EMF involving H+ ions? (1 Viewer)

[Treedom]

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]

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

 

[Treedom]

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

So I disregard the H2O part??

 

[trecex1]

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]

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?

 

[eyeseeyou]

Trecex1, out of curiosity did u do industrial chem?

 

[trecex1]

How did you get 1.22V for MnO2 + 4H+ + 2e- → Mn2+ + 2H2O?? This equation isn't present on the standard potentials data sheet?

It's not but you can look it up - http://www.ed.gov.nl.ca/edu/k12/evaluation/chem3202/standardreductionpotentials.pdf

Trecex1, out of curiosity did u do industrial chem?

yes