What are the coefficients of the reactants and products in the balanced equation above? Remember to include H2O(l)H2O(l) and H+(aq)H+(aq) in the appropriate blanks. Your answer should have six terms. Enter the equation coefficients in order separated by commas (e. g., 2,2,1,4,4,3). Include coefficients of 1, as required, for grading purposes.

The balanced equation is then given as

BrO₃⁻ + 3Sn²⁺ + 6H⁺ → Br⁻ + 3Sn⁴⁺ + 3H₂O

The coefficients written in a bracket separated by commas is then given as

(1, 3, 6, 1, 3, 3)

Explanation:

The equation for the question is

BrO₃⁻ (aq) + Sn²⁺ + __ → Br⁻ + Sn⁴⁺ + __

The H⁺ and H₂O presented in the question shows that this redox reaction takes place in acidic medium.

We first identify which reactant specie is being oxidized or reduced.

Oxidation is the increase in oxidation number of reactant species due to the loss of electrons while Reduction is the reduction in oxidation number of reactant species due to the gain of electrons.

From the reaction given, it is evident that the Tin ion is oxidized as its oxidation number increases from +2 to +4.

And the Br in BrO₃⁻ undergoes reduction to have its oxidation number change from +5 to -1.

Note that +5 was calculated for as thus.

If the oxidation number of Br in BrO₃⁻ is unknown and called x, the oxidation number of the O in BrO₃⁻ is -2.

x + (-2)(3) = -1

x = -1 + 6 = +5

So, back to the question, since we know which reactant species are oxidized and reduced, we can then write the reduction and oxidation half reactions. Balanced reduction and oxidation half reactions at that showing the number of electrons lost or gained.

Reduction half reaction

BrO₃⁻ → Br⁻

This isn't possible, so we add the spectator compounds/elements/ions provided in the form of H⁺ and H₂O

BrO₃⁻ + H⁺ → Br⁻ + H₂O

We then balance this reaction stoichiometrically,

BrO₃⁻ + 6H⁺ → Br⁻ + 3H₂O

Now, we can check the oxidation numbers on both sides to know the number of electrons gained.

-1 + 6 → -1 + 0

+5 → -1

Hence, it is evident that 6 electrons are gained in this reduction half reaction.

So, we rewrite the reduction half reaction finally to be

BrO₃⁻ + 6H⁺ + 6e⁻ → Br⁻ + 3H₂O

We then repeat this for the oxidation half reaction.

Sn²⁺ → Sn⁴⁺

This is possible, So, no need to add the spectator compounds/elements/ions provided in the form of H⁺ and H₂O

The equation is also balance stoichiometrically, So, we just proceed to balance the charges/oxidation numbers.

+2 → +4

This shows that 2 electrons are lost, So, we rewrite our oxidation half reaction as

Sn²⁺ - 2e⁻ → Sn⁴⁺

It is more appropriate to write it in the form

Sn²⁺ → Sn⁴⁺ + 2e⁻

So, we can then write the two balanced half reactions (stoichiometrically and charge balance) on top of each other.

BrO₃⁻ + 6H⁺ + 6e⁻ → Br⁻ + 3H₂O

Sn²⁺ → Sn⁴⁺ + 2e⁻

In order to add, we need the number of electrons on both of these to be the same, so We multiply the reduction half reaction by 1 and multiple the oxidation half reaction by 3

[BrO₃⁻ + 6H⁺ + 6e⁻ → Br⁻ + 3H₂O] × 1

[Sn²⁺ → Sn⁴⁺ + 2e⁻ ] × 3

We then get

BrO₃⁻ + 6H⁺ + 6e⁻ → Br⁻ + 3H₂O

3Sn²⁺ → 3Sn⁴⁺ + 6e⁻

We can then add the two half reactions now

BrO₃⁻ + 6H⁺ + 6e⁻ + 3Sn²⁺ → Br⁻ + 3H₂O + 3Sn⁴⁺ + 6e⁻

Taking out the 6 electrons that appear on both sides, we have

BrO₃⁻ + 6H⁺ + 3Sn²⁺ → Br⁻ + 3H₂O + 3Sn⁴⁺

Written more properly as

BrO₃⁻ + 3Sn²⁺ + 6H⁺ → Br⁻ + 3Sn⁴⁺ + 3H₂O

Hope this Helps!!!

