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