I need question 4 all parts answered. Ubiqui none (reduced) (c Compare the E for
ID: 518427 • Letter: I
Question
I need question 4 all parts answered. Ubiqui none (reduced) (c Compare the E for the oxygen half O: 2H reaction to the same halfreaction in Table 18.1. Under which conditions, high pH or low pH is oxygen a stronger oxidizing agent? 3. When you are recharging your cell phone battery, is this an example ofa or a spontaneous redox reaction? Explain. 4. Use information from Table 18Iand Appendix ILB to answer the following questions dealing with the (a Use half-cell potentials from Table 18.1 to determine the standard potential for this reaction. aluminum predicted to react with acids under standard conditions? (b)Use information from Appendix ll.B to determine the standard free enery change for the reaction (e)Now clear fractions for this reaction wo give 2Aloso Iraq) 2Al r(aa) 3Hs(g Determine AG for this altered version of the reaction using your answer in Now calculate E for this altered reaction using your AG and the relation between these two thermodynamic quantities. (dy When one multiplies a reaction by a factor clear fractions what happens lo. i, the moles of electrons transferred? i) the grants of reactant(s consumed? iii) the moles of product(s produced? the free energy change for the reaction? v) the potential for the reaction?Explanation / Answer
4.a. anode: Al3+ + 3e -----> Al E0L = -1.66 V
cathode: H+ + e --------> 1/2 H2 E0R = 0 V
Cell reaction : 3 H+ + Al --------> Al3+ + 3/2 H2
standard potential, E0 = E0R - E0L = 0 - (-1.66) = 1.66 V
Since E0 > 0 so Al will reacct with acids.
b) standard free energy change, delta G0 = - nFE0 where F = fardays constant = 96500 C; n = number of electrons transferred.
Here n =3
delta G0 = - 3 x 96500 x 1.66 = 480570 C.V = 480570 J = 480.6 kJ
c) i) delta G0 = 2 x 480.6 kJ = 961.2 kJ we have to multiply with 2 because n = 6 for this reaction.
ii) E0 = 1.66 V
d) i) the moles of electrons transferred is also multiplied by the factor
ii) the grams of the reactant consumed is also multiplied by the factor
iii) the moles of product produced is also multiplied by the factor
iv) the free energy change is also multiplied by the factor
v) The standard electrode potential remains same.