A voltaic cell that uses the reaction
Tl3+(aq) + 2 Cr2+(s) �¨ Tl+(aq) + 2 Cr3+(aq)
has a measured standard cell potential of +1.19 V.
determine E�‹red for the reduction of Tl3+(aq) to Tl+(aq).
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To determine the reduction potential (E�‹red) for the reduction of Tl3+(aq) to Tl+(aq) in the given voltaic cell, we need to use the Nernst equation.
The Nernst equation relates the cell potential (E�‹cell) to the standard cell potential (E�‹cell°), the reaction quotient (Q), the gas constant (R), the temperature (T), and the number of electrons transferred (n) in the balanced equation.
The Nernst equation can be written as:
E�‹cell = E�‹cell° - (0.0592 V/n) * log(Q)
Where:
- E�‹cell is the cell potential
- E�‹cell° is the standard cell potential
- Q is the reaction quotient
- 0.0592 V/n is the Nernst constant (at 298 K)
In this case, we are given the standard cell potential (E�‹cell°) of +1.19 V. The balanced equation shows that the number of electrons transferred (n) is 3 (because three electrons are involved in the reduction of Tl3+(aq) to Tl+(aq)).
Now, we need to calculate the reaction quotient (Q). The reaction quotient is calculated using the concentrations of the species involved in the reaction.
For the given reaction:
Tl3+(aq) + 2 Cr2+(s) �¨ Tl+(aq) + 2 Cr3+(aq)
The reaction quotient (Q) is given by:
Q = [Tl+][Cr3+]^2 / [Tl3+][Cr2+]^2
Since the species Tl3+ and Cr2+ are not mentioned to have a concentration, it can be assumed that their concentrations are 1 M (standard state).
Thus, the reaction quotient simplifies to:
Q = [Tl+][Cr3+]^2
Now, let's substitute the given values into the Nernst equation:
E�‹cell = E�‹cell° - (0.0592 V/3) * log(Q)
E�‹cell = 1.19 V - (0.0197 V) * log([Tl+][Cr3+]^2)
To solve for E�‹red, we need to rearrange the equation and isolate E�‹red:
E�‹red = E�‹cell - E�‹cell° + (0.0592 V/3) * log(Q)
By substituting the given values, you can calculate the reduction potential (E�‹red) for the reduction of Tl3+(aq) to Tl+(aq).