Calculate the E of the following cell if [CIO3] = 1.5M, [H]= 0.5M, [CI]=1.0M and [CU2]= O.25M:
CI03 + 3CU + 6H = CI- +3Cu2+ +2h20
To calculate the cell potential (E) of the given cell, you need to use the Nernst equation. The Nernst equation is given as:
E = E° - (0.0592/n) * log(Q)
Where:
- E is the cell potential
- E° is the standard cell potential
- n is the number of electrons transferred in the balanced chemical equation
- Q is the reaction quotient
First, we need to determine the balanced chemical equation and the values of E° and n.
The balanced chemical equation of the given cell is:
CI03 + 3CU + 6H → CI- + 3CU2+ + 2H2O
In this equation, 3 electrons are transferred from CU to CU2+. Therefore, n = 3.
The standard cell potential, E°, can be found by looking up the standard electrode potential values for the half-reactions involved.
The half-reactions involved are:
CI03 + 6H+ + 6e- → CI- (reduction)
CU2+ + 2e- → CU (oxidation)
By looking up the standard electrode potential values, we find that:
E°(reduction) = 1.28 V
E°(oxidation) = -0.34 V
The standard cell potential, E°, can be calculated as:
E° = E°(reduction) - E°(oxidation)
E° = 1.28 V - (-0.34 V)
E° = 1.62 V
Now we can calculate the reaction quotient, Q, using the given concentrations:
Q = ([CI-]/[CI03]) * ([CU2+]^3/[CU]^3) * ([H+]^6)
Substituting the given values:
Q = (1.0M/1.5M) * ([0.25M]^3/[0.5M]^3) * ([0.5M]^6)
Simplifying further:
Q = (2/3) * (0.125/0.125) * (0.015625) * (0.03125)
Q = (2/3) * (0.015625) * (0.03125)
Q = 0.0003268
Now we can plug in the known values into the Nernst equation to calculate the cell potential, E:
E = E° - (0.0592/n) * log(Q)
E = 1.62 V - (0.0592/3) * log(0.0003268)
Calculating further:
E = 1.62 V - (0.01973) * log(0.0003268)
E = 1.62 V - (-3.685)
E = 1.62 V + 3.685 V
E = 5.305 V
Therefore, the cell potential (E) of the given cell is approximately 5.305 V.