Question

CHM 112 Electrochemical Cells and Thermodynamics Section A. Constructing a Small-Scale Electrochemical Cell Woodbridge Campus Objective To construct a small-scale electrochemical cell using a redox system, to measure the cell potential and derive thermodynamic quantities. Procedures 1. Connect the red and black alligator clips to the multimeter to read voltage. Set up the multimeter to read DC (direct current) voltage using the tab at the top. 2. Cut the filter paper into 15 strips of 1-inch length and 0.4-inch width. Two of these will be used in Section A. Set aside the rest for Section B. 3. Obtain three 10 mL beakers. Half-fill one beaker with 1M CuSO4 solution, a second beaker with 1M KNO3, and a third beaker with 1M SnCl2 solution. 4. Place a cleaned Cu wire (clean by using sand paper) in the beaker containing the 1M CuSO4 solution and a clean strip of Sn foil in the beaker containing the 1M SnCl2 solution. 5. Place the three beakers in a row next to each other with the beaker containing the 1M KNO3 solution in the middle. Place one end of the cut filter paper strip in the beaker containing the 1M CuSO4solution and the other end in the 1M KNO3 solution in the adjacent beaker. 7. Place another filter paper strip such that one end dips into the beaker containing the KNO3 solution and the other end dips into the beaker containing the 1M SnCl2 solution. NOTE: The filter paper strips should be placed in the wells just prior to measuring the voltage. M filter paper strips Cuso4 KNO Sncl r proe (nam ned probe (4) cathade tep KNO, Cu Sn Cat 8 Switch the multimeter on and attach the red alligator clip to the piece of Cu wire and the black ep alligator clip to the piece of Sn metal foil. Record the voltage on Data Table 1. Do not forget the sign!!! 9. Reverse the alligator clips. Attach the black one to the Cu metal and the red one to the Sn metal. Record what happens to the sign and switch the meter off. 2 yeversed NOTE: The voltage (potential) that you recorded first is the cell potential (Ecel) of a cell, which is made up of the two half-cells joined electrically by wires (the probes) and the salt bridge solution (KNO). The red probe is the positive terminal and the black probe is the negative terminal. The probes are touched to the electrodes in the cell (Cu metal and Sn foil) and the measured voltage will have a positive sign (+) if the black probe is on the anode and the red probe is the cathode. REMEMBER: The anode is the electrode at which oxidation occurs. The cathode is the electrode at which reduction occurs.

Page 1 and 2 are instructions. Please help me solve K for page 3 and page 4 and please check the other work on Page 3.

Thanyou very much. Will Rate!

Answer 1

part A :

Eq. constant (K) :

we have , Ecell = E^ocell - (RT/nF) ln K

at eq. Ecell = 0

we have , ln K = E^ocell / (RT/nF)  

we have , at 25 C ;  E^ocell = +0.48 V   (RT/nF) = 0.0592 (J /C) /n

Since , n = 2 (electron stiochiometry )

ln K = 0.48 V / (0.0592(J /C) /2) = 16.21 ( 1J = 1 C*V)

so, K = 1.1 *10⁷

Part B :

Mg :

cell notation Mg(s)|Mg²⁺(aq)||Cu²⁺(aq)|Cu(s)

Cathode (reduction ) :  Cu²⁺  (aq)+ 2e $\rightarrow$Cu(s) (E^o(red) = +0.34 V)

Anode (oxidation ) : Mg(s)  $\rightarrow$Mg²⁺(aq) + 2e (E^o(ox) = +2.37 V)

Pb :

cell notation Pb(s)|Pb²⁺(aq)||Cu²⁺(aq)|Cu(s)

Cathode (reduction ) :  Cu²⁺  (aq)+ 2e $\rightarrow$Cu(s) (E^o(red) = +0.34 V)

Anode (oxidation ) : Pb(s)  $\rightarrow$Pb²⁺(aq) + 2e (E^o(ox) = +0.13 V)

Al :

cell notation Al(s)|Al³⁺(aq)||Cu²⁺(aq)|Cu(s)

Cathode (reduction ) :  Cu²⁺  (aq)+ 2e $\rightarrow$Cu(s) (E^o(red) = +0.34 V)

Anode (oxidation ) : Al(s)  $\rightarrow$Al³⁺(aq) + 3e (E^o(ox) = +1.66 V)

Zn :

cell notation Zn(s)|Zn²⁺(aq)||Cu²⁺(aq)|Cu(s)

Cathode (reduction ) :  Cu²⁺  (aq)+ 2e $\rightarrow$Cu(s) (E^o(red) = +0.34 V)

Anode (oxidation ) : Zn(s)  $\rightarrow$Zn²⁺(aq) + 2e (E^o(ox) = +0.76 V)

Fe :

cell notation Fe(s)|Fe²⁺(aq)||Cu²⁺(aq)|Cu(s)

Cathode (reduction ) :  Cu²⁺  (aq)+ 2e $\rightarrow$Cu(s) (E^o(red) = +0.34 V)

Anode (oxidation ) : Fe(s)  $\rightarrow$Fe²⁺(aq) + 2e (E^o(ox) = +0.44 V)

Increasing order of reduction potential :

• Cu²⁺ (aq) + 2e $\rightarrow$Cu(s) (E^o(red) = +0.34 V)
• Pb²⁺(aq) + 2e $\rightarrow$  Pb(s)  (E^o(red) = -0.13 V)
• Fe²⁺(aq) + 2e $\rightarrow$  Fe(s)  (E^o(red) = -0.44 V)
• Zn²⁺(aq) + 2e $\rightarrow$  Zn(s)  (E^o(red) = -0.76 V)
• Al³⁺(aq) + 3e $\rightarrow$  Al(s)  (E^o(red) = -1.66 V)
• Mg²⁺(aq) + 2e $\rightarrow$  Mg(s)  (E^o(red) = -2.37 V)

- Copper is strongest oxidizing agent :  most positive standard potential .

- Mg is strongest reducing agent :  most negative standard potential .

- Battery with potential ~ 1V can be made using this cell :

Zn(s)|Zn²⁺(aq)||Cu²⁺(aq)|Cu(s) E^ocell (calculated ) = 1.1 V