A galvanic cell at a temperature of 25.0°C is powered by the following redox reaction: →+2IO−3aq+12H+aq5Cos+I2s+6H2Ol5Co+2 aq Suppose the cell is prepared with 6.64 M IO−3 and 1.54 M H+ in one half-cell and 7.82 M Co+2 in the other. Calculate the cell voltage under these conditions. Round your answer to 3 significant digits.

E = 1.47

Explanation:

To do this, you need to apply the Nerst equation which is the following:

E = E° - RT/nF lnQ (1)

Where:

E: cell voltage

E°: Standard potential reduction

R: universal constant

T: temperature of the system

n: number of electrons transfered during the reaction

F: Faraday constant.

Q: Equilibrium constant

However, as the reaction is taking place at 25 °C, and R and F have constant values, we can reduce the above expression to the following:

E = E° - 0.05916/n lnQ  (2)

We can get the value of Q because it has to do with the reaction which is the following:

2IO₃⁻(aq) + 12H⁺(aq) + 5Co(s) > I₂(s) + 5Co²⁺(aq) + 6H₂O(l)

Now, using only the aqueous state the expression of Q will be:

Q = [Co²⁺]⁵ / [H⁺]¹² [IO₃⁻]²

Replacing the values we have:

Q = (7.82)⁵ / (1.54)¹² * (6.64)²

Q = 3.728

Knowing this, all we need to know now is the standard potential reduction of the reaction. To do so, we need to write the two semi equations of reduction and oxidation:

2IO₃⁻ + 12H⁺ + 10e⁻ > I₂ + 6H₂O       E₁° = 1.20 V

5Co > 5Co²⁺ + 10e⁻                          E₂° = 0.28 V

E° = 1.2 + 0.28 = 1.48 V

Now that we have all the values (n = 10) we can write now the nernst equation to calculate the cell voltage:

E = 1.48 - 0.05916/10 ln (3.728)

E = 1.48 - 0.005916 (1.315872)

E = 1.47 V

This will be the cell voltage