Show the equation you will use to calculate the volume of 1 M Cu(NO3)2 (aq) needed to prepare a set of solutions that have concentrations in the range of 1 M to 1x10-4 M in a 10-mL volumetric flask. Write the reduction half-cell reaction for the copper(II) ion. What is the standard potential for an electrochemical cell that is prepared from a copper half-cell and a zinc half-cell

The equation we use to calculate the volume needed to prepare other [tex](C_1,V_1)[/tex] the solution that has a concentration [tex]C_2[/tex] and volume [tex]V_2[/tex] is:

[tex]C_1V_1 =C_2V_2[/tex]

[tex]V_1=\dfrac{C_2V_2}{C_1}[/tex]

where;

[tex]C_1[/tex]= concentration of the first solution

[tex]V_1[/tex] = volume of the first solution

[tex]C_2[/tex] = concentration of the second solution

[tex]V_2[/tex] = volume of the second solution

2) Reduction half cell reaction for the copper (II) ion is:

[tex]Cu^{2+} + 2e^- \to Cu[/tex]

3) [tex]Cu^{+2} + 2e^- \to Cu \text{ \ \ \ E = 0.3370}[/tex]

[tex]Zn^{+2} + 2 e^- \to Zn \ \ \ \ \ \ E = -0.763[/tex]

[tex]Zn \to Zn^{+2} + 2 e^- \ \ \ \ \ \ E = +0.763[/tex]

Since the reduction potential of Cu is more; it means copper will go into reduction and zinc will undergo oxidation.

Standard Potential =[tex]E^0_{left} - E^0_{right}[/tex]

[tex]= -0.763 -0.337[/tex] ( since both are reduction potential)

[tex]\mathbf{E^0_{cell} = -1.100 volt}[/tex]