Question

 

OBSERVATIONS

Part 1 : Electrochemical Cell Voltages (0.1 M & 0.2 M Solutions)

Trial	Right Cell Metal	Right Cell Solution	Left Cell Metal	Left Cell Solution	Total Voltage
#1	Lead	4 mL of 0.1 M Lead Nitrate [Pb(NO3)2]	Copper	40 mL of 0.1M Copper Nitrate [Cu(NO3)2]	0.47 v
#2	Lead	4 mL of 0.1M Lead Nitrate [Pb(NO3)2]	Silver	40 mL of 0.1M Silver Nitrate [AgNO3]	0.90 v
#3	Lead	4 mL of 0.1MLead Nitrate [Pb(NO3)2]	Platinum	20 mL of 0.1M Iron III Sulfate [Fe2(SO4)3] & 20 mL of 0.2M Iron II Sulfate [FeSO4]	0.91 v
#4	Copper	4 mL of 0.1M Copper Nitrate [Cu(NO3)2]	Silver	40 mL of 0.1M Silver Nitrate [AgNO3]	0.43 v
#5	Copper	4 mL of 0.1M Copper Nitrate [Cu(NO3)2]	Platinum	20 mL of 0.1M Iron III Sulfate [Fe2(SO4)3] & 20 mL of 0.2M Iron II Sulfate [FeSO4]	0.44 v
#6	Silver	4 mL of 0.1M Silver Nitrate [AgNO3]	Platinum	20 mL of 0.1M Iron III Sulfate [Fe2(SO4)3] & 20 mL of 0.2M Iron II Sulfate [FeSO4]	-0.02 v

Part 2 : Electrochemical Cell Voltages : (1.0M Solutions)

Trial	Right Cell Metal	Right Cell Solution	Left Cell Metal	Left Cell Solution	Total Voltage
#1	Lead	4 mL of 1.0M Lead Nitrate [Pb(NO3)2]	Copper	40 mL of 1.0M Copper Nitrate [Cu(NO3)2]	0.47 v
#2	Lead	4 mL of 1.0M Lead Nitrate [Pb(NO3)2]	Silver	40 mL of 1.0M Silver Nitrate [AgNO3]	0.93 v
#3	Lead	4 mL of 1.0M Lead Nitrate [Pb(NO3)2]	Platinum	20 mL of 1.0M Iron III Sulfate [Fe2(SO4)3] & 20 mL of 2.0M Iron II Sulfate [FeSO4]	0.91 v
#4	Copper	4 mL of 1.0M Copper Nitrate [Cu(NO3)2]	Silver	40 mL of 1.0M Silver Nitrate [AgNO3]	0.46 v
#5	Copper	4 mL of 1.0M Copper Nitrate [Cu(NO3)2​]	Platinum	20 mL of 1.0M Iron III Sulfate [Fe2(SO4)3] & 20 mL of 2.0M Iron II Sulfate [FeSO4​]	0.44 v
#6	Silver	4 mL of 1.0M Silver Nitrate [AgNO3]	Platinum	20 mL of 1.0M Iron III Sulfate [Fe2(SO4)3] & 20 mL of 2.0M Iron II Sulfate [FeSO4​]	-0.02 v

Part 3 : Variation of Concentration

Trial	Right Cell Metal	Right Cell Solution	Left Cell Metal	Left Cell Solution	Total Voltage
#1	Silver	4 mL of 0.1M Silver Nitrate [AgNO3]	Silver	40 mL of 1.0M Silver Nitrate [AgNO3]	0.06 v
#2	Silver	4 mL of 0.1M Silver Nitrate [AgNO3​]	Silver	40 mL of 0.1M Silver Nitrate [AgNO3]	0.00 v
#3	Silver	4 mL of 0.1M Silver Nitrate [AgNO3​]	Silver	40 mL of 0.1M Silver Nitrate [AgNO3] & 400 mL Water	-0.06 v

CONCLUSIONS:

1. When comparing the Part 1 and Part 2 data, how did the voltages differ when varying concentrations?

4 of the trials voltage charge remained the same, while 2 of the trials went up in voltage, increased by 0.03 v.

2. Based off your data, what are the standard half-cell potentials for Silver and Iron/Platinum? Explain your reasoning.

3. When comparing your data in Part 3, how did varying the concentration affect the voltage? Which trial had the largest voltage? Which trial had the lowest voltage?

The voltage decreased with each variation. Trial #1 had the largest voltage and trial #3 had the lowest voltage.

4. In the following reactions, determine which element is oxidized and which one is reduced.

- Sn + 4 HNO₃ ------- SnO₂ + 4 NO₂ + 2 H₂O

Oxidized Element: Sn ------- SnO₂

Reduced Element: 4HNO₃ ------ 4NO₂

- Mg + Br₂ ---------- MgBr₂

Oxidized Element: Mg ----- Mg

Reduced Element: 2Br ------ 2Br

Answer 1

2. The standard half-cell reduction potential can be known only when you connect it to a standard hydrogen electrode(SHE); in your case there is no SHE so some considerationas are to be made.

For silver with data obtained from table 3, the standard reduction potential can be deduced since the redox reaction is

and

Since the reaction involve sthe same metal and its ions, there should not be any change in its potentail ideally if it were SHE. However, for Ag, the potential(voltage) changed and the change in voltage can be accounted as its reduction potential. In table 3, the first trial is increasing concn and the third trial made it diluted and a change in the voltage showed +0.06 and -0.06 respectively. The half-cell standard reduction potentail of Ag will be +0.12V

, from trial 2, we know the left cell has zero voltage, so right cell of trial 1 has voltage, 0.06V

This when accounted for trial 3 will be the standard reduction potential for Ag half-cell, the reduction potential will be 0.06V -(-0.06V) = 0.12V

Using this value, the reduction potential of Pt/Fe can be calculated taking trial 6 of table 1 or 2,

, E_right is Ag-potential and E_left is Pt/Fe-potential

Reduction potential of Pt/Fe hal-cell is 0.14V.

4. Mention only the elements while showing oxidation and reduction, that would make it easier.

a) , this is equivalent to

; oxidised element - Sn. reduced element - N

b) Only two elemnts involved so what you did is fine.