That second equation is on the standard reduction potentials. MnO2 is reduced, I- is oxidised so just calculate it as usual.
I can do basic questions asking to calculate EMFs such as
Sn + Mg^{2+} → Sn^{2+} + Mg
where you'd just write the 2 half equations and then use the table of standard potentials but the ones involving water make no sense to me at all. For example,
MnO_{2} + 2I^{-} → Mn^{2+} + I_{2} + 2H_{2}O
I can get as far as writing the 2 half equations as
2I^{-} → I_{2} + 2e^{-} and
MnO_{2} + 4H^{+} + 2e^{-} → Mn^{2+} + 2H_{2}O
but then I'm unsure as to how you'd get a value for the second half equation involving H+ ions. Help is greatly appreciated ^^
Edit: Oh the answers state the final answer is +0.68V btw
That second equation is on the standard reduction potentials. MnO2 is reduced, I- is oxidised so just calculate it as usual.
Trecex1, out of curiosity did u do industrial chem?
It's not but you can look it up - http://www.ed.gov.nl.ca/edu/k12/eval...potentials.pdf
yes
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