Electrochemistry (Other (Not Listed) Sample)

Electrochemistry
Name
Tutor
Course
College
Date
Segment 9 – Checkpoint #2
A certain process is carried out at constant pressure. The heat flow (measured by a temperature change in a calorimeter) is found to be -2.20x103 kJ. In other experiments, the change in internal energy is found to be -2.19x103 kJ and the change in entropy is - 3.50x103 J/K.
1 What is the value of ΔH?
ΔH = qp = ΔE + PΔV
Where
ΔE = internal energy (associated with constant pressure)
qp = constant pressure
P = pressure
V = Volume
ΔH = Enthalpy heat flow at constant pressure
ΔE = -2.19x103 kJ
ΔH = qp = -2.20x103 kJ
2 What is the value of w?
Solution
ΔE = q+ w
Internal energy is given by heat flow to the system plus the work done on the system.
Therefore,
w = ΔE - q
= -2.19x103 + 2.20x103 = 0.01 x 103 = 10kJ
3 Assuming the heat flow to the surroundings is reversible and the temperature of the surroundings is 25.0 oC, show that ΔS (univ.) = ΔS(sys.) + ΔS(surr.) is > 0 (process is spontaneous.)
ΔS (univ.) = ΔS(sys.) + ΔS(surr.)
ΔE = Efinal - Einitial ΔE = ΔEproducts - ΔEreactants
ΔE system +ΔEsurroundings = 0
ΔEsystem = -ΔEsurroundings
Hence the process is spontaneous
(Intro.chem.okstate.edu, 2015)
Segment 10
A spoon is being coated with gold from a solution of K3Au(CN)6. What mass of gold (aw = 197.0) will be deposited on the spoon by a current of 0.150 amps in 3.00 hours?
Solution
The equation for deposition at the cathode is;
Au+ (aq) + e- => Au (s)
1F = 96500C
Q= Quantity of charge = It
I= current, t = time
I = 0.150A, t = 3x 60x60 = 10800s
Q = 0.150x10800 = 1620C
1 mole = 96500C
?= 1620C = 1620x1/96500
= 0.01679 moles
Mass= moles x atomic weight = 0.01679 x 197
= 3.3071g
(Wadhawan & Compton, 2013)
Segment 11-Checkpoint
NiO2(s) + 4H3O+ + 2e– => Ni2+ + 6 H20 Eo = +1.70 vNi2+ + 2e– => Ni Eo = - 0.25 vCd2+ + 2e– => Cd Eo = - 0.40 vWhen the switch is closed:
1 Which electrode (Cd or Ni) is positive?
The direction of the galvanometer shows the flow of current from cathode to anode. Therefore the positive electrode is Nickel (Ni), anode. Cd is the cathode and Nickel is the anode.
2 Which electrode (Cd or Ni) is the cathode?
The direction of the galvanometer shows opposite to the flow of electrons. The electrons flow from anode to cathode, Cd is the negative electrode, cathode.
3 What value will the voltmeter read?
At the anode Ni is oxidized and at the cathode Cd is reduced therefore,
E.m.f = Ereduced – Eoxidized
= -0.40+0.25 = -0.15 V
4 Which direction (left to right or right to left) does electrons move across the switch?
Electrons flows opposite to the flow of current, electrons flows from right to left, that is, from the anode to cathode, from Nickel electrode to the Cd electrode.
5 Which direction (left to right or right to left) do cations move in the salt bridge?
Cations from the salt bridge move from the anode to the cathode. Therefore, they move from Cd electrolyte to Ni electrolyte, left to right.
6 What will happen to the voltmeter reading if water is added to the Cd cell?
The reading will increase since water has a higher hydrogen concentration which increases the potential of the cell.
(Wadhawan & Compton, 2013)
Segment 11- Check point#2
1 Write the overall balanced equation corresponding to this abbreviated cell notation:
Cr / Cr3+ // Pb4+, Pb2+ / Pt
Solution
Most common soluble salt solution used in electrochemistry are nitrates
Therefore, the resulting overall balanced reaction is;
2Cr(s) + 3Pb(NO3)4 (aq) = > 2Cr(NO3)3 (aq) + 3Pb(NO3)2(aq)
Explanation
In the above abbreviated cell notation, Cr metal is oxidized to Cr ions while Pb metal is reduced from Pb4+ to Pb2+ ions. This is because Cr has more affinity for electrons than lead and hence displaces some of the lead ions from its solution.
2 Write the abbreviated cell notation for the reaction:
2Al(s) + 3 Cd (NO3)2 => 3 Cd + 2 Al(NO3)3
Solution
The common electrode used in this electrolysis is platinum since it is inert (it does not react with electrolytes or products of electrolysis). Therefore, the abbreviated cell notation for the above complete equation is;
Cd3+ /Cd//Al, Al3+ /Pt
Explanation
In the above abbreviated cell notation, cadmium ions are reduced to cadmium solid while Al metal is oxidized from solid Al to Al ions. This is because Al is more reactive and hence displaces Cd ions from its solution.
Segment 11-Checkpoint #3
Calculate the cell voltage (NOT standard voltage) of the electrochemical cell shown below:
2 Al + 3 Cd(NO3)2(a = 3.20 M) => 3 Cd + 2 Al(NO3)3(a =1 .00 x...