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3 years ago

Match the element or group to the rule assigning its oxidation state.

Chemistry

2 answers:

slamgirl [31]3 years ago

6 0

It would go B. A. E. D. C.
Hope I helped!

Anna11 [10]3 years ago

5 0

C. Lone elements and atoms in gases : 0

Elements in the free elemental have an oxidation number of zero

E. Elements in groups 1, 2, and 17 and polyatomic ions : Ionic charge

Group 1, 2, and 17 ions are formed by the alkali metals, alkaline metals and halogens respectively. Alkali metals, alkaline metals and halogens form +1, +2 and -1 ions respectively. Polyatiomic ions have -1, -2, -3 charges such as OH-, CO32-, PO43- respectively.

B. Hydrogen : +1, – 1 if bonded to a diatomic metal

H has an oxidation number of +1. H in metal hydrides such as NaH (sodium hydride) has an oxidation number of -1

A. Oxygen : Almost always –2

O has an oxidation number of -2.

D. Elements with multiple oxidation states : Determined by other elements in the compound

V (vanadium) has different oxidation which depends on the compound in which V is found.

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2 years ago

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3 years ago

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<em>Check the image below:</em>

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3 years ago

Given the balanced ionic equation representing the reaction in an operating voltaic cell: zn(s) + cu2+(aq) → zn2+(aq) + cu(s) th

soldi70 [24.7K]

Answer: from the Zn anode to the Cu cathode

Justification:

1) The reaction given is: Zn(s) + Cu₂⁺ (aq) -> Zn²⁺ (aq) +Cu(s)

2) From that, you can see the Zn(s) is losing electrons, since it is being oxidized (from 0 to 2⁺), while Cu²⁺, is gaining electrons, since it is being reduced (from 2⁺ to 0).

3) Then, you can already tell that electrons go from Zn to Cu.

4) The plate where oxidation occurs is called anode, and the plate where reduction occus is called cathode.

So you get that the electrons flow from the anode (Zn) to the cathode (Cu).

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2 years ago

Read 2 more answers