1. Hi all,

Finding this question very hard.

Here is a cell:
I2(aq), I-(aq) | Pt +0.54V
Fe2+(aq) | Fe(s) -0.44V

Here is my cell diagram and emf value:
Fe2+(aq) | Fe(s) ⁞ I2(aq), I-(aq) | Pt

emf = E(cell) = E(RHS) – E(LHS)
emf = E(cell) = (+0.54) –(-0.44)
emf = E(cell) = +0.98V

I am now tasked with writing balanced ionic equations that occur when the cell is discharged.

Any help would be very much appreciated. Thanks.

2.

3. Well what is going on is:

Fe = Fe2+ + 2e-

I2 + 2e- = 2I-

Combining them:

Fe + I2 = Fe2+ + 2I- = FeI2

is this what you are meaning?

4. Yes, thanks. What is the Platinum doing?

5. It doesn't seem to be doing much...

Is it what the electrodes are made of?

(My electrochemistry is very rusty, I studied this ~20 years ago and haven't thought about it much since...)

6. Originally Posted by JackAnimated
Yes, thanks. What is the Platinum doing?

Nothing.
It functions as an inert electrode since it does not take part in the chemical reactions provided in post #2.

7. Originally Posted by Cogito Ergo Sum
Originally Posted by JackAnimated
Yes, thanks. What is the Platinum doing?

Nothing.
It functions as an inert electrode since it does not take part in the chemical reactions provided in post #2.
I was pretty sure that was the case but I'm a bit rusty on this stuff...

8. Thanks you folks. So this equation represents the discharge of the cell?

Fe + I2 = Fe2+ + 2I- = FeI2

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