but know think of it in terms of bond breaking and bond making.

h–h + h–h + o=o ==> h–o–h + h–o–h

as you can see from the diagram, theoretically, more energy is released by forming water, than was absorbed in breaking up the original hydrogen and oxygen molecules.

the energy level of the fuel + oxygen is higher than that of water.

therefore the fuel + oxygen mixture has the potential to release potential chemical energy as heat energy or, crucially, electrical energy, if the reaction can be made to go via  oxidation and reduction reactions involving electron transfer.

see this page for more details of the full  bond energy calculations

a  fuel cell  is a device to make electricity that functions like an electrical cell or battery that you supply with a fuel (liquid or gas) and an oxidant (oxidising agent, usually oxygen gas) which react together in a redox reaction (oxidation + reduction) to release energy as a flow of electrons i.e. an electrical current capable of doing useful work.

the oxidation of the fuel and the reduction of the oxidant reactions take place on electrodes (see diagram on right, all fully explained further down the page).

fuel cells essentially a device to release the chemical potential energy of combustible fuels as electrical energy  at a much lower temperature and without the flame!

so, the idea of a  hydrogen–oxygen fuel cell  is to release the energy from hydrogen reacting with oxygen, not as heat as in normal combustion in air, but as  useful electrical energy  i.e. a practical electricity supply.

hydrogen-oxygen fuel cells are much efficient than conventional power stations or batteries (e.g. zinc-carbon) because the electrical energy is directly generated from the chemical reaction between the oxidant and the fuel - there are no complications with boilers, turbines and generators.

in fossil fuel power stations, motor vehicles, coal/gas fires etc. quite a percentage of energy is lost as waste heat.

with a fuel cell there are fewer stages in producing the useful energy, so there is less opportunity to lose potentially useful energy e.g. waste heat, friction from moving parts etc.

a hydrogen–oxygen fuel cell is a non–polluting clean fuel since the only combustion product is water.

fuel cells  do not produce pollutants like carbon monoxide, sulfur dioxide and nitrogen oxides - gases you get from fossil fuel combustion in power stations or burnt road vehicles..

cars powered by fuel cells would be quite an environmental advantage in cities, where electric cars are already beginning to be significantly used in developed countries.

fuel cells could replace larger batteries which are not easily recycled and contain highly toxic metal compounds.

it would be an ideal fuel on this basis e.g. for motor vehicles, but that's not the only factor to consider!

it would be ideal if the hydrogen fuel could be manufactured by electrolysis of water e.g. using solar cells.

hydrogen can be used to power fuel cells.

it all sounds wonderful but, still technological problems to solve for large scale manufacture and distribution of 'clean' hydrogen gas or use in generating electricity and its rather an inflammable explosive gas!

explanation:

fuel cells e.g. the hydrogen-oxygen fuel electrical cell

fuel cells have to be supplied by an external source of fuel (e.g. hydrogen) and an oxidant e.g. oxygen or air.

the hydrogen or any other fuel is oxidised electrochemically inside the fuel cell to produce a potential difference i.e. a voltage capable of producing a working current.

the overall chemical reaction in a hydrogen fuel electrochemical cell involves the oxidation of hydrogen by oxygen to produce only water.

hydrogen fuel cells offer an alternative to rechargeable cells and batteries.

a fuel cell will produce a potential difference  ('voltage')  and a workable electric current until one of the reactants is used up.

hydrogen gas can be used as fuel.

it burns with a pale blue flame in air reacting with oxygen to be oxidised to form water.

hydrogen + oxygen  ==>   water

2h2(g)  + o2(g)  ==>   2h2o(l) 

i'm pretty sure its b i might be wrong though!