Calculate the initial emfa of this cell. Cl2|cl-|pt||cr3+|cr
To calculate the initial electromotive force (emf) of the cell, we need to know the standard reduction potentials of the half-reactions involved.
The half-reaction for the Cl2/Cl- couple is:
Cl2 + 2e- -> 2Cl- E° = +1.36 V
The half-reaction for the Cr3+/Cr couple is:
Cr3+ + 3e- -> Cr E° = -0.74 V
The emf of the cell can be calculated using the Nernst Equation:
Ecell = E°cell - (0.0592 V/n) * log(Q)
where E°cell is the standard cell potential, n is the number of moles of electrons transferred, and Q is the reaction quotient.
Here, the reaction quotient (Q) for the cell is given by:
Q = ([Cr3+]/[Cl-]^2)
Since the concentration of Cl- is 1M (because it is a standard state), we only need to determine the concentration of Cr3+.
Assuming the concentration of Cr3+ is 1M, we can calculate the emf of the cell as follows:
Ecell = E°cell - (0.0592 V/3) * log(1/[Cl-]^2 [Cr3+])
= E°cell - (0.0592 V/3) * log(1/(1)^2 [1])
= E°cell - (0.0592 V/3) * log(1/1)
= E°cell - (0.0592 V/3) * log(1)
= E°cell
Therefore, the initial emf of this cell is equal to the standard cell potential (E°cell), which needs to be provided in order to calculate the exact value.
To calculate the initial electromotive force (emf) of a cell, we need to determine the standard electrode potentials of the half-reactions involved.
The half-reactions involved in this cell are:
1. Anode oxidation half-reaction: Cl2 (g) -> 2Cl- (aq) (E° = +1.36 V)
2. Cathode reduction half-reaction: Cr3+ (aq) + 3e- -> Cr (s) (E° = -0.74 V)
The overall cell reaction is the sum of the half-reactions:
Cl2 (g) + 2Cl- (aq) + Cr3+ (aq) -> 2Cl- (aq) + Cr (s)
To calculate the initial emf (Ecell), we subtract the reduction half-reaction potential from the oxidation half-reaction potential:
Ecell = E°cathode - E°anode
Ecell = (-0.74 V) - (+1.36 V)
Ecell = -2.10 V
The initial emf of this cell is -2.10 V. Since the emf is negative, it indicates a non-spontaneous reaction under standard conditions.