If the value of E_cell^0 is 2.10 V for the reaction F_2 (g) + 2Fe^2+ (aq) righta

Question

If the value of E_cell^0 is 2.10 V for the reaction F_2 (g) + 2Fe^2+ (aq) rightarrow 2Fe^3+ (aq) + 2F^- (aq), what is the value of E_cell^0 for F^+ (aq) + Fe^3+ (aq) rightarrow Fe^2+ (aq) + 1/2 F_2 (g)? a. -4.20 V b -1.05 V c. 2.10V d. 1.05 V e. -2.10 V Given. Al^3+ (aq) + 3e^- Al(s); E^0 = -1.66 V Cl_2 (g) + 2e 2Cl^- (aq); E^0 = 1.36 V What is Delta G degree for the following cell reaction? 2AICl_3 (aq) 2Al(sl) + 3Cl_2 (g) a. -5.8 times 10^5 J b. 5.8 times 10^5 J c. -6.4 times 10^5 J d. -1.7 times 10^4 J e. 1.7 times 10^6 J What is E of the following cell reaction at 25 degree C? E_cell^0 = 0.460 V. Cu(s)|Cu^2+ (0.017 M)||Ag^+ (0.18 M)|Ag(s) a. 0.468 V b 0.282 V c. 0.460 V d. 0.490 V e. 0.479 V What mass of chromium could be deposited by electrolysis of an aqueous solution of Cr_2 (SO_4)_3 for 160 mm using a constant current of 15.0 A? (F = 96485 C/mol) a. 0.431 g b. 25.9 g c. 232.8 g d. 0.187 g e. 38.8 g

Explanation / Answer

18. Reverse the reaction and divide by 2,
   Eocell = (-2.10V)/2 = -1.05V , so option b) is the answer
19. 2 Al+3 + 6e- -------> 2Al ; Eo = 2*(-1.66) = -3.32V
6 Cl- ---------> 3Cl2 + 6e- ; Eo = 3*(-1.36) = -4.08V
   Add these 2 above equations to get the overall 9eqn given in the question)
Eocell = Ecathode - Eanode
Go=-nFEocell , use these formulas to get the answer
20) At room temperature,
   Ecell = Eocell - 0.0592/n log [Ox]/[Red]
   Ecell = 0.460 - 0.0592/2 log (0.017 / 0.18)
   Ecell = 0.4903 V option d)
21) electrons deposited = it /F = (15*160*60 sec) / 96485 C/mol = 1.49245 mole of electrons
   1.49245 * (1 mole of Cr / 3moles of elcrons) = 0.49748 mol of Cr
   mass of Cr deposited = 0.49748 mol of Cr * 51.99 gm/mol = 25.86 gms = 25.89 gms approx

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