Consider the two half-reactions. You wish to use these reactions to produce curr

Question

Consider the two half-reactions. You wish to use these reactions to produce current in a galvanic cell. Which would you choose to be the anode (oxidation reaction) and what would be the cell potential? 6. Sn2+ 2e Sn (s) Al3+ + 3e- Al (s) E"=-0.14 V E" =-1.66V b. Sn(s) /Sn2q)cl +1.80V c, Al(s) /A134(aq) Eocell = + 1.52V d. Al(s) /Al+aq) e. Al(s) /Alaq) E"cell =-1.80V E"cell = + 0.52V Which of the following reactions or processes show an increase in entropy? 7. a. KClO3(s) 2KCI (s) + 302(g) b. N2(g) + 3H2(g)2NH3(g) c. N2H2(g) + H2(g) N2H4(g) d. 2CO(g) +02(g) 2CO2(g) e, H2O(l) H2O (s) 8. The enthalpy of vaporizaon(AHy)of liquid L is 77 kJ/mol and Aspe is 175 J/K/mol. What is the boiling point of liquid Lat 1 atm pressure (standard condition)? a. 250 K b. 440 K C. 400 K d. 3.3K e. 0.44K 9. Use Hess's law to calculate G°rxn using the following information: 7N2(g) + 12H,O(I) 6Txn =? 5N2H,(i) 50.6 4 HNO3(g) + Hof (kJ/mol) -133.9 S°(J/mol-K) 266.9 -285.8 70.0 121.2 191.6 a. +4.90×103 kJ b. -2.96 × 103 kJ C. -2.04 × 103 kJ d. +3.90 x 103 kJ e. -3.29 × 103 kJ

Explanation / Answer

Q6

Remember that each species will have a specific reduction potential. Remember that this is, as the name implies, a potential to reduce. We use it to compare it (numerical) with other species.

Note that the basis if 2H+ + 2e- -> H2(g) reduction. Therefore E° = 0 V

All other samples are based on this reference.

Find the Reduction Potential of each reaction (Tables)

Sn2+ + 2e- = Sn(s) E = -0.14 V

Al+3 + 3e- = Al(s) E = -1.66

The most positive has more potential to reduce, it will be reduced

The most negative will be oxidized, since it will donate it selectrons

For total E°cell potential:

E°cell = Ered – Eox

Eox = -Ered of the one being oxidized

for this exercise

Al is going to Oxidize, since it has a lower electrode potential

Sn is going to reduce

so

E°cell = Ecathode - Eanode = -0.14 - -1.66 = 1.52 V

from the list, choose C

Q7

increase in entropy -->

Entropy is a measure of chaos; therefore, the more chaotic it becomes, the more entropy increases.

In general:

Entropy increases with the number of particles/moles in the system:

1 mol of AB has less entropy than 1 mol of A and 1 mol of B; due to higher amount of microstates

Entropy increases as the volume of particles increases. This can be compared with physical states

Volume of Solid < Volume of Liquid << Volume of Gas

Therefore, Gases have higher entropy than liquids and solids. Liquids have higher entropy than solids.

As T increases, the entropy increases, by definition, since it is a state function which depends on Temperature.

Mixture typically increases Entropy, so ordered states get higher amount of microstates.

Example could be Gas A + Gas B separated by a membrane, then eventually mixing A + B, microstates increases, meaning that entropy increases

now..

a) increases, since O2 gas is produced

b)

decreases, sine 1+3 = 4 mol of gases produces only 2

c) decreses since 2 mol of gases produce onyl 1

d)

decreases, since only 2 mol of gas are produced

e)

decreases, since solid is much ordered than liquid

Q8

dG = 0

dG = dH - T*dS

dH - T*dS = 0

dH/dS = T

T = 77*1000/175

T = 440

Q9

GRxn --> dh - t*ds

dH = (7*0+12*-285.8) - (-133.9*4+5*50.6) = -3147

dS = (7*191.6+12*70) - (266.9*4+5*121.2) = 507.6

d G= -3147 - 298*507.6/1000

dG = -3298.2

dG = -3.3*10^3 kJ

choose e

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