Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta

Slightly confused by oxidation states/rules.... watch

    • Thread Starter
    Offline

    17
    ReputationRep:
    And now for the products......

    NO ( N = +2; O = -2)

    H3AsO4 =

    (H = (3x1 =3)
    (O4=4x-2=-8)
    (As=+5)


    H2O = H2;
    O = -2


    Is this correct?

    Therefore, As is oxidised, and N is reduced.
    Offline

    19
    ReputationRep:
    (Original post by apronedsamurai)
    And now for the products......

    NO ( N = +2; O = -2)

    H3AsO4 =

    (H = (3x1 =3)
    (O4=4x-2=-8)
    (As=+5)

    H2O = H2;
    O = -2

    Is this correct?

    Therefore, As is oxidised, and N is reduced.
    yeah N goes from +5 to +2 so it's reduced
    As goes from +3 to +5 so it's oxidised
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by z33)
    yeah N goes from +5 to +2 so it's reduced
    As goes from +3 to +5 so it's oxidised
    So...net changes

    As (+2) x3
    N (-3) x2

    Therefore ratio of Oxidation to Reduction: 3:2?
    Offline

    19
    ReputationRep:
    (Original post by apronedsamurai)
    So...net changes

    As (+2) x3
    N (-3) x2

    Therefore ratio of Oxidation to Reduction: 3:2?
    u wot m8? why do you need the ratio of oxidation to reduction never worked with it before xD
    • Thread Starter
    Offline

    17
    ReputationRep:
    (Original post by z33)
    u wot m8? why do you need the ratio of oxidation to reduction never worked with it before xD
    Apparently you plug it into the original equation to determine coeffecients to balance the equation :s
    Offline

    19
    ReputationRep:
    (Original post by apronedsamurai)
    Apparently you plug it into the original equation to determine coeffecients to balance the equation :s
    there's an equation?
    http://preparatorychemistry.com/Bish...cing_Redox.htm i found that <---
    never used this before tbh

    we use half equations and stuff to balance
    like you have the e- on either side and them times them so they both have the same amount of e- on either side and then you cancel the e- and the equation is balanced!


    like that ^
    Offline

    2
    ReputationRep:
    (Original post by Serine Soul)
    Yeah, it's right.

    Could you work out the oxidation states of each atom in SO42-?
    Hey, I don't know whether your question has been answered or not but im gonna answer it anyways If you're familiar with the rules of oxidation numbers, we know that the overall oxidation state of a compound is the ionic charge which in this case is 2-. Also, we know that one Oxygen atom usually has a 2- oxidation state and in this case we have 4 oxygen atoms so its -2 x 4 = -8. How we get that -8 to increase to -2? We have the sulphur and one atom of the element to get -8 to -2 we add 6. Therefore the oxidation state of sulfur will be +6 and the oxidation state of oxygen (one oxygen not all 4) will be -2.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: May 22, 2016
Poll
Do I go to The Streets tomorrow night?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.