Question

Based on the solution concentrations we will be using in lab (0.10 M, 0.30 M, 1.0 M, and 3.0 M copper (II) nitrate solution and 1.0 M zinc (II) nitrate AND 50 mL 1.0 M potassium nitrate), calculate the theoretical Ecell values and plot a calibration curve using Excel. Include the equation for the calibration curve. Attach the calibration curve to this prelab assignment. Tip: save this file and it will make your postlab a lot easier! (5 pts)

Answer 1

The theoretical value for the given cells can be calculated using Nernst's equation:

$\Delta E=\Delta E^{o}-\frac{RT}{nF}\cdot ln(Q)$

Where R is the universal constant, T is the temperature (in Kelvin), n is the number of electrons involved in the reaction, F is Faraday's constant and Q is the ratio between the concentration of products and of reactants.

Our cell will be the combination of the following two half-reactions:

$Zn_{(s)}\rightarrow Zn_{(aq)}^{2+}+2e^{-}$

$Cu_{(aq)}^{2+}+2e^{-}\rightarrow Cu_{(s)}$

So the total reaction is:

$Cu_{(aq)}^{2+}+Zn_{(s)}\rightarrow Cu_{(s)}+Zn_{(aq)}^{2+}$

This reaction has a standard potential of:

$\Delta E^{o}=E_{Cu^{2+}/Cu}^{o}-E_{Zn^{2+}/Zn}^{o}=0.34V-(-0.76V)=1.10V$

We can now write the complete Nernst equation:

$\Delta E=1.10V-0.0129V\cdot ln(\frac{[Zn^{2+}]}{[Cu^{2+}]})$

This yields the following results for the given combinations (please bear in mind that commas are decimal separators):

[Zn(11)] (M) [Cu(11)] (M) delta E (V) 0,1 1,07044148 0,3 1,08454448 1 1,1 3 1,114103

Which are here plotted in a delta E vs ln[Cu(II)] graph, with the added linear fit:

1,115 1,11 1,105 1,1. y = 0,0128x + 1.1 R2=1 delta E 1,095 1,09 1,085 1,08 1,075 -2,5 -2 -1,5 -1 1,07 1,065 -0,5 In[Cu(11)] 0 0,5 1 1,5