Calculate the cell potential E at 25 °C for the reaction below given that [Fe2+] = 0.020?
2Al(s) + 3Fe2+ (aq) --> 2Al3+ (aq) + 3Fe(s)
A) +1.03 V
B) +1.17 V
C) +1.18 V
D) +1.20 V
Update:Calculate the cell potential E at 25 °C for the reaction below given that [Fe2+] = 0.020
M, [Al3+] = 0.10 M, and the standard reduction potential is -1.66 V for Al3+/Al
and -0.45 V for Fe2+/Fe.
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The answer is D. You have to find the half reactions and determine which is oxidized and which is reduced and to do this u have to look at the SRP table. You find that
Al3+(s)-----> Al(s) +3e- E cell= -1.66
you will realize that we need the opposite which so we need +1.66.
This is oxidation. To find reduction we see on the table that:
2e- + Fe2+ --------> Fe(s) E cell= -.44
We need this as is. And to calculate the cell potential we add the 2 values we have:
(+1.66) + (-.44) = 1.22 ======> answer is D. (1.20 V)