Table 1. Half-cells Mg (anode) Cu (anode) Zn (anode) Al (anode) Mg (cathode) ***
ID: 522590 • Letter: T
Question
Table 1.
Half-cells
Mg (anode)
Cu (anode)
Zn (anode)
Al (anode)
Mg (cathode)
*****
875
Cu (cathode)
1460
*****
840
570
Zn (cathode)
*****
Al (cathode)
850
260
*****
2. Use the data above to fill in the following table (show all calculations for G and Keq):
Table 2.
Balanced Net reaction
G
Keq
#1 Highest
positive voltage
#2 Highest
positive voltage
#3 Highest
positive voltage
Lowest
positive
voltage
Half-cells
Mg (anode)
Cu (anode)
Zn (anode)
Al (anode)
Mg (cathode)
*****
875
Cu (cathode)
1460
*****
840
570
Zn (cathode)
*****
Al (cathode)
850
260
*****
Explanation / Answer
1) Magnesium can not be cathode in this combination, it only can be anode. so all values are = zero
2) Mg+ Cu+2 ---------------> Mg+2 + Cu
delta G (free energy change) = - nFE
= - 2x 96500 x 1450 = - 281780000 J / mole = - 281780 kj/mole
calculation of Kc
delta G (free energy change) = - 2.303 x RT logKc
log Kc = - delta G / 2.303 x RT = - 281780/ 2.303 x 0.8314 KJ/Kelv. mol x273
logKc = 2.303 x 0.8314 x 273/ 281780
Kc = 4.7622 x 10-28
3) here ZINC CAN BE CATHODE AND Mg CAN BE ANODE BUT VALUES ARE NOT SERVED.
4) This equation is calculated for highest possible potential only , the given combinations are not possible for lowest potentials.
3 Mg+ 2Al+3 ---------------> 3Mg+2 + 2Al
delta G (free energy change) = - nFE
= - 6x 96500 x 850 = - 492150000 J / mole = - 492150kj/mole
log Kc = - delta G / 2.303 x RT = - 492150kj/ 2.303 x 0.8314 KJ/Kelv. mol x273K
logKc = 2.303 x 0.8314 x 273/ 492150kj
Kc = 1.8249 x 10 -30
please note that the table data given is not quite clear, the units of the digits given in the table are not mentioned
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