Question

A galvanic cell at a temperature of 25.0°C is powered by the following redox reaction:

→+2IO−3aq+12H+aq5Cos+I2s+6H2Ol5Co+2aq

Suppose the cell is prepared with 6.40M IO−3 and 0.445M H+ in one half-cell and 1.34 M. Co+2 in the other. Calculate the cell voltage under these conditions. Round your answer to 3 significant digits.

Answer 1

For any reaction

aA + bB ---> cC + dD

equilibrium constant, Q= [ C ]^c * [ D ]^d / [ A ]^a * [ B ]^b ----(1)

The Given reaction is

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

The given concentration are

[ IO₃⁻ ] = 6.40 M

[ H⁺ ] = 0.445 M

[ Co²⁺ ] = 1.34 M

In this question we are going to use Nernst Equation

i.e. E = E^o - RT/nF lnQ ---(2)

where

Gas constant, R = 8.314 J / K * mole

Faradat constant, F = 9.648 * 10⁴ J / V* mole

Temperature, T = 25^oc = 298 K

after putting these value in above equation (2) we will get

E = E^o - 0.05916/n lnQ ---(3)

Now the calculation for Q

from equation (1)

Q =  [ Co²⁺ ]⁵ / [ H⁺ ]¹² * [ IO₃⁻ ]²

Q = ( 1.34 )⁵ / ( 0.445 )¹² * ( 6.40 )²

Q = 1749.06

Now we need to find the E^o , by reducing the reaction equation into half equation

2IO₃⁻ + 12H⁺ + 10e⁻ ---> I₂ + 6H₂O

5Co ---> 5Co²⁺ + 10e⁻

E_I⁻/I₂^o = 1.20 V

E_{Co / Co²⁺}^o = 0.28 V

E^o = 1.20 + 0.28 = 1.48 V and n = 10

Now putting all these values in equation (3)

E = 1.48- 0.05916/10* ln(1749.06)

E = 1.43 V ( This is the required answer )