The standard potential of this cell
Sn|Sn^+2 || Cr^+3|Cr is -0.60 V , what is the reductional standard potential of
Cr^+3|Cr ? if we know that reduction potential of Sn is
Sn^+2 + 2e- <==> Sn : -0.14 V
I wonder if this is a made up problem?
Sn^2+ + 2e ==> Sn E = -0.14
Cr --> Cr^3+ + 3e E = ?
----------------------
Ecell = -0.60 then Cr must be -0.46 written as an oxidation so as a reduction it would be +0.46 but that isn't what it gives in the tables.
But the standard Eo of Cr electrode is -0.74
You repeated my line but didn't answer if this was a made up problem or not. I don't like that answer until I have more information about the question.
Of course it's not a made up problem , it's only a question in our homework
Sometimes , I'm just confused on some fine and tiny problems , just it is..
I'm sorry for any mistake
and , I'm very grateful
To find the reduction standard potential of Cr^+3|Cr, we can use the Nernst equation. The Nernst equation relates the reduction potential (E) of a half-cell to the standard reduction potential (E°) and the concentration of the species involved.
The Nernst equation is given as follows:
E = E° - (0.0592/n) * log(Q)
Where:
- E is the reduction potential of the cell under non-standard conditions.
- E° is the standard reduction potential of the cell.
- n is the number of electrons involved in the half-reaction (in this case, it is 1 since Cr^3+ gains 1 electron to form Cr).
- Q is the reaction quotient, which is the ratio of product concentrations to reactant concentrations, each raised to the power of their stoichiometric coefficients.
In this case, we have the reduction potential of the cell (E) as -0.60 V and the reduction potential of Sn (Sn^+2 + 2e- <==> Sn) as -0.14 V.
To find the reduction potential of Cr^+3|Cr, we need to rearrange the Nernst equation to solve for E°:
E° = E + (0.0592/n) * log(Q)
Since Sn is on the left side of the cell notation, its concentration should be in the numerator of Q. Cr is on the right side, so its concentration should be in the denominator. The stoichiometric coefficients are determined by the balanced half-reactions. The equation for the cell reaction can be written as:
2Cr^3+ + 3Sn <==> 2Cr + 3Sn^2+
Now we can determine the reaction quotient Q. Since we only have standard potentials given, we can assume that the concentrations of Sn^2+ and Cr^3+ are both 1 M. So Q = [Sn^2+]^3 / [Cr^3+]^2.
Now, plug in the values into the rearranged Nernst equation:
E° = -0.60 V + (0.0592/1) * log(([Sn^2+]^3) / ([Cr^3+]^2))
Since both Sn^2+ and Cr^3+ are at 1 M concentration, the equation can be simplified to:
E° = -0.60 V + (0.0592/1) * log((1^3) / (1^2))
E° = -0.60 V + (0.0592/1) * log(1)
E° = -0.60 V + 0
Therefore, the reduction standard potential of Cr^+3|Cr is -0.60 V.