EMF involving H+ ions? (1 Viewer)

Treedom

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I can do basic questions asking to calculate EMFs such as

Sn + Mg2+ → Sn2+ + 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,

MnO2 + 2I- → Mn2+ + I2 + 2H2O

I can get as far as writing the 2 half equations as

2I- → I2 + 2e- and
MnO2 + 4H+ + 2e- → Mn2+ + 2H2O

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
 

trecex1

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That second equation is on the standard reduction potentials. MnO2 is reduced, I- is oxidised so just calculate it as usual.
 

trecex1

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So I disregard the H2O part??
What do you mean disregard? That's part of the half reaction.
The reduction potential of MnO2 + 4H+ + 2e- → Mn2+ + 2H2O is 1.22 V
The potential for 2I- → I2 + 2e- is -0.54 V (assuming it's solid otherwise you won't get your answer)
If you add these you get 0.68 V
 

Treedom

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What do you mean disregard? That's part of the half reaction.
The reduction potential of MnO2 + 4H+ + 2e- → Mn2+ + 2H2O is 1.22 V
The potential for 2I- → I2 + 2e- is -0.54 V (assuming it's solid otherwise you won't get your answer)
If you add these you get 0.68 V
How did you get 1.22V for MnO2 + 4H+ + 2e- → Mn2+ + 2H2O?? This equation isn't present on the standard potentials data sheet?
 

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