Question

How does photosynthesis use oxidation and reduction?

Answer 1

Photosynthesis involves oxidation and reduction by oxidizing the oxygen in water and reducing the carbon in carbon dioxide.

Explanation:

Photosynthesis is the process by which a plant converts the energy of sunlight into the chemical energy of sugars such as glucose.

The process consists of several steps, but the overall reaction is

`"6CO"_2 + "6H"_2"O" stackrel("sunlight")(→) "C"_6"H"_12"O"_6 + "6O"_2`

The process makes more sense if we write the formula of glucose `"C"_6"H"_12"O"_6` as `"(H-C-OH)"_6`.

Photosynthesis can be separated into two parts.

The Photo Part (Photolysis)

The photo part of photosynthesis involves the oxidation of the oxygen from water.

`"2H"_2"O" stackrel("energy")(→) "4H"^+ + "O"_2 + 4"e"^-`

Each `"O"` atom loses two electrons, so the oxygen in water is oxidized.

The electrons released during photolysis are picked up by a carrier molecule called nicotinamide adenine dinucleotide phosphate (`"NADP"`), changing it from its oxidized state (`"NADP"^+`) to its reduced state (`"NADPH"`):

`"2NADP"^+ + "2H"^+ + "4e"^(-) → "2NADPH"`

The Synthesis Part

Here, the `"NADH"` gives up its electrons and reduces the carbon in carbon dioxide.

`"2NADPH" → "2NADP"^+ + "2H"^+ + "4e"^-`

`"CO"_2 + "4H"^+ + "4e"^(-) → "(H-C-OH)" + "H"_2"O"`

The overall reaction is

`cancel(2)"H"_2"O" → cancel("4H"^+) + "O"_2 + cancel(4"e"^-)`
`"CO"_2 + cancel("4H"^+) + "4e"^- → "(H-C-OH)" + cancel("H"_2"O")`
`bar("CO"_2 + "H"_2"O" → "(H-C-OH)" + "O"_2)`

Multiply by 6, and we get the equation for photosynthesis

`"6CO"_2 + "6H"_2"O" → "C"_6"H"_12"O"_6 + "6O"_2`