Where do negative ions migrate from the salt bridge to which electrode in the voltaic cell?

In the voltaic cell with $$\ce{Mg/Mg^2+}$$ and $$\ce{Ag+/Ag}$$, I am not sure whether the ions move from the salt bride to the $$\ce{Mg}$$ cathode or the $$\ce{Mg^2+}$$ cathode. I distantly recall my teacher saying the cathode (in this case) is a metal, and so it can't be $$\ce{Mg^2+}$$? I am confused, and please help ASAP. Final is tomorrow.

Somewhat related question: At the cathode during electrolysis, what is its charge and what type of reaction occurs?

Is the correct answer: positive charge, and reduction?

• Cathode or anode, $\ce{Mg^2+}$ is not a electrode. Jun 18 '19 at 23:59
• So is Mg2+ the cathode? Isn't the cathode the site of reduction, where Mg2+ would reduce? Jun 19 '19 at 1:03
• Mg is the cathode, you were already told Mg2+ is not. But Mg2+ is NOT reduced, at least not in water environment. There reduction 2 H+ + 2 e- -> H2. See also cathode=catabasis of electrons = reduction by incoming electrons Jun 19 '19 at 4:03
• "what is its charge and what type of reaction occurs?"the negative charge, and reduction Jun 19 '19 at 22:02

Oxidation half reaction (happens at the anode)

$$\ce{Mg -> Mg^2+ + 2e-}$$

The electrons get removed via the wire, and the magnesium cations enter the aqueous solution, giving it a net positive charge.

Reduction half reaction (happens at the cathode)

$$\ce{Ag+ + e- -> Ag}$$

The electrons are supplied via the wire, and the silver cations leave the aqueous solution to be deposited on the cathode, giving the solution a net negative charge.

Salt bridge

The salt bridge provides anions (negative charged particles) to the solution of the oxidation half cell, and cations (positively charged particles) to the solution of the reduction half reaction.

Silly mnemonics

Oil rig: oxidation is loss, reduction is gain (of electrons)

Red cat: reduction occurs at the cathode. An ox: oxidation occurs at the anode.

Where do negative ions migrate from the salt bridge?

They don't migrate to any electrode (neither cathode nor anode). Instead, they migrate to the solution of the solution of the oxidation half cell.

A voltaic cell is an electrochemical cell that uses a chemical reaction to produce electrical energy. An electrochemical cell consists of two electrodes (anode and cathode), two electlolyte solutions, and a salt bridge: The anode is an electrode where oxidation occurs and the cathode is an electrode where reduction occurs. The oxidation and reduction reactions are separated into compartments, which are called half-cells (each half-cell consists of electrode and the appropriate electrolyte solution). The salt bridge connecting half-cells is a chamber of electrolytes, which are needed to maintain the neutrality of the solutions in both half-cells. The external circuit is used to conduct the flow of electrons between electrodes, which usually go through a load (e.g., a light bulb or voltmeter) to close the circuit (see the diagram).

Now, let's see what chemical reactions are happening in your voltaic cell. Your two half cells are $$\ce{Mg/Mg^2+}$$ (suppose your $$\ce{Mg^2+}$$ solution is an aqueous $$\ce{Mg(NO3)2}$$ solution) and $$\ce{Ag/Ag+}$$ (suppose your $$\ce{Ag+}$$ solution is an aqueous $$\ce{AgNO3}$$ solution). Suppose your salt bridge is $$\ce{NaNO3}$$. To find where oxidation and reduction happening, you may need to look table at the end of your textbook, which is called electrochemical series, giving reduction potentials of many elements such as following:

$$\ce{Ag+ + e- <=> Ag} \qquad \mathrm{E}^\circ = \pu{0.7996 V} \tag{1}$$ $$\ce{Mg^2+ + 2e- <=> Mg} \qquad \mathrm{E}^\circ = \pu{-2.372 V} \tag{2}$$

To have a positive electron flow (or spontaneous redox reaction), $$\mathrm{E}_\mathrm{cell}^\circ$$ should be positive. So, you can rearrange equation (2) such that complete redox reaction gives positive $$\mathrm{E}_\mathrm{cell}^\circ$$:

$$\ce{Mg <=> Mg^2+ + 2e-} \qquad \mathrm{E}^\circ = \pu{+2.372 V} \tag{3}$$

Now see, in equation (1), $$\ce{Ag+}$$ gains $$\ce{e-}$$s, so that it reduces. Therefore, $$\ce{Ag/Ag+}$$ half-cell is the cathode where reduction happens.

In equation (3), $$\ce{Mg^2+}$$ gives away $$\ce{e-}$$s, so that it oxidizes. Therefore, $$\ce{Mg/Mg^2+}$$ half-cell is the anode where oxidation happens.

Solving (1) and (3) in order to cancel $$\ce{e-}$$s, you'll get total redox reaction:

$$\ce{Mg + 2Ag+ -> Mg^2+ + 2Ag} \qquad \mathrm{E}_\mathrm{cell}^\circ = \pu{+3.172 V} \tag{3}$$

Overall, at anode, the $$\ce{Mg}$$ electrode slowly dissolve, and the magnesium cations enter the aqueous solution, giving it a net positive charge. The electrons left at the anode will travel to the cathode through the external circuit (connected to the load).

At the same time, $$\ce{Ag+}$$ travel to the cathode, and gains $$\ce{e-}$$s comes through the wire to become $$\ce{Ag}$$. $$\ce{Ag}$$ deposits on cathode, and thus, the aqueous solution loses the silver cations, giving it a net negative charge. The salt bridge, then does his duty by neutralizing these two aqueous solutions (see the diagram).

Note: The galvanic cell is courtesy of ELI JONES. The original one was modified in order to suit to the current question in hand.

• I just looked more closely (drinking coffee) and noticed that the galvanic cell is actually a modified Cu and Ag cell! It even still says copper (II) nitrate under the left beaker! That also explains the curious cell voltage font, the copper-colored anode, blue solution in the left beaker, etc. I am quite puzzled by this.
– Ed V
Jun 19 '19 at 12:40
• @ Ed V: Made the correction. So, you can enjoy your coffee! :-) Jun 19 '19 at 16:14
• Excellent! I just upvoted your answer and will link to it when I encounter questions that need are in need of a very clear explanation! I will also leave up my “answer” and delete the last sentence in it. Then the mods can decide its ultimate fate.
– Ed V
Jun 19 '19 at 16:23
• @ Ed V: I also included the link to the original source of my modified galvanic cell. Jun 19 '19 at 16:28
• chem.libretexts.org/Bookshelves/General_Chemistry/… Figure 17.2.2 from OpenStax chemistry Jun 20 '19 at 15:11

Both answers are very good, but Mathew Mahindaratne's answer uses a galvanic cell figure that needs some work. I respectfully request that his galvanic cell figure be replaced and I hereby offer up my handmade figure (see below), either as a temporary filler or for whatever period Mathew may decide. It is entirely up to him. I do not claim my handmade figure is wonderful, but I can truthfully say it is my work and anyone can use it. • Nice cartoon! I like your version. But, I'd make changes to my version in few min. Please don't erase yours. Thanks for your remarks. Jun 19 '19 at 15:59