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18. If the value of E is 2.10V for the reaction what is the value of Eo ee for a, -4.20V 1.05 V c. 2.10V d, 1.05 V -2.10V 19. Given: Al (aq)+3e 1.66 V Cl,(g) +2e What is AG for the following cell reaction? 2AICiadaq) a. 5.8 x 10 J b, 5.8 x 10 J c. -6.5 x 10 J 1.7 x 1.7 10 J 20. What is E of the following cell reaction at 25°C? Eocell 0.460 V. cus cu (0.017 Ag (0.18 MIAgos 0.468 V b. 0.282 V c, 0.460 V d. 0.490 V e. 0.479 V 21. What mass of chromium could be deposited by electrolysis of an aqueous solution of Crz(souh for 160 min using a constant current of 15.0 A? (F-96485 C/mol) a. 0.431 g 25.9 g c. 232.8 g d, 0.187 g e, 38.8 g
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Answer #1

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
ΔGo=-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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