balance Ag+(aq) + Fe(s) = Fe2+ (aq) + Ag(s)?

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  • 1 decade ago
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    Ag+(aq) + Fe(s) --> Fe2+ (aq) + Ag(s)

    2(Ag+ + 1e- --> Ag(s))

    Fe(s) --> Fe2+ + 2e-

    ---------------------------------------

    2Ag+ + Fe(s) --> Ag(s) + Fe2+

    The redox reaction must be charged balanced as well as mass balanced.

    ======== Follow up =========

    Some responders are making this way more difficult that it needs to be. You don't need to introduce hypothetical anions in order to balance the equation. Simply use the "half-reaction" method.

    And this statement, "...there is no need to balance an ionic equation, unless you are given the anion itself." is utter nonsense. Of course you need to balance ionic equations.

  • 1 decade ago

    I'm not entirely sure but... i think it might be

    2Ag+(aq) + Fe(s)= Fe2+(aq) + 2 Ag(s)

    I only say that because if you have 2 lots of + charges you would be able to fill the gap left by the 2+ iron... If that makes sense

  • 1 decade ago

    (1) Calculate E0

    cell for the reaction Ag+(aq) + Co2+(aq) → Ag(s) + Co3+(aq)

    Half reactions:

    Ag+ + e- → Ag reduction E0

    red = +0.80 V

    Co2+ → Co3+ + 1 e- oxidation E0

    ox = -E0

    red = -(+1.80) = -1.80 V

    E0

    cell = Eox + Ered = +0.80 V + (-1.80 V) = -1.00 V

    (2) Calculate E0

    cell for the reaction Fe(s) + 2 H+(aq) → Fe2+(aq) + H2(g)

    Half reactions:

    Fe → Fe2+ + 2 e- oxidation E0

    ox = -E0

    red = -(-0.41 V) = +0.41 V

    2 H+ + 2 e- → H2 reduction E0

    red = 0.00 V

    E0

    cell = Eox + Ered = +0.41 V + 0.00 V = +0.41 V

    (3) Calculate E0

    cell for the reaction 2 I-(aq) + 2 H+(aq) → I2(s) + H2(g)

    Half reactions:

    2 I- → I2 + 2 e- oxidation E0

    ox = -E0

    red = -(+0.54 V) = -0.54 V

    2 H+ + 2 e- → H2 reduction E0

    red = 0.00 V

    E0

    cell = Eox + Ered = -0.54 V + 0.00 V = -0.54 V

    (4) Calculate E0

    cell for the reaction 2 Fe3+(aq) + 3 Pb(s) → 2 Fe(s) + 3 Pb2+(aq)

    Half reactions:

    2 Fe3+ + 6 e- → 2 Fe reduction E0

    red = -0.04 V

    3 Pb → 3 Pb2+ + 6 e- oxidation E0

    ox = -E0

    red = -(-0.13 V) = +0.13 V

    E0

    cell = Eox + Ered = +0.13 V + (-0.04 V) = +0.09 V

    Cell diagrams/cell notation

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