Answer:
Explanation:
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]
