going to I minus? So that means that we would need to flip the sign of that second reaction when we go to combine them. So we keep the first reaction and flip the 2nd 1 So now we can rake that up. We keep the first reaction, which is Eman 04 minus plus eight. H plus was five electrons goes to mn two plus plus four h +20 And again, the standard reduction potential for that half reaction is 1.51 volts. And now we have to flip the second reaction in order to get it in the form that the overall reaction is written. So that is to I minus goes to I to plus two electrons and then we flip the value for that standard reduction potential since we flip the equation. So that is now negative 0.54 bolts. And then we add these two equations Together we have to multiply the first reaction by two in the second reaction by five so that we get 10 electrons to cancel out on either side. And that leaves us with the overall balanced cell equation which comes out to 16 each plus Equus Waas to Eman 04 minus acquis plus 10 I minus equally ists goes to five I to Equus was two mn two plus Equus plus eight each to a liquid. And now, in order to calculate the overall cell potential, we add this to standard reduction potentials for the half reactions together. So that's 1.51 minus 0.54 which comes out to 0.97 volts. Now. We also want to determine whether or not this is spontaneous. Since the value for the standard cell potential is positive, that means that this reaction is spontaneous. If it were negative, then it would be non spontaneous. And now for the reaction given in part B, these were the two reduction half reactions that are taking place now we need to see we need to adjust them based on how the overall reaction is written. We have Emma No. Four minus going to mn two, plus I m in 04 minus going to mn two plus. So that means that we keep that second reaction as it is for the first reaction we have f minus going to have to hear. We have F two going to F minus, so we need to flip the value of that standard reduction potential as well as the half reaction itself for that first reaction. And so that will come out to negative 2.87 And now we can combine them. So he flipped that first half reaction. So it's now two F minus goes to F two plus two electrons in the standard reduction potential. After we flip, it is negative 2.87 volts, and we keep the second reaction, as is so that's M N 04 minus waas eight h plus plus five. The electrons goes to Eman two plus plus four waters, and we keep that standard reduction potential as it is. So 1.51 volts. And now we combine these two equations and we add their cell. There's reduction potentials. Together we multiplied the first equation by five in the second equation by to get 10 electrons cancelling out on either side. Now we can combine the remaining terms to get the overall balanced sell equation that comes out to 16 each plus acquis waas to M and 04 minus Equus waas 10 F minus Equus goes to five F to gas was to mn two plus in cui ists plus eat h 20 liquid. So that's the overall balance cell equation and how to get the overall standards cell potential. We had the two, the two Santa reduction potentials together. So negative 2.87 plus 1.51 comes out to an overall cell potential of negative one point 36 volts. And because this time the the overall cell potential is negative, that means that this reaction is non spontaneous.

For a reaction in which zinc oxidizes and toe certain two plus an aqueous solution when we're U three plus an aqueous solution reduces toe you two plus in aqueous solution at platinum electrodes. So in this reaction, in order, the zinc oxidizes and toe didn't two plus air it release two electrons. The e oxidation for the sink is 0.76 world. We're in the electrochemical. Siri's reduction value for the think is minus 0.76 world to change the reduction. Where you into oxidation? We will just change the sign. So the EU exploration for the sink is 0.76 Or on the other hand, at get or the reduction of the you three plus signs get tau electron changes and toe you to bless science. The e reduction value for this is Toby determined while the e not off the sill. His gamble, which is equal toe 0.40 world is we know the not off the self is equal toe in our top oxidation plus be not up production to determine e not up reduction which will be equal toe not off the cell minus in not up oxidation. So in art off, the sale is 0.40 while e oxidation is 0.76 words. Subtracting the values way will get the E production for the You three plus and toe you two plus, which is equal to zero point 36 world minus 0.36 world. So the the reduction value for the U three plus sign is minus 0.36 warm, which is the required solution to the gallon question.

To figure out whether or not these reactions are spontaneous. We need to figure out what the reduction potentials are for the 2/2 reactions that comprise the overall reaction and going to a table of reduction potentials will find this reaction here with a reduction potential of 1.51 in the second half. Reaction with a reduction potential 0.54 now. Normally, what we would do is we would some of these two half reactions together changing the one that has the lower reduction potential? Because that would be the one that would not be allowed to be reduced because the other half reaction has a higher reduction potential. But because we have the written chemical reaction, we need to some these two reactions in such a way that we get the chemical reaction that was provided and whichever half reaction we have to flip. That would be the one that is serving as the an ode. The other half reaction will be the one serving as the cathode, it turns out, in this case we have to flip the second half reaction. We'll play the first half reaction by two in the second half. Reaction by five. And then we can get the balanced Redox reaction that was given to us because we are flipping the second half reaction that is serving as the an ode. So we'll take the Catholic potential 1.51 minus theano potential 0.54 and we'll get the cell potential of 0.97 because this is a positive self potential than this is a spontaneous chemical reaction. We'll do the same thing with a second chemical reaction identifying the 2/2 reactions. One is the same as before, with a reduction potential of 1.51 and then flooring going to fluoride with the reduction potential of 2.87 in order to get the electrons to counsel you to multiply the second half reaction by five in the first half, reaction by two. Then, in order to get the overall reaction that was given to us, we have to flip the second half reaction. The second half reaction then will be serving a piano. The first half reaction will service the cathode and will stay as written, but we some these up. We then get the overall balance Redox reaction, and then what's interesting with this one is because we flip the second half reaction with the higher reduction potential, we're going to end up with a negative cell potential, meaning the reaction has written is not spontaneous.


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