if the [Zn2]-2.0 and Using the relationships ?G \"J\' = 2.0 M and [Ap.)-0.50 M,
ID: 704513 • Letter: I
Question
if the [Zn2]-2.0 and Using the relationships ?G "J' = 2.0 M and [Ap.)-0.50 M, this would be a non-standard cen calculate non-standard cell potentials, the Nernst equation --nF? and aG.mFe we can come upwi where nF n # moles of electrons in the balanced half-reactions Fe faradays constant 96,000 C/mol e. Q - reaction quotient, [products] I [reactants) Calculate non-standard ell potential (e) at 298 K when the Zn2j - 2.0 M and [AP) 0.50 M. 4. Calculate ?G and ?G° and discuss the difference in the work being done, in terms of [products] and [reactants]. 5. 6. What is the effect on $ of doubling the size of the aluminum electrode? 7, What is the effect on ? of adding 500 mL of water to the cathode, assuming there was 1 L of the solution to begin with?Explanation / Answer
4. Non standard cell potential
cell reaction,
2Al(s) + 3Zn2+(aq) --> 2Al3+(aq) + 3Zn(s)
a) Q = [Al3+]^2/[Zn2+]^3
feeding given concentration values,
Q = (0.5)^2/(2.0)^3 = 0.03125
b) Eo = Ecathode - Eanode
= -0.763 - (-1.660)
= 0.897 V
c) n = 6
d) E = 0.897 - (8.314 x 298/6 x 96500) ln(0.03125)
= 0.912 V
6. Doubling the size of Al electrode would produce more electrons by oxidation of Al to Al3+ and therefore, more of Zn2+ would get reduced. Ratio Q would become larger and E also increases accordingly.
7. Diluting the solution of the cathode (Zn2+) by addition of 500 ml water wuld reduce the [Zn2+] concentration and therefore, Q would becomes larger. The E value increases in this case.
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