One drop has a volume of about 0.05 mL, thus the concentration of Cu^2+ in the d

Question

One drop has a volume of about 0.05 mL, thus the concentration of Cu^2+ in the dilute half-cell is about (0.05 mL/25 mL)(0.1 M) = 2 times 10^-4 M. Use Equation 16.15 to calculate it at 298 K and compare it with your observed value? Why might they differ? After adding 11 drops (0.5 mL) of Na_2S solutions, [S^2] is about (0.5 mL/25 mL)(0.1 M) = 2 times 10^-3 M a. Calculate [Cu^2+] from the solubility product constant: K_sp = 8 times 10^-37 = [Cu^2+][S^2] b. Then calculate E and again compare it with the compare with the E observed after you added Na_2S Solution to dilute Vu^2+ half-cell. Suppose you had solutions 1 M in Hg(NO_3)_2 and Na_2 available. K_sp HgS = 1.6 times 10^-34. What is the highest E you could achieve with a Hg-Hg^2+ concentration cell? What would be the constituents of each half-cell? Neglect the fact that S^2- hydrolyzes extensively in water.

Explanation / Answer

Solution:

2. Given, E= -(0.0592/n) x log([(An+)(M_dil)]/[(An+)(M_conc]

Or, E= -(0.0592/n) x log([(Cu2+)(M_dil)]/[(Cu2+)(M_conc]; Here, n = 2

Or, E= -(0.0592/2) x log([2x10-4]/[0.1]

Or, E= -(0.0592/2) x log([2x10-4]/[0.1]

Or, E= -(0.03) x log([2x10-4]/[0.1])

Or, E= -(0.03) x [log(2x10-3)]

Or, E= -(0.03) x (-2.7) = 0.081 V

3. (a) Given, Ksp = 8x10-37 = [cu2+][S2-]

Therefore, [cu2+] = [S2-]/8x10-37

Or, [cu2+] = 2x10-3/8x10-37 = 2.5x1033 M

(b) The reaction involved in the half-cell reaction is: Cu2+ + 2e ----------> Cu(s)

According to Nernst electrochemical equation at STP we have:

E = E0 – 0.0591/n log [Cu]/[Cu2+]   

= 0.34 - 0.0591/2 log 1/(2.5x1033) ; E0(cu) = +0.34 (from electrochemical series), also we know [Cu(s)] = 1

= 0.34 – 0.03 log(4x10-34)

= 0.34 – 0.03 (0.602 - 34)

= 1.34 V

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