# Chemistry - Copper electrode in a Daniel cell

## Solution 1:

If viewing these as charged objects, we're using a physics/electrostatics viewpoint, not usually done in electrochem. Or at best, it's a topic for postdoc courses in surface nano-chemistry (yet I think if it were discussed in undergrad classes, it would clear up much student misconception!) The metal/electrolyte interface forms a capacitor. For this reason we must deal with actual potentials, NOT the shifted values shown in tables of Standard Electrode Potentials.

You're right to be confused. As with the zinc rod, the copper rod will also acquire a negative charge.

From a physics standpoint, the two electrodes in contact are 1) the electrolyte and 2) the metal. A half-cell then has two conductive electrodes, not one. A complete electrolysis cell has four. In a half-cell, a genuine voltage appears between the metal and the electrolyte. The voltage is not undefined or somehow abstract or "ghostly." It's just difficult to measure. A simple voltmeter cannot do it.

These concepts go back over a century, to when chemistry and physics weren't fully separate. Alessandro Volta and Michael Faraday studied electrolysis from this viewpoint. I've seen "single" electrode potentials discussed in 1930s undergrad textbooks, but not in the modern ones. (Well, Bockris electrochem has it.)

Search: Galvani Potential, also Volta Potential, both found at surfaces and junctions between differing materials. Perhaps also look for Helmholtz layer as well as Helmholtz Double Layer.

.

Whenever metal first contacts electrolyte, a brief oxidation reaction takes place which establishes a static potential appearing across the double-layer located at the metal surface. The net reaction halts once these potentials appear. The copper is then left with a negative charge, same as the zinc, with the electrolyte having net positive charge.

However, if either a salt-bridge or a common electrolyte is connecting the half-cells, the zinc and copper rods will always acquire two different amounts of negative surface-charge, leading to a well-known potential-difference measured between them.

For working with these physics/nanochem concepts, it's useful examine the SHE reference electrode. The relative potential-difference of the Standard Hydrogen Electrode is roughly 4.44V, not zero. This is the Galvani potential between metal and electrolyte, also called "Absolute electrode potential," although that term comes from chemistry, and a physicist would insist that this is a genuine potential-difference, not "absolute." (The two electrodes after all are the conductive electrolyte on one side, and conductive metal on the other!) If the SHE potential-difference is 4.44V, then all the values in the table of Electrodes need to be shifted by that amount. In that case, the polarity of every electrode is negative. Zinc and copper will both end up with the same polarity. Copper is not positive-charged.

However, once these concepts are grasped, there's no major need for them, since measurements in electrochemistry are always biased by the inherent Galvani potential of various Reference Electrodes (SHE, or SCE, or Ag/AgCl.) If we ignore electrostatics considerations, and deny that the electrolyte is itself a conductive electrode, then we can take the reference electrode potential-difference to be zero. In that case, voltage measurements become trivially easy. No vibrating non-contact probes needed, no quadrant electrometers or Faraday-cup charge-detectors.

But in that case, surface nanochemistry questions such as @Sharad 's cannot be answered unless we first discuss the zero SHE-potential being a "convenient lie," and the table of Standard Electrode Potentials must be shifted by 4.44V to reflect electrostatic reality. (All the potentials in that table have the same polarity. Oxidizing of solid metal always removes positive ions, even for the metals listed on the table "below" the SHE's zero-voltage entry.)

## Solution 2:

Suppose I have a beaker containing $$\ce{CuSO4}$$ solution and I dip $$\ce{Cu}$$ rod in it. Now my question is what reactions are going to take place?

On a microscopic level, copper ions can approach the rod and pick up two electrons. Or copper atoms can detach from the rod, leaving two electrons behind. On a macroscopic level there will be no net reaction because there is no source or sink for electrons.

$$\ce{Cu^2+ + 2e^-<=> Cu}$$

If the liquid is inhomogeneous, with high concentrations of copper ions touching part of the rod and low concentrations of copper ions touching another part of it, you might get a situation resembling a concentration cell, and have a current carry electrons along the rod.

Without special tricks, though, this would correspond to a half-reaction, and you need another half-reaction connected to it (via a wire and a salt-bridge, for example), to get a net reaction.

Is the $$\ce{Cu}$$ rod going to acquire any positive charge? I am asking this question to clear my doubt regarding a Daniell cell as to why $$\ce{Cu}$$ rod used in the cell should be positively charged.

The rod is not positively charged. The (+) and (-) in diagrams indicate the potential difference or the flow of electrons. The conventions for (+) and (-) vary, so it is easier to understand the chemistry by looking for the anode and cathode labels.

I don't get it, though $$\ce{Zn}$$ rod dipped in $$\ce{ZnSO4}$$ solution acquires negative charge due to oxidation of $$\ce{Zn}$$ to $$\ce{Zn^2+}$$ is clear to me, but how can $$\ce{Cu}$$ rod dipped in $$\ce{CuSO4}$$ be positively charged.

Again, they don't acquire substantial charges, both the solutions and the electrodes are roughly neutral.