Can someone please help my with this Electrochemistry lab from chemistry Lab 121
ID: 591252 • Letter: C
Question
Can someone please help my with this Electrochemistry lab from chemistry Lab 1212!
Chem 1212 Lab Manual- Revised 11/2016 0.1 M ZnSOs and unknown CuSO4 code and potential (2 pts. each)) Unknown concentration of CuSO4 (10 pts.) (Use "observed potential for the Zn/Zn2 0.1 MCu0.M Cu cell as the E in the calculation and note the number of significant figures given on reagent bottles.) Procedure II.pH meter (1/2 pt. for each blank in table) pH of Buffer mV reading H01log 10HO'1 0058 500 01 OD 1-13 le 3 ,da (2 pts.) Efficiency of Calibrated pH meter (divide by 100, if efficiency given in percent) Slope of graph of buffer yoltage as a function of logio [H3O] Slope above divided by the pH meter efficiency In the space below, based on the magnitude of this value, tell what part of the Nernst equation (6 pts.) (3 pts.) this value represents. (3 pts) What is the value of "n" from the Nernst equation for the equation represented by your graph? (2 pts.) Attach the plot of voltage (in units of volts) vs log[HsOl to the report for grading. (10 pts.)Explanation / Answer
In this question, firstly we calculate theoritical cell potential of the reaction from Nernst equation
Ecell = E0cell - 0.0059 / 2 log [Zn2+ ] / [Cu2+ ]
E0cell = 1.10 V
Ecell = 1.10 - 0.0059 / 2 log [0.1] / [0.1]
Ecell = 1.10 - 0.0059 / 2 log 1
Ecell = 1.10 - 0.0059 / 2 X 0
Ecell = 1.10 V
Percent error = 1.10 - 1.06 / 1.10 = 0.0363 X 100 = 3.63 %
Now we calculate value of Q when Ecell = 0
Then equation becomes
E0 = 0.059 / 2 log Q
Log Q = E0 X 2 / 0.059
Log Q = 1.06 X 2 / 0.059
Log Q = 2.12 / 0.059
log Q = 35.93
Now we have to calculate the unknown concentration of ZnSO4
1.055 = 1.06 - 0.059/2 log [Zn2+ ] / 0.1
-0.005 = 0.059/2 log [Zn2+ ] / 0.1
0.005 X 2 / 0.059 = log [Zn2+ ] / 0.1
0.17 = log [Zn2+ ] - log 0.1
0.17 = log [Zn2+ ] + 1
0.17 - 1 = log [Zn2+ ]
-0.83 = log [Zn2+ ]
[Zn2+ ] = Anti (-0.83)
[Zn2+ ] = 0.15 M
Now do same calculation for Unknown Cu conc
1.049 = 1.06 - 0.059/2 log [0.1] /[Cu2+ ]
-0.011 = 0.059/2 log [0.1] /[Cu2+ ]
0.011 X 2 / 0.059 = log [0.1] /[Cu2+ ]
0.37 = log [0.1] - log [Cu2+ ]
0.37 = -1 - log [Cu2+ ]
0.37 - 1 = log [Cu2+ ]
-0.63 = log [Cu2+ ]
[Cu2+ ] = Anti (-0.63)
[Cu2+ ] = 0.23 M