Exercise 20.21 Five part question. However, I have already done two parts. If yo

Question

Exercise 20.21 Five part question. However, I have already done two parts. If you are unable to answer all the remaining parts of this question, please allow someone who will. Thanks.

Use the data in Appendix D in the textbook to calculate the standard cell potential for each of the following reactions.

Exercise 20.21 Part A Use the data in Appendix D in the textbook to calculate the standard cell potential for each of the following reactions. H2(g) + F2 (g) 2H+ (aq) + 2F-(aq) Express your answer using four significant figures. Ecell: 2.87 Submit My Answers Give Up Incorrect; Try Again; 5 attempts remaining Part B Cu(s) + Ba2 + (aq) Cu2+ (aq) + Ba (s) bull= -3.26 V Submit My Answers Give Up Correct

Explanation / Answer

Answer – We are given the redox reaction we need to calculate the Eocell for each reaction –

Part A) H2(g) + F2(g) -----> 2H+(aq) + 2 F-(aq)

So in this one there is H2 gets oxidized and F2 gets reduced

H2(g) -----> 2H+(aq) + 2e- Eo = 0.00 V

F2(g) + 2e- -----> 2 F-(aq), Eo = +2.87 V

H2(g) + F2(g) -----> 2H+(aq) + 2 F-(aq) , Eocell = +2.870 ( Four significant )

Part D) Hg(l) + HgCl2(aq) -----> Hg2Cl2(s)

So,

Hg2+ + 2e- -----> Hg (l)    Eo = +0.854 V

Hg2Cl2 (s) + 2e- ----> 2Hg(l) + 2Cl-   Eo = +0.2676 V

Now we need Hg(l) in reactant side and Hg2Cl2(s) in product side ,so

Hg2+(aq) + 2e- -----> Hg (l)    Eo = +0.854 V
2Hg(l) + 2Cl- (aq) ----> Hg2Cl2 (s) + 2e- Eo = -0.2676 V

Hg(l) + HgCl2(aq) -----> Hg2Cl2(s) Eo = 0.59 V

Part E) We are not given part C, but the value for the redox reaction Eo cell is positive that one will occur spontaneously, so form here AD will occur spontaneously

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