Electrochemistry. Calculate E cell and determine if spontaneous Calculate the E_

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Electrochemistry. Calculate E cell and determine if spontaneous

Calculate the E_cell degree for the following reactions and determine if they would be spontaneous. Ca2+(aq) + Zn(s) rightarrow Ca(s) + Zn2+(aq) 2H_2O_(l) + SO_2(s) + Sn2+(aq) rightarrow SO_4 - 2(aq) + Sn(s) + 4H_+(aq) A voltaic cell employs the reaction: 2Fe_3+(aq) + 3Mg_(s) rightarrow 2Fe_(s) + 3Mg_2+(aq) Calculate the E_cell at 25 degree C under each of the following conditions Fe^+3 + 3e rightarrow Fe E_red degree = -0.036V [Fe^3+] = 1.00 M; [Mg^2+] = 1.00M [Fe^3+] = 1.8 Times 10^-3M; [Mg^2+]= 2.50M [Fe^3+] = 2.50M; [Mg^2+] = 1.8 Times 10^-3M Which of the 3 sets of conditions (a, b or c) is the most spontaneous? Does this support Le Chatelier's principle (why/why not)?

Explanation / Answer

(1) Note: All E° were read from Petrucci's Book.

Solution:

Ca+2   + 2e-   ----> Ca (s)      E° = -2.84 V

Zn (s)   ------> Zn+2   +   2e-     E° = 0.763

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Ca+2 + Zn(s) --->   Ca (s) + Zn+2      E° = -2.84 + 0.763 = -2.077 V

dG° = -nFE° = - 2 * 96500 * -2.077 = 400861 > 0 non-spontaneuos

Solution:

2H2O (l) +   SO2(g)   ---> 2e- + SO4-2(aq)   + 4H+      E° = -0.17 V

Sn+2(aq) + 2e- ----> Sn (s)                                       E° = - 0.137

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2H2O (l) +   SO2(g) + Sn+2(aq)   ---->   SO4-2(aq) + Sn (s)   + 4H+          E° = -0.17 - 0.137 = -0.307 V

dG° = - n F E° = - 2* 96500 * -0.307 = 59251 > 0 non-spontaneous

(2)  

(Fe+3   + 3e- ----> Fe(s) ) x 2 E° = -0.036

(Mg(s) ------> Mg+2 + 2e- ) x3    E° = 2.356 V

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2Fe+3   +   3Mg(s) ----> 2Fe(s) +3 Mg+2        E° = -0.036 + 2.356 = 2.32 V

From Nernst equation:

E = E° - (RT/nF) * lnQ

and

dG = dG° + RT ln Q or

dG = -nFE° + RT ln(Q)

Where:

n = moles of electrons (6 in your case)

T = 25 °C + 273 = 298 K

F = 96500 C/mol

Q = [products]a / [reagents]b (ionic compounds) or = [Mg+2]3 / [Fe+3]2

Solving each case:

(a) [Fe+3] = 1 M and [Mg+2] = 1 M

dG = -nFE° + RT ln(Q) = (- 6 * 96500 * 2.32) + (8.314*298*ln(13/12) = - 1 343 280 J or -1343 kJ

(b) [Fe+3] = 1.8*10^-3 M and [Mg+2] = 2.50 M

dG = -nFE° + RT ln(Q) = (- 6 * 96500 * 2.32) + (8.314*298*ln(2.503/0.00182) = - 1 305 153 J or -1305 kJ

(c) [Fe+3] = 2.50 M and [Mg+2] = 1.8*10^-3 M

dG = -nFE° + RT ln(Q) = (- 6 * 96500 * 2.32) + (8.314*298*ln(0.00183/2.502) = - 1 394 795 J or - 1394 kJ <<<---- This is the more spontaneous.

Yes, it supports Le Chatelier's principle because in case (c) there's more reagent than product, so the reaction is more likely to go in the forward direction (->) than others.

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