# 1. 3 Sn2+ (aq) + 2 Al(s) --> 2 Al3+ (aq) + 3 Sn(s). What is the voltage for this cell?

2. What is the voltage for the cell in question 1 if the Sn2+ (aq) is [.15] and Al3+ (aq) is [1.80]?
3. Using proper line notation, describe the cell described in question 1.

## Use Ecell = EoCell - (0.05916/n)log(M^x+)

Eocell is the reduction potential you look up in the table in your text/notes. M^x+ is the concentration of the ion in the problem. (You will note that if you substitute 1 for the ion the log M is 0 and Ecell = Eocell from the tables). All this is is a way of adjusting the voltage because the molarity of the ion is not 1.0M.
After you have each reduction potential adjusted, reverse the equation with the larger negative number (the equation is reversed and the number becomes +), then add in thee other half cell and the sum of the two will give you the new Ecell.

The last part is
Al|Al^3+||Sn^2+ |Sn
I'm sure you've seen this notation before, perhaps you just didn't recognize it from the problem.

## 1. To find the voltage for this cell, you will need to use the standard reduction potentials of the half-reactions involved and apply the Nernst equation.

First, identify the half-reactions taking place in the cell:
- Oxidation half-reaction: Al(s) → Al3+(aq) + 3e-
- Reduction half-reaction: 3Sn2+(aq) + 6e- → 3Sn(s)

Next, find the standard reduction potentials for these half-reactions. You can find these values in a standard reduction potential table. Let's assume the standard reduction potentials are:
- E°(Al3+/Al) = -1.66 V
- E°(Sn2+/Sn) = -0.14 V

The voltage for this cell can be calculated using the Nernst equation:
E = E°(cathode) - E°(anode)

Substituting the values:
E = 0.00 V - (-1.66 V)
E = 1.66 V

Therefore, the voltage for this cell is 1.66 V.

2. To find the voltage for the cell in question 1 with given concentrations of Sn2+ (aq) and Al3+ (aq), you will need to use the Nernst equation. Let's assume the concentrations are [Sn2+] = 0.15 M and [Al3+] = 1.80 M.

The Nernst equation can be written as:
E = E° - (0.0592/n) log(Q)

First, determine the reaction quotient (Q) by dividing the concentrations of the reactants by the concentrations of the products, raised to their stoichiometric coefficients:
Q = ([Al3+]^2/[Sn2+]^3)

Now, substitute the given values into the equation and calculate the voltage:
E = E° - (0.0592/6) log(([Al3+]^2/[Sn2+]^3))
E = 1.66 V - (0.0592/6) log((1.80^2)/(0.15^3))

Calculate the value using a calculator:
E ≈ 1.66 V - (0.0592/6) log(64)
E ≈ 1.66 V - (0.0592/6) * 1.806

Final Calculation:
E ≈ 1.66 V - 0.0099
E ≈ 1.6501 V

Therefore, the voltage for the cell in question 1 with the given concentrations is approximately 1.6501 V.

3. The line notation of the cell described in question 1 can be written as follows:

Al(s) | Al3+(aq) || Sn2+(aq) | Sn(s)

- The "|" symbol indicates a phase boundary, and it separates the oxidation half-cell (on the left) from the reduction half-cell (on the right).
- The "||" symbol represents the salt bridge or the ion-conducting bridge, which helps maintain charge balance during the cell reaction.
- The letters inside the brackets represent the phase of the species involved. In this case, (s) represents the solid phase, and (aq) represents the aqueous phase.

So, in summary, the line notation for the cell described in question 1 is: Al(s) | Al3+(aq) || Sn2+(aq) | Sn(s).