ENGR161 Lab 2: Determination of the EMF of an Electrochemical Cell 1. Experimental results a. Solutions concentrations F

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Engr161 Lab 2 Determination Of The Emf Of An Electrochemical Cell 1 Experimental Results A Solutions Concentrations F 1 (36.33 KiB) Viewed 67 times

Engr161 Lab 2 Determination Of The Emf Of An Electrochemical Cell 1 Experimental Results A Solutions Concentrations F 2 (65.89 KiB) Viewed 67 times

ENGR161 Lab 2: Determination of the EMF of an Electrochemical Cell 1. Experimental results a. Solutions concentrations Ferric alum: [Fe"] = 0.1 mol L' Ferrous ammonium sulphate: [Fe] -0.1 mol L b. Measurement of EMFs ferrous:ferric 10:15:12:11:21:5 1:10 Vm from V Total volume EMF/V 2. Calculations a. Plot of EMF vs. (Note: feel free to insert a computer-generated plot here] mV 3. Question Calculate from the standard electrode potentials given below: = 0.22 V = 0.77 V Compare and comment on the differences between you have measured and you have calculated.
The volume of the solutions should be 20-25 mL, I'll select 20 mL Fe3+= X and Fe 2+ = Y -Ratio 10:1 (10 mol Fe+3 : 1 mol Fe +2) X/Y=10 (10:1), then X= 10 Y Y= 2 mL X= 10x 2mL = 20 ml Vtotal= 20 mL + 2 mL = 22 mL ----> (20-25) mL -Ratio 5:1 (5 mol Fe3+:1 mol Fe2+) X/Y=5 X= 5Y Y= 4 mL X= 5*4 mL = 20 mL Vtotal= 20 mL + 4 mL = 24 mL ---> (20-25) mL -Ratio 2:1 X/Y=2 X= 2Y Y= 7 mL X= 2*7 mL = 14 mL Vtotal= 14 mL + 7 mL = 21 mL Ratio 1:2 X/Y= 1/2 Y= 2X X= 7 mL Y= 2*7 mL= 14 mL Vtotal= 14 mL + 7 mL= 21 mL -Ratio 1:5 X/Y= 1/5 Y= 5X X= 4 mL Y= 5*4 mL= 20 mL Vtotal= 20 mL + 4 mL= 24 mL -Ratio 1:10 X/Y= 1/10 Y= 10X X= 2 mL Y= 10*2 mL= 20 mL Vtotal = 20 mL + 2 mL = 22 mL Note that in this case I didn't use the concentration for the calculations because the concentration in both stock solutions of the cations is the same (0.1 M). But if such concentrations were different, we should do the calculations with the moles, because in that case the ratio of volumes is different than the ration in moles, but in this specific case, both ratios are equal, due to the concentration of the cations are identical.