Question

Write a balanced chemical equation describing the oxidation of the bromide ion by solid xenon trioxide to produce bromine liquid and xenon gas in an acidic aqueous solution. Use the smallest whole number coefficients possible.

How many electrons are transferred in the redox reaction?

Answer 1

Bromide ion is Br^-

Xenon trioxide is XeO₃

Bromine liquid is Br₂(l)

Xenon gas is Xe(g)

So in first place the statement says that the reaction has this form:

Br^- + XeO₃   Br₂(l) + Xe(g), of course this reaction is unbalanced we need to consider the acid medium

The first thing to do is to identify the reduction and oxidation species

Br- goes from -1 to Br₂ with 0 oxidation state

an electron is lost by the Br- so this is an oxidation

for the XeO₃, the Oxygen has an oxidation state of -2 in most cases so for the XeO3 to have a total oxidation state of zero

-2*3 = -6

x - 6 = 0, x must be +6, oxidation number of xenon is +6

Xe goes from +6 to zero, it gains 6 electrons this is a reduction so, lets start to balance the equation,

Lets balance the equation

Br-   Br₂ , first balance the Br and add the electrons

2Br-    Br₂ + 2e^- this 2 electrons balance the charges on both sides

For the other half reaction (reduction)

XeO₃ Xe , Xenon is already balanced, now balance the oxygens this is by adding water

XeO₃ Xe + 3H₂O , 3 oxygens are balanced by adding 3 water molecules

The next thing to do is to balance the hydrogens, as you can see we have 6 hydrogens on the right and zero on the left so we add 6 H on the left side, remember to add the electrons to neutralize the positive charge of H+

6H⁺ + 6e^- + XeO₃ Xe + 3H₂O

put equations together

6H⁺ + 6e^- + XeO₃ Xe + 3H₂O

2Br-    Br₂ + 2e^- , we have to neutralize the electrons when wee add this two equations to do that we need 6 electrons instead of 2 to do that we need to multiply this equation by 3

6Br-   3Br₂ + 6e^- , now add this 2 equations

6H⁺ + 6e^- + XeO₃ Xe + 3H₂O

6Br-   3Br₂ + 6e^-

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6H⁺ (aq)+ 6Br^- (aq)+ XeO₃(s) Xe(g) + 3Br₂ (l) + 3H₂O (l)

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