Question

Question #e32b6

Answer 1

`I_(3(aq))^(-) + 2S_2O_(3(aq))^(2–) -> 3 I_((aq))^(–) + S_4O_(6(aq))^(2–)`

Explanation:

• Oxidation:

`2S_2O_(3(aq))^(2–) -> S_4O_(6(aq))^(2–) + 2e^-`

• Reduction: `I_(3(aq))^(–) + 2e^(-) -> 3 I_((aq))^(–)`

the overall reaction then:

• Redox:

`I_(3(aq))^(–) + 2S_2O_(3(aq))^(2–) -> 3 I_((aq))^(–) + S_4O_(6(aq))^(2–)`

• Stoichiometry:

From the balanced equation, we can say that:

`(1)/"1" * n_(I_3^(–)) = (1)/"2" * n_(S_2O_3^(2–))`.

where the denominators 1 and 2 were taking from the corresponding coefficients of `I_3^–` and `S_2O_3^(2–)` respectively.

Now,

`[I_3^–]` . `V = (1)/"2" * [S_2O_3^(2–)] * V^'`

(since `n = C * V`), where

`V^' = "10.50 mL "` and `" "V = "15.00 mL"`

Therefore,

`[I_3^–] = (1)/"2" * [S_2O_3^(2–)] * (V')/"V"`

`[I_3^–] = (1)/"2" * "0.0500M" * (10.50 cancel("mL"))/(15.00 cancel("mL")) = "0.0175M"`

(rounded to 3 significant figures)

I would like to suggest the following video, that deals with a similar example in terms of stoichiometry of a redox reaction.

Answer 2

`(a)" "I_(3(aq))^(-)+2S_2O_(3(aq))^(2-)rarrS_4O_(6(aq))^(2-)+3I_((aq))^-`

`(b)" "0.0175"mol/l"`

Explanation:

Thiosulfate ions give out electrons:

`2S_2O_(3(aq))^(2-)rarrS_4O_(6(aq))^(2-)+2e`

These are taken in by `I_3^-` which are reduced:

`I_(3(aq))^(-)+2erarr3I_((aq))^(-)`

Adding both sides gives:

`I_(3(aq))^(-)+2S_2O_(3(aq))^(2-)rarrS_4O_(6(aq))^(2-)+3I_((aq))^-`

(b).

`c=n/v`

`n=cxxv`

So no. moles `S_2O_3^(2-)=0.05xx0.0105=5.25xx10^(-4)`

From the equation we can see that the number of moles of `I_3^-` must be half of this:

no. moles `I_3^(-)=(5.25xx10^(-4))/(2)=2.625xx10^(-4)`

`c=n/v`

So `[I_(3(aq))^(-)]=(2.625xx10^(-4))/(0.015)=0.0175"mol/l"`