What is the element oxidized and the element reduced in the equation:

HAsO2 + 2 Ag+ + 2 H2O 2 Ag + H3AsO4 + 2 H+

Go through the equation and assign oxidation numbers to each element. Look at the elements that have changed oxidation state. The loss of electrons is oxidation and gain of electrons is reduction.

Here is how you assign oxidation numbers.
http://www.chemteam.info/Redox/Redox-Rules.html

To determine the element that is oxidized and the element that is reduced in a redox reaction, we need to identify the changes in the oxidation states of each element.

In this equation, we can assign the oxidation states as follows:
- Hydrogen (H) typically has an oxidation state of +1, except when it is bonded to metals, where its oxidation state is -1.
- Oxygen (O) usually has an oxidation state of -2, except in peroxides (where it is -1) and in certain compounds where it can have a positive oxidation state.
- Silver (Ag) typically has a +1 oxidation state.
- Arsenic (As) is less common, but in H3AsO4, each hydrogen has +1 oxidation state, and oxygen has -2 oxidation state. The overall charge on H3AsO4 is 0, which means that the oxidation state of As must be +5.

Using these oxidation states, let's analyze the changes in oxidation states:

In HAsO2 (vanadium's oxidation state is +4), vanadium has an oxidation state of +4.

In 2 Ag+, each silver ion (Ag+), has an oxidation state of +1.

In Ag (the product formed when silver ions lose electrons), each silver atom has an oxidation state of 0.

In H3AsO4 (arsenic's oxidation state is +5), arsenic has an oxidation state of +5.

In H+ (hydrogen's oxidation state is +1), hydrogen has an oxidation state of +1.

Based on the changes in oxidation states, we can see that:
- The element vanadium goes from an oxidation state of +4 in HAsO2 to an oxidation state of +5 in H3AsO4. It is therefore oxidized.
- The silver ions (Ag+) go from an oxidation state of +1 to an oxidation state of 0 in Ag. They are therefore reduced.

Therefore, in the given equation: HAsO2 + 2 Ag+ + 2 H2O -> 2 Ag + H3AsO4 + 2 H+, vanadium (V/As in HAsO2) is oxidized, and silver ions (Ag+) are reduced.