Part IT: Concentration Cell Prepared by Dilution a. Write the balanced cell reac
ID: 532662 • Letter: P
Question
Part IT: Concentration Cell Prepared by Dilution a. Write the balanced cell reaction (show Cu" molarities) and an algebraic expression (.e. no numbers) for the reaction quotient, Q (see the example in your lab manual in Part Iry G and calculations based on the Nernst cquatioo. Show b. Fill in the following table based on your experimental data the calculation below the table. 0. O 2 0 0.0592 Part Copper-Ammonia system a. Show the chemical reaction (with phases) that occurs when NE,(ag) is added to 0.10 M Cu (ag). Include a relevant equilibrium constant value and circle tbe sideorthe equilibrium that is favored. b. Fill in the table below u your ab predictions, ex mental data and calculations. Part III: Copper-Sodium Hydroxide System a. Show the chemical reaction (with phases) that occurs when NaOH(ag) is added to 0.10 MCu (ag). Include a relevant equilibrium constant value and circle the side of the equilibrium that is favored. he table below usin e-lab predictions, ex mental data and calculations Free ICu 1A) Calculated OH on the "free" (Cu after addition ofoH after addition of OH decease 290 c. Show the chemical reaction that occurs (with phases when Hasoe(ag) is added to the half-cell containing Cu and NaOH. As you add acid to the half cell containing Cu and NaoH, how should the concentration of "free" cu change? How do you expect the cell potential to change with this change in concentration of"free" Cu Did your measured change in cell potential agree with your expected change in "free" [CuExplanation / Answer
answer
1.E CELL= E0 CELL-0.0592/2 log Q ====>0.017=0-0.0296 log Q ===>0.017/0.0296==>logQ===> 0.47
Q= cu+2/1.0 M cu+2===> antilog100.47 ==>2.95 so x/1.0 M==>2.95 than x [cu+2 ion ]==>2.95 M
2 E CELL= E0 CELL-0.0592/2 log Q ====> 0.0327==-0.0296logQ===>0.0327/0.0296==>logQ===> 1.10
Q= Cu+2/0.10 M cu+2===> antilog101.10 ==>12.5 8 so x/0.10 M==>12.58 than x [cu+2 ion ]==>1.25 M
3..E CELL= E0 CELL-0.0592/2 log Q ====> 0.0296==-0.0296logQ===>0.0296/0.0296==>logQ===> 1.00
Q= Cu+2/0.10 M cu+2===> antilog101.00 ==>10.0 so x/0.10 M==>10.00 than x [cu+2 ion ]==>1.00 M