Answer:
0.79 V
Explanation:
From Nernst equation:
E= E°- 0.0592/n log [red]/[ox]
Number of electrons transferred=3
[Red]=0.797 M .
[Ox]= 0.00150 M
First we obtain E°= 0.80-(-0.04)= 0.84 V
E= 0.84- 0.0592/3 log [0.797]/[0.00150]
E= 0.84-0.05
E= 0.79 V
The cell potential for the given galvanic cell is;
Ecell = 1.006 V
We are given;
Reaction as;
Fe (s) + Au^(3+) (aq) = Fe^(3+) ( aq ) + Au (s)
[Fe 3+] = 0.00150 M
[Au 3+] = 0.797 M
Now, from a table of standard reduction potentials online, we have;
Fe^(3+) + e^(-) = Fe^(2+) ; Eº = +0.77 V
Au^(3+) + 3e^(-) = Au ; Eº = +1.69 V
Thus, when combined;
Eº = 1.69 - 0.77
Eº = 0.89
Combining the 2 half cell equations gives;
3Fe^(3) + Au = 3Fe^(2+) + Au^(3+)
There is a transfer of 3 electrons.
Thus, n ; 3
Reaction quotient Q is calculated as;
Q = [Fe^(3)]³/[(Fe^(2+))³ × (Au^(3+))]
Q = 0.0015³/(0.15³ × 0.797)
Q = 0.000001254705
Formula for cell potential of the galvanic cell is;
Ecell = Eº - ((0.0591/n) log Q)
Ecell = 0.89 - (0.0591/3) × log 0.000001254705
Ecell = 0.89 - (0.0197 × -5.901)
Ecell = 1.006 V
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