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

Relevance

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

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

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

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

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

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.

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) 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