Part d 2clo2(g)+2i−(aq)→2clo−2(aq)+i2(s) drag the appropriate labels to their respective targets. e−→e superscript- rightarrow ←e−leftarrow e superscript- cathode cathode anode anode ii superscript- clo−2c l o subscript 2 superscript- request answer part e indicate the half-reaction occurring at anode. express your answer as a chemical equation. identify all of the phases in your answer. nothing

The half reaction occurring at anode is

Explanation:

The substance having highest positive potential will always get reduced and will undergo reduction reaction.

For the given chemical equation:

The half reaction follows:

Oxidation half reaction:  

Reduction half reaction:      ( × 2 )

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

Hence, the half reaction occurring at anode is

n+h→nh3

we must remember that nitrogen and hydrogen are both diatomic molecules in their standard gas form. this adjusts the equation to

n2+h2→nh3

now we need to adjust coefficients in order to balance the atoms on each side of the equation. currently we have 2 atoms of nitrogen and 2 atoms of hydrogen on the reactant side and 1 atom of nitrogen and 3 atoms of hydrogen on the product side.

we can balance the hydrogens by placing a coefficient of 2 in front of ammonia and a coefficient of 3 in front of the hydrogen.

n2+3h2→2nh3

this gives us 6 hydrogen on each side and coincidentally the nitrogens now equal 2 on each side.

d.) circle motion requires a force to give it a centripetal acceleration.