Calculate the cell potential for the galvanic cell in which the reaction Fe ( s ) + Au 3 + ( aq ) − ⇀ ↽ − Fe 3 + ( aq ) + Au ( s ) occurs at 25 ∘ C , given that [ Fe 3 + ] = 0.00150 M and [ Au 3 + ] = 0.797 M .

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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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 electrical  conductivity  of asolution  of an electrolyte is measured by determining the resistance of thesolution  between two flat or cylindrical electrodes separated by a fixed distance.

the number of protons in the nucleus of a given element. has slight luster and conducts electricity at high temperatures like metal.but like a nonmetal,boron is brittle and does not conduct electricity well at low temperatures. in graphite, carbon is a soft,brittle solid.it is reflective,ductile,or malleable.