Consider the electrochemical cell below: T| (s) | T|C|(s) | CdCI_2(0.01m) | Cd(s
ID: 1073664 • Letter: C
Question
Consider the electrochemical cell below: T| (s) | T|C|(s) | CdCI_2(0.01m) | Cd(s) The reduction potential, E degree, for Cd^2++ + 2e rightarrow Cd is -0.40V What is the cathodic half reaction? What is the anodic half reaction? What is the overall cell reaction? Next we are going to do a little Hess' law exercise... The solubility product for T|C| is 1.6 times 10^-3 at 25 degree C, write the chemical reaction and calculate delta G degree? What is delta G degree for the half reaction T|^+ + e rightarrow T| if E degree for the one electron reduction of T|+ is -0.34 V Given your answers for b, d and e above (i.e. you should be able to use reactions in steps d and e to get the half reaction), determine E degree for you anodic half reaction Calculate E and E degree for this cell at 25 degree CExplanation / Answer
Line notation: anode II cathode
Given that Cd2+ + 2e- ---------> Cd Eo = - 0.40 V
We know that Tl+ + e- ---------> Tl Eo = - 0.34 V
a) At cathode, reduction takes place.
Cd2+ + 2e- ---------> Cd
b) At anode, oxidation takes place.
Tl ---------> Tl+ + e-
c) Overall cell reaction:
2Tl + Cd2+---------> Cd + 2Tl+