For #1.
I do these this way. The dashed lines represent the voltage for the reduction reactions shown.
--------- Cl2 + 2e ==> 2Cl^-   +1.4
--------- Hg^+2 + 2e ==> Hg   +0.85
--------- 2H^+ + 2e ==> H2   0.0
The spacing is arbitrary; I just put the vertical spacing between the dashed lines to show how I do it. Normally, you see, the standard H cell is 0 voltage and the Cl2 reduction potential is +1.4 volts above H. The difference between 1.4 and 0 is of course 1.4 volts. Now if we replace that H reference cell with a mercury(II)/Mercury cell, that is +0.85 with reference to H. So the Cl2 potential measured against the Mercury as a reference cell will be 1.4 - 0.85 = 0.55 volts. These number I am using are from an OLD OLD text I had in graduate school and you need to look in your text/notes and pick up the up to date numbers. I hope this helps. I think the drawing with the dashed lines is the only way I can see how to do these. It may or may not help you.
1) Assume that the reference half-cell is changed to a standard mercury-mercury (II) half-cell.
a) What would be the reduction potential of a standard chlorine half-cell.
c) What would be the cell potential of a standard chlorine-nickel cell.
2) For each of the following standard cells, write the cell notation, label electrodes, and determine the cell potential.
a) tin(IV) zinc standard cell.
If someone could explain the questions/help me with them so I can do the remaining work it would be greatly appreciated. Thanks!
2 answers
I don't know your trouble with the Zn/Ni cell. I don't understand the last problem.
0 answers