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    The feasibility of these displacement reactions can be predicted from the E  values above.
    2KCl(aq) + Br2(l) solution changes from colourless to red-brown
    1. Cl2(g) + 2e ⇌ 2Cl (aq)

    2Br-(aq)⇌2Br(l) + 2e-

    E   = +1.36 V for 1

    E   = -1.07 V for 2

    E = +0.29 V is positive so the reaction is feasible

    how is the final value +0.29? I thought it was the negative value which had its sign changed :

    +1.36 + 1.07 as it has its sign switched?
    the answer won't give me 0.29 though.. where am I going wrong? please help

    I thought thats how you calculate e.m.f

    Hmmmm, well, to be honest, I don't think the reaction should be feasible! I mean, this is a displacement reaction but bromine is less reactive than chlorine so it won't be able to displace the chlorine from KCl!

    Using the two electrode potentials given, the only reaction that would be feasible is if the bromine was oxidised to bromide ions and the chlorine was reduced to chloride ions!

    At least this is what I think ! Hopefully someone else can shed some more light on this


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