Map t Sapling Learning macmillan learning Calculate the cell potential for the f
ID: 591741 • Letter: M
Question
Map t Sapling Learning macmillan learning Calculate the cell potential for the following reaction as written at 25.00 , given that Cr21-0884 M and [Fe21 0.0160 M. Standard reduction potentials can be found here. Cr(s) + Fe2+(aq) Cr2+(aq) + Fe(s) Number E- 1.84 -O Previous ) Give Up & View Solution *Try Again Next Exit Explanation First, calculate the standard cell potential, E°. Then use the Nernst equation RT nF where n is the number of electrons transferred, F is the Faraday constant (F 96485 J/(V mol), R is the gas constant (R = 8.3145 J/(mol-K), Tis the absolute temperature, and Q is the reaction quotient (product concentrations over reactant concentrations). Note that n is sometimes symbolized as ve or z in some textbooks.Explanation / Answer
Lets find Eo 1st
from data table:
Eo(Cr2+/Cr(s)) = -0.91 V
Eo(Fe2+/Fe(s)) = -0.44 V
the electrode with the greater Eo value will be reduced and it will be cathode
here:
cathode is (Fe2+/Fe(s))
anode is (Cr2+/Cr(s))
The chemical reaction taking place is
Fe2+(aq) + Cr(s) --> Fe(s) + Cr2+(aq)
Eocell = Eocathode - Eoanode
= (-0.44) - (-0.91)
= 0.47 V
Number of electron being transferred in balanced reaction is 2
So, n = 2
we have below equation to be used:
E = Eo - (2.303*RT/nF) log {[Cr2+]^1/[Fe2+]^1}
Here:
2.303*R*T/F
= 2.303*8.314*298.0/96500
= 0.0591
So, above expression becomes:
E = Eo - (0.0591/n) log {[Cr2+]^1/[Fe2+]^1}
E = 0.47 - (0.0591/2) log (0.884^1/0.016^1)
E = 0.47-(5.151*10^-2)
E = 0.4185 V
Answer: 0.418 V
Feel free to comment below if you have any doubts or if this answer do not work. I will edit it if you let me know