Question 3 A galvanic cell at a temperature of 25.0°C is powered by the following redox reaction: →+2VO+2aq+4H+aqFes+2VO+2aq+2H2OlFe+ 2aq Suppose the cell is prepared with 2.22 M VO+2 and 3.55 M H+ in one half-cell and 5.56 M VO+2 and 6.55 M Fe+2 in the other. Calculate the cell voltage under these conditions. Round your answer to 3 significant digits.

E = +1.430  V

Explanation:

The given cell reaction is written as:

2VO²⁺(aq) + 4H⁺(aq) + Fe(s)---> 2VO₂⁺(aq) + 2H₂O(l) + Fe²⁺(aq)

E⁰cell = E⁰_{VO₂/Vo₂⁺}  - E⁰_{Fe²⁺/Fe}

= + 1.00 - ( - 0.447) = + 1.45 V

From Nernst equation...

E = E⁰cell - (RT/nF) lnQ

where, Q = [products]/[reactants]

= [Fe²⁺][VO₂⁺]²/[VO²⁺]²[H⁺]⁴

Q= (6.55*5.56)/(2.22*3.55)

 = 36.418/7.881

 = 4.621

T = 25 + 273 = 298K

at 298K the above equation could also be written as...

E = E⁰cell - (0.0592/n) log Q

n is number of electrons transferred and it is 2. Let's plug in the values...

E = 1.45 - (0.0592/2) log 4.621

E = 1.45 – (-0.01968)

E = +1.430  V

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