# Predicting the feasibility of reactionsWatch

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#1
I'm not sure how to predict whether a reaction will be feasible or not. I assumed that if it was >0V then the reaction will be possible but this does not always seem to be the case.

There were a couple questions which confused me:
Predict whether a reaction is possible between the following pairs:
a) 2Cl- + Sn2+ > Cl2 + Sn
b) 2Fe2+ + Cl2 > 2Fe3+ + 2Cl-

Fe3+ e- > Fe2+ (0.92V)
Cl2 + 2e- > 2Cl- (1.36V)
Sn2+ + 2e- > Sn (-0.14V)

i) Fe2+ and I2
ii) Sn4+ and I-

I2/2I- (0.54). I couldn't find the one for I- but part i should be doable.

How is this worked out? Any help would be greatly appreciated. I can't seem to grasp it.
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4 years ago
#2
Do you have the electrode potential values?
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#3
(Original post by Pigster)
Do you have the electrode potential values?
I put them in just now.
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4 years ago
#4
The more positive equation tends to the right, the more negative to the left.

For (a) Cl2 is more +ve so forms Cl- and Fe2+ forms Fe3+
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#5
(Original post by Pigster)
The more positive equation tends to the right, the more negative to the left.

For (a) Cl2 is more +ve so forms Cl- and Fe2+ forms Fe3+
Thanks for the reply. Sorry to be a bother but could you elaborate a bit more? How does that then make the reaction feasible?
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4 years ago
#6
(Original post by misteltain)
I'm not sure how to predict whether a reaction will be feasible or not. I assumed that if it was >0V then the reaction will be possible but this does not always seem to be the case.

There were a couple questions which confused me:
Predict whether a reaction is possible between the following pairs:
a) 2Cl- + Sn2+ > Cl2 + Sn
b) 2Fe2+ + Cl2 > 2Fe3+ + 2Cl-

Fe3+ e- > Fe2+ (0.92V)
Cl2 + 2e- > 2Cl- (1.36V)
Sn2+ + 2e- > Sn (-0.14V)

i) Fe2+ and I2
ii) Sn4+ and I-

I2/2I- (0.54). I couldn't find the one for I- but part i should be doable.

How is this worked out? Any help would be greatly appreciated. I can't seem to grasp it.
When you calculate the overall emf of the cell it will either be positive or negative, if it is negative the position of equilibrium will lay more to the left so reaction is less feasible and if positive more feasible

Cell emf = more positive cell - the more negative cell
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#7

When you calculate the overall emf of the cell it will either be positive or negative, if it is negative the position of equilibrium will lay more to the left so reaction is less feasible and if positive more feasible

Cell emf = more positive cell - the more negative cell
Thanks for the reply. For a) I calculated the emf to be +1.5V (1.36 - (-0.14)) so this means the eq should be to the right and the reaction more feasible? The answers say that this is not a feasible reaction.
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4 years ago
#8
(Original post by misteltain)
Thanks for the reply. For a) I calculated the emf to be +1.5V (1.36 - (-0.14)) so this means the eq should be to the right and the reaction more feasible? The answers say that this is not a feasible reaction.
Ahh for a. they have reversed the reaction. So your calculation is assuming the other direction of equilibrium.

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#9

Ahh for a. they have reversed the reaction. So your calculation is assuming the other direction of equilibrium.

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Oh I see. How sneaky.

For parts i and ii, do I assume they're the products in the reaction? Just so I know which way the eq shifts. Sorry to be a bother! Thanks a lot.
0
4 years ago
#10
(Original post by misteltain)
Oh I see. How sneaky.

For parts i and ii, do I assume they're the products in the reaction? Just so I know which way the eq shifts. Sorry to be a bother! Thanks a lot.
The equations will be fe3+ + I2 FOR A.

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