Online Chem Lab - a bunch of quick answer mini questions Note: I will give 3000

Question

Online Chem Lab - a bunch of quick answer mini questions

Note: I will give 3000 points to whoever answers both questions

Activity : Concentration Electrochemical Cells

http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/electroChem/voltaicCellEMF.html

1. Start the software and construct a zinc-copper electrochemical cell using 1.0 M Zn2+ and 1.0 M Cu2+ solutions.

a. What is the EMF of this cell (include units)? ___________

b. For the voltage that you are measuring, is this E

Explanation / Answer

1)
a) 1.10v
b)E^0 CELL BEACAUSE WE TAKING THE STANDARD CONDITIONS LIKE 1M CONCENTRATION
C) Zn(S) + Cu^+2(aq)-------------.> Zn^+2(aq) + cu(s)
d) spontaneous because E^0cell is positive
E)zn(s)/zn^+2(aq)(1M) // Cu^+2(aq)(1M)1/CU(S)
F)
THE OXIDATION HALF-CELL REACTION IS

Zn(s)----------------> Zn^+2(aq)
   oxidation occures at anode
    the anode is zinc electrode

2)

example -1
zn(s)/zn^+2(aq)(1M) // Ag^+1(aq)(1M)1/Ag(S)

E^0 cell= REDUCTION POTENTIAL OF CATHODE -REDUCTION POTENTIAL OF ANODE
E^0 CELL= 0.80-(-0.76)=1.56V

example-2
zn(s)/zn^+2(aq)(1M) // Au^+3(aq)(1M)1/Au(S)
E^0 CELL= 1.40-(-0.76) = 2.16V

a)
example -1
zn(s)/zn^+2(aq)(1M) // Ag^+1(aq)(1M)1/Ag(S)

example-2
zn(s)/zn^+2(aq)(1M) // Au^+3(aq)(1M)1/Au(S)

b)example-1---- E0 CELL=1.56V
EXAMPLE-2 ------- E^0 CELL= 2.16V

C)E^0CELL FOR BOTH

D) SPONATENEOUS

E)
THE OXIDATION HALF-CELL REACTION IS

Zn(s)----------------> Zn^+2(aq)
   oxidation occures at anode
    the anode is zinc electrode

F)

EXAMPLE-1
Zn(S) + 2.Ag^+1(aq)<-------------.> Zn^+2(aq) + 2 Ag(s)

example-2

3 Zn(S) + 2Au^+3(aq)<-------------.> 3Zn^+2(aq) +2Au(s)

G)

example -1
zn(s)/zn^+2(aq)(1M) // Sn^+2(aq)(1M)1/sn(S)

E^0CELL= REDUCTION POTENTIAL OF CATHODE - REDUCTION POTENTIAL OF ANODE
E^0 CELL = -0.15-(-0.76)= 0.61V

example-2

zn(s)/zn^+2(aq)(1M) // Fe^+2(aq)(1M)1/Fe(S)

E^0 CELL=-0.44-(-0.76) =0.32V

h)

example -1
zn(s)/zn^+2(aq)(1M) // Sn^+2(aq)(1M)1/sn(S)

example-2
zn(s)/zn^+2(aq)(1M) // Fe^+2(aq)(1M)1/Fe(S)
I)
EXAMPLE-1 0.61V
EXAMPLE-2:0.32V

J)

eXAMPLE-1
Zn(S) + Sn^+2(aq)<-------------.> Zn^+2(aq) + 2Sn(s)

example-2

Zn(S) + Fe^+2(aq)<-------------.> Zn^+2(aq) +Fe(s)

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