Posted by Kelsey on Wednesday, August 25, 2010 at 10:47am.
Fe^+2 + 2e ==> Fe Eo = ??
Ag^+1 + e ==> Ag Eo = ??
Look up Eo values for each, reverse the one with the larger negative value and add the two half cells. You should obtain a + voltage.
a) oxidation occurs at the anode and reduction at the cathode; multiply the Ag reaction by 2 and add the two half cells (one oxidation and one reduction) to obtain the cell reaction.
b)If you get through a you should be able to do b on your own.
Related Questions
Chemistry - Little confusing!! I need help PLZ!!! Two half cells in a galvanic ...
grade 12 chemisstry - I have no idea what I'm doing here, please help. 2 ...
Chem -plz check work - so i've worked on this some more, now I am asking if ...
chemistry - a galvanic cell: A copper electrode is placed in a Cu(NO3)2 ...
chemistry - what changes in colour of iron nails and copper sulphate solution do...
chemistry - A galvanic cell has an iron electrode in contact with 0.20 M FeSO4 ...
chem lab - KMn04 with iron(II) ammonium sulphate hexahydrate-redox titration. 1-...
AP CHEMISTRY - A solution is prepared in which a trace or small amount of Fe2+ ...
Chemistry...please help - A galvanic cell is based on the following half-...
chemistry - Calculate the cell potential of voltaic cells that contain the ...
For Further Reading