Predict the standard emf of each of the following galvanic cells.
(a) Sn(s)|Sn2+(aq)||Sn4+(aq),Sn2+(aq)|Pt(s)
(b) Pt(s)|Fe3+(aq),Fe2+(aq)||Ag+(aq)|Ag(s)
I don't want the answer, I want the procedure/methods of how to do these types of problems.
The full equation you obtain from the set up. Read and write from left to right.
Sn==> Sn^2+ + 2e
Sn^4+ + 2e ==> Sn^2+
--------------------
Add the two half equations to obtain
Sn + Sn^4+ ==> Sn^2+ + Sn^2+ Eo = ?
Look up the Sn ==> Sn^2+ + 2e Eo value. You will find it in your table of reduction potentials as
Sn^2+ + 2e ==> Sn. My table shows -0.136 but my set is very old and you should use your value. Then reverse the sign to make it read +0.136 because your reaction is the reverse of what you looked up.
Then look up the other half equation. That one is a reduction potential written as a reduction and my table has +0.15. Again, you should use the value in your set.
Then add the oxidation half to the reduction half to find the Eo for the cell as you have it written.
That will be 0.136 + 0.15 = ? volts = Eo cell.
I did the same thing for part b:
Flipped the E for the lesser value (Fe half reaction)
0.80 - 0.771 = 0.029, but it wasn't correct.
Did you get the right answer for part a? You should have.
I don't think you followed directions for part b. Why did you flip it? The way the cell is set up it is non-spontaneous. So
Fe^3+ + e ==> Fe^2+ Eo = 0.77
Ag ==> Ag^+ + e Eo = -.8
So Ecell should be 0.77+(-0.80) = -0.03 volts which means it will not go as written but will proceed in the opposite direction spontaneously.
To predict the standard emf of galvanic cells, you can follow these steps:
Step 1: Identify the half-cell reactions
In each galvanic cell, there are two half-cell reactions taking place at the electrode surfaces. Identify these reactions by looking at the species involved in the cell setup.
Step 2: Write the balanced half-cell reactions
Write the balanced half-cell reactions for each electrode. Make sure to balance both mass and charge in each equation.
Step 3: Look up the standard reduction potentials
Consult a table of standard reduction potentials to find the standard reduction potentials for each half-cell reaction. These values are measured under standard conditions (1 M concentration and 1 atm pressure at 25°C).
Step 4: Identify the oxidation and reduction reactions
Determine which half-cell reaction is the oxidation reaction and which is the reduction reaction. The oxidation reaction involves the loss of electrons, while the reduction reaction involves the gain of electrons.
Step 5: Calculate the standard emf
The standard emf (E°) of the galvanic cell is the difference between the standard reduction potentials of the reduction and oxidation reactions. The reduction reaction's potential should be positive, while the oxidation reaction should be negative. To find the standard emf, subtract the reduction potential of the oxidation reaction from the reduction potential of the reduction reaction.
For example, let's go through the first galvanic cell:
(a) Sn(s)|Sn2+(aq)||Sn4+(aq),Sn2+(aq)|Pt(s)
Half-cell reactions:
Oxidation: Sn(s) → Sn2+(aq) + 2e-
Reduction: Sn4+(aq) + 2e- → Sn2+(aq)
Standard reduction potentials (from tables):
Sn2+(aq) + 2e- → Sn(s) E° = -0.14 V
Sn4+(aq) + 2e- → Sn2+(aq) E° = 0.15 V
The reduction reaction has the higher potential (0.15 V), and the oxidation reaction has the lower potential (-0.14 V). To calculate the standard emf, subtract the oxidation potential from the reduction potential:
Standard emf (E°) = 0.15 V - (-0.14 V) = 0.29 V
Repeat the same process for the second galvanic cell (b) Pt(s)|Fe3+(aq),Fe2+(aq)||Ag+(aq)|Ag(s) to calculate the standard emf.
Remember, these calculations assume standard conditions. If the concentrations of the electrolytes are different or the temperature is not at 25°C, additional calculations using the Nernst equation may be required to adjust the standard emf.