 # Calculations on charges

Electrochemistry

## The flow of electric charges

When the circuit is closed, the electrons start to move, and you can detect a current, i.e. the number of ampere moving through the system. This means that if you have measured the current for a time, you can calculate e.g. how much EMF has changed for a cell or how much of an electrode is removed, over time.

The current running is a shift in the reaction quotient towards equilibrium, i.e. the number of electrons being moved is proportional to the number of atoms getting oxidized and reduced in the electrical cell's redox reaction.

For calculating the charges we use the formula

 q = F · n(e) = I · Δt = E(elec) E

 q = charge n(e) = the amount of electrons I = current Δt = time E(elec) = electric energy E = electromotive force (EMF)

From the formula, you can now calculate changes in EMF, how much material is precipitated/dissolved, how much energy is used, etc.

## An example

If we take a look at the cell having the cell diagram

Zn(s) │ Zn2+ (aq, 0.05 M) ║ Ag+ (aq, 0.25 M) │ Ag(s)

then we have the reactions:

Zn(s) + 2 Ag+(aq) 2 Ag(s) + Zn2+(aq)

Let us say that we have the circumstances/informations:
• the poles are connected with a wire for 180 s
• the current was measured as 0.75 A at 25 °C
• the volume of liquid in each half-cell is 100 ml
The total number of electrons being moved we can calculate:

F · n(e) = I · Δt
 n(e) = I · Δt = 0.75 A · 180 s = 1.40 · 10-2 mol F 9.65·104 C/mol

From the reaction equation we ca see that 2e are transferred per Zn and 1e per Ag, and as we now know n(e), we can calculate

 Δn(Zn2+) = ½n(e) = 6.99·10-4 mol Δn(Ag+) = n(e) = 1.40·10-3 mol

We can now calculate the changes for the reaction that produces a current:

 Zn(s) + 2 Ag+(aq) 2 Ag(s) + Zn2+(aq) Start 2.50·10-3 mol 5.00·10-3 mol Change -6,99·10-4 mol -1.40·10-3 mol 1.40·10-3 mol 6.99·10-4 mol End 1.10·10-3 mol 4.30·10-3 mol

The volume for the two half cells do not change while the reaction is taking place, there simply isn't enough time, so now we can calculate the new concentrations of Zn2+ and Ag+:

 [Zn2+] = n(Zn2+) = 4.30·10-3 mol = 4.30·10-3 M V 0.100 l

 [Ag+] = n(Ag+) = 1.10·10-3 mol = 1.10·10-3 M V 0.100 l

Had the volume changed, you would just use the two new volumes.

The electromotive force can now be determined directly by inserting the values in the formula for EMF:

 E = E°(Ag/Ag+) - E°(Zn/Zn2+) - R · T · ln ⎛⎝ [Zn2+] ⎞⎠ z · F [Ag+]2
 E = 0.723 V - (-0.762 V) - 8.31451 J·mol -1·K -1 · 298 K · ln ⎛⎝ 4.30·10-3 M ⎞⎠ 2 · 9.65·104 C·mol -1 (1.10·10-3 M)2
E = 1.59 V