How do you get this answer?
A concentration cell is constructed using two Fe3+/Fe half-cells. What is the cell potential, Ece for this concentration cell given the following two half-cell Fe3+ion concentrationsFeau 5.87 x 106M and [Fe3+]c3.74 x 10-2M? 14. a) 0.173 V ) 0.075V b0103 V c) 0.225 v
Solution
Following are the half cell reactions
Fe(s)(dil) - - - - - - > Fe 3+ (aq) + 3e
Fe 3+ (aq)(conc) + 3e - - - - - - > Fe(s)
standard reduction potential for both half reaction is same and E° cell is zero
Therefore
Nernst equation for concentration cell is
E cell = (0.0592/n)log([Fe 3+ ]conc/[Fe 3+ ]dil)
where,
n = number of electron transfer, 3
Q=[Fe 3+ ]conc/[Fe 3+ ?]dil = 3.74×10 -2 M /5.87×10 -6 M = 6.37×10 3
Therefore
E cell = (0.0592/3)log(6.37×10 3 )
E cell = 0.075V
.
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How do you get this answer- A concentration cell is constructed using.docx
1. How do you get this answer?
A concentration cell is constructed using two Fe3+/Fe half-cells. What is the cell potential, Ece
for this concentration cell given the following two half-cell Fe3+ion concentrationsFeau 5.87 x
106M and [Fe3+]c3.74 x 10-2M? 14. a) 0.173 V ) 0.075V b0103 V c) 0.225 v
Solution
Following are the half cell reactions
Fe(s)(dil) - - - - - - > Fe 3+
(aq) + 3e
Fe 3+
(aq)(conc) + 3e - - - - - - > Fe(s)
standard reduction potential for both half reaction is same and E° cell is zero
Therefore
Nernst equation for concentration cell is
E cell = (0.0592/n)log([Fe 3+
]conc/[Fe 3+
]dil)
where,
n = number of electron transfer, 3
Q=[Fe 3+
]conc/[Fe 3+
?]dil = 3.74×10 -2
M /5.87×10 -6
M = 6.37×10 3
Therefore
E cell = (0.0592/3)log(6.37×10 3
)
E cell = 0.075V
![How do you get this answer?
A concentration cell is constructed using two Fe3+/Fe half-cells. What is the cell potential, Ece
for this concentration cell given the following two half-cell Fe3+ion concentrationsFeau 5.87 x
106M and [Fe3+]c3.74 x 10-2M? 14. a) 0.173 V ) 0.075V b0103 V c) 0.225 v
Solution
Following are the half cell reactions
Fe(s)(dil) - - - - - - > Fe 3+
(aq) + 3e
Fe 3+
(aq)(conc) + 3e - - - - - - > Fe(s)
standard reduction potential for both half reaction is same and E° cell is zero
Therefore
Nernst equation for concentration cell is
E cell = (0.0592/n)log([Fe 3+
]conc/[Fe 3+
]dil)
where,
n = number of electron transfer, 3
Q=[Fe 3+
]conc/[Fe 3+
?]dil = 3.74×10 -2
M /5.87×10 -6
M = 6.37×10 3
Therefore
E cell = (0.0592/3)log(6.37×10 3
)
E cell = 0.075V](https://image.slidesharecdn.com/howdoyougetthisanswer-aconcentrationcellisconstructedusing-230203012714-dadcd933/85/How-do-you-get-this-answer-A-concentration-cell-is-constructed-using-docx-1-320.jpg)
