So, I'm confused about finding E cell value Please explain me !
Mn(s) ∣ Mn2+(aq) ∥ Ag+(aq) ∣ Ag(s)
overall reaction is Mn(s)+Ag+(aq) -> Mn2+(aq)+Ag(s)
Oxidation: Mn(s)-> Mn2+(aq)+2e- E: -1.18 I think here it should be 1.18 to calcuate because on the table, Reduction Half reaction is Mn2+(aq)+2e- -> Mn(s) and the order has been changed. WHY IT IS -1.18???
Reduction: Ag+(aq)+e- -> Ag(s) E: 0.80 This makes sense following by the table.
For another example, Sn(s) | Sn2+(aq) || Ag+(aq) | Ag(s)
Overall rxn: Sn(s)+ Ag+(aq) -> Sn2+(aq)+Ag(s)
Oxidation rxn: Sn(s)-> Sn2+(aq)+e- E=-0.14 Here, I think it should be 0.14 but why is it -0.14?????
Reduction rxn: Ag+(aq)+e- -> Ag(s) E= 0.80
E'cell=Ecatode-Eanode= 0.80-(-0.14)=0.94 Can I also calculate like E'cell= Ered-Eoxi?
People calcuated like E'cell= Ered+Eoxi. Why is that?
Please Please explain me !!
Get Answers For Free
Most questions answered within 1 hours.