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lorijessica16 · May 20, 2009, 03:56 PM

Cr^3+ oxidized to Cr207^-2
S208^2- is reduced to sulfate ion

How do you write the balanced net ionic equation for the reaction? :confused:

Perito · May 20, 2009, 08:24 PM

1. Balance the half reactions
a. Balance metals first.
b. Add water to balance O
c. Add H+ to balance hydrogen
d. Add e- to balance charge.

$2Cr^{+3} +7H_2O \rightleftharpoons Cr_2O_7^{-2} + 14 H^+ + 6e^-$

$S_2O_8^{-2} +2e^- \rightleftharpoons 2SO_4^{-2}$

2. Multiply half reactions by values so that the electrons will cancel. Electrons must be on opposite sides of the arrows (one half reaction is oxidation, the other is reduction). Add the half reactions to get the net ionic equation.

3. If this is in alkaline solution and any H+ is left, add OH- to both sides of the equation for each H+ present. The H+ + OH- will form H2O.