A. Use the Cu/Cu2+ half cell your team has already prepared in the previous part
ID: 1021880 • Letter: A
Question
A. Use the Cu/Cu2+ half cell your team has already prepared in the previous part of the lab. Your team will also use the same Zn/Zn2+ half cell for the first measurement. Transfer the the voltage reading obtained for the galvanic cell created by these two half cells to Table 2.
B. Your team will prepare another Zn/Zn2+ cell with 0.020 M Zn(NO3)2 solution. The new solution is obtained by diluting 1.0 mL 0.20 M Zn(NO3)2 to a volume of 10.0 mL. Clean and place the Zn strip into the volume of this this solution needed to ¾ fill a small vial and measure its voltage vs. the Cu/Cu2+ half cell. Record the positive voltage value in Table 2.
C. Your team will prepare another Zn/Zn2+ cell with 0.0020 M Zn(NO 3 ) 2 solution. The new solution is obtained by diluting 1.0 mL 0.020 M Zn(NO 3 ) 2 to a volume of 10.0 mL. Clean and place the Zn strip into the volume of this this solution needed to ¾ fill a small vial and measure its voltage vs. the Cu/Cu2+ half cell. Record the positive voltage value in Table 2.
D. Your team will prepare a 0.020 M Cu(NO 3 ) 2 solution. The new solution is obtained by diluting 1.0 mL 0.20 M Cu(NO 3 ) 2 to a volume of 10.0 mL. Clean and place the Cu strip into the volume of this solution needed to ¾ fill a small vial and measure its voltage vs. the original 0.20 M Zn/Zn2+ half cell your team first used at the beginning of the lab. Record the positive voltage value in Table 2. Table 2. Electrolyte Concentration Effects Cell Voltage Zn|0.20 M Zn2+ ||0.20 M Cu2+|Cu Zn|0.020 M Zn2+ ||0.20 M Cu2+|Cu Zn|0.0020 M Zn2+ ||0.20 M Cu2+|Cu Zn|0.20 M Zn2+ ||0.020 M Cu2+|Cu
Key Questions
9. Based on the overall reaction that occurs in the galvanic cells in Table 2, which chemical species are the products and which are the reactants?
10. What would be the equilibrium expression for the overall reaction above?
11. Based on the measured voltages in Table 2, what was the effect of reducing the reactant ion concentration? What was the effect of reducing the product ion concentration?
12. Based on these observations, what should be the effect on the cell voltage by increasing the concentration of the reactant ion? What should be the effect on the cell voltage by increasing the concentration of the product ion?
P/s: Please help me with these key question, I'm writing the lab report but I'm stucking on these questions. Please, please, help me with these 4 key questions, I'm desperate.
Cell Voltage Zn|0.20 M Zn2+ ||0.20 M Cu2+|Cu 0.938V Zn|0.020 M Zb2+ ||0.20 M Cu2+|Cu 0.889V Zn |0.0020 M Zn2+||0,20 M Cu2+|Cu 0.804V Zn|0.20 M Zn2+|| 0.020 M Cu2+|Cu 0.924VExplanation / Answer
Cell reaction
In this cell reaction oxidation occur at one electrode namely Zn here Zn give Zn2+ ion and two electcron
And reduction occur at the another electode namely Cu electrode. Here Cu2+ take 2e- to form Cu
9. Answer is
Reactant – Zn and Cu2+ ion
Product – Zn2+ and Cu
10. The equllibrium expression for over all cell reaction is
11.Based on the table
When reduced the reactant ion concentration the voltage of the cell will almost same compared with both in the same concentration. For this the reactant is Cu2+ ion the concentration reduced from 0.2 to 0.02 the obsverd voltage is 0.938 & 0.924
When reduced the Product ion concentration the voltage of the cell is decreases compared with both in the same concentration. For this the product is Zn2+ ion the concentration of change from 0.2 to 0.02 the observed voltage is 0.889 and further reduced to 0.002 the voltage is further reduced to 0.804V.
12. When increasing the reactant ion concentration the voltage of the cell will almost same as compared with both in the same concentration.
When increasing the Product ion concentration the voltage of the cell is increases as compared with both in the same concentration
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