Consider the following unbalanced equation.

H1+(aq) + Fe(s) H2(g) + Fe2+(aq)

What is the balanced oxidation half-reaction?
What is the balanced reduction half-reaction?
What is the balanced overall cell reaction?

redn half equation is

2H^+ + 2e ==> H2
oxidation half equation is
Fe ==> Fe^2+ + 2e

2H^+ + Fe --> Fe^2+ + H2 is the total balanced equation.

Well, this equation is as unbalanced as a one-legged ballerina attempting a pirouette! But fear not, for Clown Bot is here to bring balance to the chemical world.

Let's first tackle the oxidation half-reaction, where a species loses electrons. In this case, it's hydrogen going from a +1 oxidation state to a 0 oxidation state. So we can write it like this:

H1+ (aq) H2 (g) + e-

Now, for the reduction half-reaction, we have iron going from a 0 oxidation state to +2. So we can balance it like this:

Fe (s) + 2e- Fe2+ (aq)

Now, let's add these two halves together and see what kind of chemical performance we get!

Multiply the oxidation half-reaction by 2 to balance the electrons:

2H1+ (aq) 2H2 (g) + 2e-

Now combine it with the reduction half-reaction:

2H1+ (aq) + Fe (s) 2H2 (g) + Fe2+ (aq)

And we're done! We've balanced both halves to create the balanced overall cell reaction:

2H1+ (aq) + Fe (s) 2H2 (g) + Fe2+ (aq)

Remember, in the world of chemistry, balance is key! Unless you're a clown balancing plates on your nose – that's just for fun.

To balance the oxidation half-reaction, we need to determine the changes in oxidation states for the elements involved in the reaction. In this case, iron (Fe) is being oxidized from 0 to +2 oxidation state.

The balanced oxidation half-reaction is:

Fe(s) Fe2+(aq) + 2 e-

Next, we need to balance the reduction half-reaction. Hydrogen (H) is being reduced from +1 to 0 oxidation state.

The balanced reduction half-reaction is:

2 H1+(aq) + 2 e- H2(g)

To find the overall cell reaction, we need to make sure that the number of electrons produced in the oxidation half-reaction equals the number of electrons consumed in the reduction half-reaction. In this case, we need to multiply the reduction half-reaction by 2 to balance the electrons.

Balancing the electrons:

2[(2 H1+(aq) + 2 e- H2(g))]

Overall balanced cell reaction:

2 Fe(s) + 4 H1+(aq) 2 H2(g) + 2 Fe2+(aq)

To balance the oxidation half-reaction, we need to separate the reactant and product species and balance the atoms and charges. Let's start with the unbalanced oxidation half-reaction:

H1+(aq) H2(g)

In this reaction, hydrogen is being oxidized from a +1 oxidation state to a 0 oxidation state. To balance the atoms, we need to add one water molecule (H2O) to the reactants side:

H1+(aq) + H2O(l) H2(g)

Now, let's balance the charges. The reactant side has a +1 charge from the H1+ ion, and the product side has a 0 charge from the H2 molecule. To balance the charges, we need to add one electron (e^–) to the reactants side:

H1+(aq) + H2O(l) + e^– H2(g)

Now the oxidation half-reaction is balanced.

To balance the reduction half-reaction, we need to separate the reactant and product species and balance the atoms and charges. Let's start with the unbalanced reduction half-reaction:

Fe(s) Fe2+(aq)

In this reaction, iron (Fe) is being reduced from a 0 oxidation state to a +2 oxidation state. To balance the atoms, we need to add two hydrogen ions (H1+) to the product side to balance the charges:

Fe(s) Fe2+(aq) + 2H1+(aq)

Now, let's balance the charges. The product side has a +2 charge from the Fe2+ ion and a +2 charge from the two H1+ ions. The reactant side has no charge. To balance the charges, we need to add two electrons (2e^–) to the reactants side:

Fe(s) + 2e^– Fe2+(aq) + 2H1+(aq)

Now the reduction half-reaction is balanced.

To write the overall cell reaction, we need to ensure that the number of electrons transferred is equal in both the oxidation and reduction half-reactions. In this case, one electron is transferred in both half-reactions. Multiply the oxidation half-reaction by 2 to balance the number of electrons:

2(H1+(aq) + H2O(l) + e^– H2(g))

Now write the reduction half-reaction:

Fe(s) + 2e^– Fe2+(aq) + 2H1+(aq)

The electrons will cancel out, so we can simply add these two half-reactions together:

2H1+(aq) + 2H2O(l) + Fe(s) 4H1+(aq) + 2H2(g) + Fe2+(aq)

Now we have the balanced overall cell reaction:

2H1+(aq) + 2H2O(l) + Fe(s) 4H1+(aq) + 2H2(g) + Fe2+(aq)