Visualize a Galvanic cell with a Mg electrode submerged in a 1.0 mol L-1 Mg(NO3)2 solution and a silver Ag electrode in a separate 1.0 mol L-1 AgNO3 solution, connected by a salt bridge. The magnesium solution is a vibrant light green color, while the AgNO3 solution has a faint cream tinge. The two solutions are housed in clear glass containers placed on a wooden desk. The Mg electrode is a silver-grey rod and the Ag electrode is a shiny silver rod. The exterior ambient environment is a standard chemistry laboratory with a temperature gauge showing 25 Celsius degrees.

A Galvanic cell consists of Mg electrode in a 1.0 mol L -1 Mg(NO3)2 solution and a Ag electrode in a 1.0 mol L-1 AgNO3 solution. Calculate standard emf for the spontaneous reaction of this electrochemical cell at 25 C, given that the standard reduction potential Eo for Ag+ is +0.80 V and Eo for Mg2+ is -2.37 V.

Answer is 3.17 V.

I know that Cathode - Annode = Eo cell
How can I find out which is the Cathode and Annode since this is written in a question and not in an equation. The annode is usually on left of equation and cathode on right.

Thanks

A voltaic cell consist of a Mg electrode in 1.0 M Mg(NO3)2 and a Ag electrode in 1.0 M AgNO3. Calculate the standard cell potential at 25oC.

Mg →Mg2+ +2e-
Ag → Ag+ +e-

E°=-2.37-0.80

E°=-3.17

Ah, the eternal electron dance in electrochemistry. Let's see if we can solve this with a touch of humor!

In the given setup, we have a magnesium (Mg) electrode and a silver (Ag) electrode. Now, let's imagine these electrodes at a party. The Mg electrode, being a bit of a social butterfly, loves to mingle and meet new friends. It bumps into the solution of Mg(NO3)2, where there are plenty of Mg2+ ions happily swimming around.

On the other hand, the Ag electrode, being a bit shy and reserved, prefers to keep to itself. It finds solace in the company of the AgNO3 solution, where Ag+ ions roam around, making it feel right at home.

Now, let's determine who's who in this electrochemical party. The electrode where oxidation occurs is the anode, while the electrode where reduction occurs is the cathode. And as you rightly mentioned, the cathode is usually on the right side of the equation.

Remember that "ANode" starts with an "A," just like "Awfully Shy" Ag, so Ag is our anode. The Mg electrode takes up the role of the cathode, proudly standing on the right side of the equation, ready to make friends with those free-roaming Mg2+ ions.

With that sorted, let's calculate the standard emf of this electrochemical cell! The standard emf of the cell, Eo cell, is equal to the potential of the cathode (Eo cathode) minus the potential of the anode (Eo anode).

Eo cell = Eo cathode - Eo anode

Using the given values of Eo for Ag+ (+0.80 V) and Mg2+ (-2.37 V), we can plug them into our equation:

Eo cell = (+0.80 V) - (-2.37 V)
= +0.80 V + 2.37 V
= 3.17 V

And there you have it, my friend! The standard emf for this electrochemical cell is indeed 3.17 V. Hopefully, this electrochemical party has sparked a few laughs along the way. Cheers!

To determine which electrode is the cathode and which is the anode, you can follow a few steps:

1. Write down the half-cell reactions:
- Mg electrode: Mg2+ + 2e- ⇌ Mg (reduction half-reaction)
- Ag electrode: Ag+ + e- ⇌ Ag (reduction half-reaction)

2. Look at the standard reduction potential (Eo) values given for each half-reaction. The larger (more positive) value corresponds to the reduction half-reaction at the cathode, and the smaller (more negative) value corresponds to the oxidation half-reaction at the anode.

In this case:
- Eo for Ag+ is +0.80 V (more positive)
- Eo for Mg2+ is -2.37 V (more negative)

Therefore, Ag+ (silver) is reduced and acts as the cathode, while Mg2+ (magnesium) is oxidized and acts as the anode.

3. Now you can write the balanced overall cell reaction:
- Mg + Ag+ ⇌ Mg2+ + Ag

4. Finally, use the given standard reduction potentials to calculate the standard electromotive force (Eo cell) for the cell by subtracting the reduction potential of the anode from that of the cathode:
- Eo cell = Eo cathode - Eo anode
- Eo cell = +0.80 V - (-2.37 V) = +3.17 V

Therefore, the standard emf for the spontaneous reaction in this galvanic cell is 3.17 V.

The more positive reduction potential will always be the cathode. That is, Ag+ will be the cathode, and Mg2+ will be the anode.