17. The emf of the cell comsisting of the ZnZn half-cell with 1Zn- SHE with P (H

Question

17. The emf of the cell comsisting of the ZnZn half-cell with 1Zn- SHE with P (H2)-2atm and [H'] ?.8Mís. _dZn +21r Two half-cell reactions: -0.45 M and the 211. +2e. ? H2 (g) (E%0.00V) A) 0.78 V B) 0,84 V C) 0.58 V D) 0.64 V E) 0.94 V 18. Using the standard reduction potential Er 0.54V) of reaction Br2() + 2raq)?2B'(aq) + 1ds) to find the equilibrium constant at 25C is A) 5.2 x 10 B) 1.7 x 101 C) 2.7 x 10 D) 1.75 x 10 19. What mass of Ti(s) may be deposited from an aqueous TiCI: solution if a current of 2.50 A is applied to the solution for 365 s? (The atomic mass of Ti is 47.867, F- 96500 C/mol) A) 0.453 g B) 0.226g C) 0.369 g D) 0.906 g E) 0.139 g 20, what's -1? particle? A) Proton B) Neutrorn C) Electron D) Positron 21. X, inthe equation 27G+ In--SMn. X is A):? B) Ip

Explanation / Answer

17) Oxidation= Zn----> Zn2+ + 2e- Eo = +0.76V

Reducion = 2H+ + 2e- ---> H2(g) Eo = +0.0 V (standard hydrogen electrode)

Overall: Zn + 2H+----> Zn2+ + H2

Eo cell =+0.76 V

Now, the emf can be calculated using the Nernst equation E cell

= Eo cell - (0.059 / n) log Q

Qcell = ([Zn2+] (PH2) / [H+]^2

Q = ((0.45)(2)) / (1.8)^2 = 0.278

n = 2 as 2 electrons are transferred

E cell = 0.76 V - 0.59 / 2) log (0.278) = 0.776 V= 0.78 V

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