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    write the redox equation for the reaction of copper (ii) ion cu2+ and the iodide ion i-
    i do not answer how the answer is: 2cu2+ +4i- --> 2CuI +I2
    i got cu2+ +2i- ---> i2 + cu

    how do you get this answer?

    I- isn't a good enough reducing agent to reduce the Cu2+ all the way to Cu (o.n. +2 -> 0)

    It can reduce the Cu2+ to Cu+, though (o.n. +1 -> +1)

    You need 2I- in order to make an I2, hence why 2Cu2+ is needed (each Cu2+ can steal 1e- off each I-)

    But since Cu+ ions don't exist in solution, CuI must ppt out, hence the need for the 3rd and 4th I- (which aren't oxidised).
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