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    • Thread Starter
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    Hey,

    I'm after the half equations for

    Cr_2O_7^{2-} + I^- --> Cr^{3+}

    One must be Cr_2O_7^{2-} + 14H^+ + 6e^- --> 2Cr^{3+} + 7H_2O

    but what is the other?
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    The iodide ion is going to be oxidised and loose an electron to form iodine.

    So...

    2I- -> I2 + 2e-

    Then you just have to balance out the electrons on either side so

    6I- -> 3I2 + 6e-
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    So overall reaction is:

    6I- + Cr2O7 2- + 14H+ -> 3 I2 + 2Cr3+ + 7H2O
    • Thread Starter
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    Cool, cheers

    What about with H3ASO3 --> H3AsO4?

    I've never seen one like that before... My instinct would be to add H2O on the left... (This is in 4-9M HCl alongside IO3- in an Andrew's titration)
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    (Original post by vector)
    Cool, cheers

    What about with H3ASO3 --> H3AsO4?

    I've never seen one like that before... My instinct would be to add H2O on the left... (This is in 4-9M HCl alongside IO3- in an Andrew's titration)
    H3AsO3 + H20 --> H3AsO4 + 2H + 2e- (i think)
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    Yeah, that was my guess... Thank you!
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    (Original post by vector)
    Yeah, that was my guess... Thank you!
    No problem
 
 
 
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