I have a lot of questions actually.
it's electrochemistry (which is obviously a bunch of fun)
2 electrodes Cr(s)/Cr3+ and Sn(s)/Sn2+ are combined to afford a spontaneous electrochemical reaction. the standard reduction potention in V for Cr3+ and Sn2+ are -.74 and -.14 V respectably. E in V=.....
a. .88 v b. -.88 v c. .60 v d. -.60 v e 2.5 v
calculate delta G for reaction of elemental bromine with chloride ion
Br-->E=1.09
Cl-->E=1.36
a. 2.1x10^5 J b. -2.62x10^5 J c. .27 J d. 5.21x10^4 J
consider the half reaction below. what will happen to E of the half cell if pH increases
2IO3-(aq)+10e-+12H+(aq)->I2(s)+6H2O(l) E=1.195 V
a. increases b. decreases c. no change
standard reduction potential in V for Zn2+ and Cu2+ are -.76 and .34 respectably, what is the potential of the cell below
Zn | Zn2+ (1.0 M) || Cu2+ (.10 M) | Cu
a. -1.14 v b. -.42 v c. 1.0 v d. 1.10 v
standard reduction potential in V for Ag+ and Fe2+ to Fe2+ are .80 and .77 calculate K (equilibrium constant) for the following reaction (R=8.314 J/k mol, F=96800 J/V)
Ag+(aq)+Fe2+(aq)->Ag(s)+Fe3+(aq)
a. .10 b. 2 c. 3.2 d. 1.0
thanks sosomuch if you can help =/ =D
1. Write both half cells as reductions potentials. Locate the most negative of the two, reverse it (and change the sign of the voltage), then add the two half equations to obtain the cell reaction and add the two potentials to obtain the cell voltage.
That post is for #2 (the first question).
For IO3^- problem.
Use
E = Eo-(0.059/n)*log (red/ox). Write in the reduced species and oxidized species, and see how an increase of pH will change E.
Zn and Cu problem.
Calculate E for Zn ==> Zn^+2 + 2e. At 1 M concn Zn^+2, that will be the reverse of the standard reduction potential.
Calculate E for Cu^+2 ==> Cu. For this you must use
E = Eo-(0.059/n)*log(Cu/Cu^+2)
n is 2, Cu = 1, and (Cu^+2) = 0.1 M.
Then add the two E values to obtain the cell value.
Last problem.
Calculate E for the cell as in the first problem. Then use nEocell*F = RT*ln K
A note: Long posts like this usually go unanswered. You will do well in the future to limit your posts to one problem per.