Sodium thiosulfate, na 2s 2o 3, is used as a "fixer" in black and white photography. identify the reducing agent in the reaction of thiosulfate with iodine. 2s 2o 3 2–( aq) + i 2( aq) → s 4o 6 2–( aq) + 2i –( aq)

Reducing agent in the given reaction is .

Explanation:

A reducing agent is defined as an element which tends to lose electrons to other element leading to an increase in its oxidation number.

In the given reaction, oxidation state of sulfur in is +2 and has 0 oxidation state.

In oxidation state of S is 2.5 and in oxidation state of I is -1.

Since, an increase in oxidation state of S is occurring from +2 to +2.5. Hence, it is acting as a reducing agent.

Thus, we can conclude that reducing agent in the given reaction is .

the best answer is d because the strong cohesion is caused by the ability of water molecules to undergo hydrogen bonding.

a is incorrect because when water solidifies, it becomes   ice, which expands and occupies a bigger volume for the same mass of water in its liquid state.

b is incorrect because water has high surface tension. besides bonding to itself it hydrogen bonds to other molecules at the top decreasing possibility for breaking of attractive forces.

c is incorrect because the water molecule has oxygen which is highly electronegative and so pulls the electron pair towards itself creating a dipole moment, making the molecule polar.

e is incorrect because water has strong adhesion, ability to stick to other molecules such as glass because of this molecules having stronger hydrogen bonding than water.