For the following half reactions below a. Balance the overall reaction and calcu

Question

For the following half reactions below a. Balance the overall reaction and calculate E degree cell for the reaction under standard conditions b. Write the equilibrium expression for the reaction (K_c or Q_c) c. Determine E_cell for each i. [Ag^+] = 100 M and [Al^3+] = 0.200 M ii. [Ag^+] = 0.0100 M and [Al^3+] =0.300 M Ag^+(aq) + e^- rightarrow Ag(s) E degree = 0.799 V Al^3+(aq) + 3e^- rightarrow Al(s) E degree = -1.66 V Determine how much time it would take to plate out 10.0 g of chromium (Cr) from a solution of CrCl_3 using 12.0 A. Determine what mass of Titanium would be plated out of a solution of TiCl_2 using a current of 10.0 A for 5.00 hours

Explanation / Answer

a. Al (s) + 3Ag+ -------> 3 Ag (s) + Al3+

Given, Ag+ + e ------> Ag E0R= 0.799 V

Al3+ + 3e -------> Al    E0L = -1.66 V

E0 = E0R - E0L = 0.799 - (-1.66) = 2.459 V where E0R , E0L are  electrode potential of cathode and anode resectively.

b) Kc = [Al3+]/[Ag+]3   where Al and Ag will not appear in the expression because they are pure solid.

c) using Nernst equation,

E cell = E0 - (2.303RT/nF) log [Al3+]/[Ag+]3 here n = 3 and 2.303RT/F = 0.059V

i) [Ag+]= 1.00 M and [Al3+] = 0.200 M

E cell = E0 - (2.303RT/nF) log [Al3+]/[Ag+]3

   = 2.459 - (0.059/3)log (0.200/13) = 2.459 - (0.059/3) x (-0.7) = 2.459 + 0.014 = 2.473 V

ii) [Ag+]= 0.010 M and [Al3+] = 0.300 M

E cell = E0 - (2.303RT/nF) log [Al3+]/[Ag+]3

   = 2.459 - (0.059/3)log {0.300/(0.01)3} = 2.459 - (0.059/3) x (5.477) = 2.459 - 0.108 = 2.351 V

2. Ti3+ + 3e --------> Cr

current = 12 A, amount of Cr deposited = 10 g

3F charge is required to deposit = 1 mole of Cr = 52 g Cr (from reaction) atomic mass of Cr = 52 g

or, To deposit 52 g Cr charge required = 3F = 3 x 96487 C (1 F = 96487 C)

To deposit 10 g Cr charge required = 3 x 96487 x 10/52 = 55665 C

charge = current x time

time = charge/current = 55665C / 12A = 4639 s = 77 min

The requird time = 77 min

3. Ti2+  + 2e --------> Ti

current = 10 A, time = 5 h = 5 x 60x 60 s= 18000 s

charge = current x time = 10 A x 18000 s = 180000 C

2F charge is required to deposit = 1 mole of Ti = 48 g Cr (from reaction) atomic mass of Ti = 48g/mol

180000 C charge will deposit = 48 x 2F/180000 = (48 x 2 x 96487)/180000 = 51.5 g

The amount Ti deposited = 51.5 g

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