3 Sn^2+ (aq) + 2 Al (s) <--> 2 Al3+(aq) + 3 Sn(s). What is the voltage for this cell?
What is the voltage for the cell if the Sn2+(aq) is [.15] and Al3+(aq) is [1.80]?
Using proper line notation, describe the cell described in the question.
To determine the voltage for this cell, we can use the Nernst equation. The Nernst equation is given by:
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 equation, and
Q is the reaction quotient.
First, let's identify the specific half-cell reactions happening in the cell:
1. Reduction half-reaction (cathode):
Al3+(aq) + 3e- -> Al(s)
2. Oxidation half-reaction (anode):
3Sn2+(aq) -> 3Sn(s) + 6e-
Now, we need to find the standard cell potential, E°. The standard potentials can be found in a table of standard electrode potentials. Let's assume that we find the following values:
E°(cathode) = +1.66 V
E°(anode) = -0.14 V
Notice that the anode half-reaction is already balanced in terms of electrons, so n is equal to 6.
Now, we can calculate the reaction quotient Q using the concentration values given in the question:
[Sn2+] = 0.15 M
[Al3+] = 1.80 M
Q = [Al3+]^2 / [Sn2+]^3
Plugging in the values, we get:
Q = (1.80)^2 / (0.15)^3
Now, let's calculate the voltage using the Nernst equation:
E = E° - (0.0592/6) * log(Q)
Substituting the values, we get:
E = 1.66 - (0.0592/6) * log[(1.80)^2 / (0.15)^3]
Calculate the log and simplify the expression to find the voltage for this specific cell.
As for the line notation, it can be written as:
Al(s) | Al3+(aq) || Sn2+(aq) | Sn(s)