Is the current through the electrolyte double that through the circuit?

[etotheipi]

Suppose, just for explanation purposes, we have an electrolyte of Na⁺ and Cl^- ions connected to an external circuit with an ammeter which reads 96500A.

In 1 second, we expect 96500C of charge to flow out of the cathode as 1 mole of Na⁺ ions are reduced, and the same amount of charge to flow into the anode as 1 mole of Cl^- ions are oxidised.

Now consider an arbitrary point not in the external circuit, but in the electrolyte, through which the ions are flowing. Since, in 1 second, 1 mole of Na⁺ ions move in one direction past this point and 1 mole of Cl^- ions move the other direction, we have a situation where +96500C has moved one way and -96500C of charge has moved the other way across this arbitrary point.

This is equivalent to 193000C of positive charge (or negative, if we define in the opposite direction) flowing past this point in 1 second, which equates to a current of 193000A.

So it seems as if, in this case, the current due to the ions flowing both ways in the electrolyte is double the size of the current through the external circuit. Is this correct analysis?

 

[phinds]

So it seems as if, in this case, the current due to the ions flowing both ways in the electrolyte is double the size of the current through the external circuit. Is this correct analysis?

It can't possibly be, as it would imply a serial circuit section where the current is different in one part than in another part.

 

[hutchphd]

I enjoyed this question.
The ionic nature of the solution is maintained from electrochemistry independent of the external current. Think about the cathode. Excess electrons there can either:

1. Turn Cl into Cl^-
2. Turn Na⁺ into Na

The anode analysis is obvious.
So in equilibrium you get half as many new ions as electrons used and the current is conserved. In the real world it is not half and half but the principal holds.

 

[Herman Trivilino]

Since, in 1 second, 1 mole of Na+ ions move in one direction past this point and 1 mole of Cl- ions move the other direction, we have a situation where +96500C has moved one way and -96500C of charge has moved the other way across this arbitrary point.

You're assuming that the number of reductions (or oxidations) is equal to the number of charges moving through the electrolyte.

 

[anorlunda]

It makes a good multidisciplinary question. You can look at it from the chemistry point of view, or from the electrical continuity equation point of view as @phinds mentioned to come to the same result.

Consider any small control volume, such as one containing the junction between wire and the cell. If the sum of currents through the walls of the volume do not sum to zero, charge (plus or minus) will build up and oppose further current.

$\nabla{J} = -\frac{d\rho}{dt}$

 

[etotheipi]

I enjoyed this question.
The ionic nature of the solution is maintained from electrochemistry independent of the external current. Think about the cathode. Excess electrons there can either:

1. Turn Cl into Cl^-
2. Turn Na⁺ into Na

The anode analysis is obvious.
So in equilibrium you get half as many new ions as electrons used and the current is conserved. In the real world it is not half and half but the principal holds.

This does indeed seem to sort out everything from the current perspective.

What if we now wanted to calculate the mass of e.g. sodium deposited? Chemistry texts generally instruct you to determine the number of moles of electrons that pass through the cathode (in this case, if we run the cell for 1 second, 1 mole) and equate this to the number of moles of Na from the half equation which is, again, 1 mole.

However, now we have determined that only about half of the reductions are for the Na⁺ to Na reaction, so we would end up with only about 0.5 moles of sodium?

 

[hutchphd]

I think you will still get a mole of Na because of the solution chemistry: it is a dynamic process. I point out that absent the chemistry the Na and Cl ions could recombine and produce no current and a pile of salt! I'll let you or others ponder this for now..