Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Mg2 ] = 0.763 M and [Sn2 ] = 0.0180 M
Mg(s)+Sn^2+(aq)-->Mg^(2+)+Sn(s)
I keep getting 2.57 but it says I'm wrong. Please help me!
If you post your work so we know what you did perhaps we can find the error. Especially important is what are you using for Eo values.
To calculate the cell potential for the given reaction, you need to use the Nernst equation. The Nernst equation relates the concentration of reactants and products in the cell to the cell potential (Ecell).
The Nernst equation is given by:
Ecell = E°cell - (RT/nF) * ln(Q)
Where:
- Ecell is the cell potential
- E°cell is the standard cell potential (at standard conditions)
- R is the gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin (25.00 °C = 298.15 K)
- n is the number of electrons transferred in the reaction (in this case, 2)
- F is the Faraday constant (96,485 C/mol)
- Q is the reaction quotient.
First, let's calculate the reaction quotient (Q). The reaction quotient is the ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of their stoichiometric coefficients.
Q = [Mg2+]/[Sn2+]
Plugging in the given concentrations, we have:
Q = 0.763/0.0180 = 42.39
Now, we need to look up the standard cell potential (E°cell) from a table of standard reduction potentials. The standard reduction potential for the reduction of Mg2+ to Mg is -2.37 V, and the standard reduction potential for the reduction of Sn2+ to Sn is -0.136 V.
Next, we substitute the values into the Nernst equation:
Ecell = (-2.37 V) - (0.0592 V/n) * log10(42.39)
Ecell = -2.37 - (0.0592/2) * log10(42.39)
Ecell = -2.37 - (0.0296) * log10(42.39)
Ecell = -2.37 - (0.0296) * 1.628
Ecell = -2.37 - 0.0480
Ecell = -2.418 V
Therefore, the cell potential for the given reaction at 25.00 °C is -2.418 V.