I need to balance this redox reaction using the half-reactionmethod. The spectat

Question

I need to balance this redox reaction using the half-reactionmethod.
The spectator ion (Na) needs to be first taken out, then added backin at the end.
Please show all work.

NaClO(aq) + NaCrO2(aq) NaCl(aq) + Na2CrO4(aq)

Explanation / Answer

ClO- + CrO2+ -> Cl- + CrO42- So on the left, ClO-, we know O usually has a charge of 2-, so Clin ClO- has charge of +1. But on right Cl- has charge of -1; thus achange in 2 for charge ClO- + 2e- => Cl- Balance O with H2O ClO- + 2e- => Cl- + H2O Balance H with H+ ClO- + 2H+ + 2e- => Cl- + H2O This is the reduction equation For the oxidation, we note that on th eleft,CrO2- has Cr with a charge of 3+, (2 oxygenswith 2- each, so 3 + (2*-2) = -1, the overall charge of the ion).ON the right, in CrO42-, Cr has a charge of6+ (6 + 4*(-2) = -2, the overall charge) Hence there is a change of 3 charge (3 to 6) CrO2- => CrO42- +3e- Balance O with H2O CrO2- + 2H2O=>CrO42- + 3e- Balance H with H+ CrO2- + 2H2O=>CrO42- + 3e- + 4H+ The oxidation reaction Now we add both oxidation and reduction reactoins, and eliminateelectrons (ClO- + 2H+ + 2e- => Cl- +H2O)*3                             (we note the common multiple 6 (CrO2- + 2H2O=>CrO42- + 3e- + 4H+)*2                                                                              3ClO- + 6H+ + 6e- +2CrO2- + 4H2O => 3Cl- +3H2O + 2CrO42- + 6e- + 8H+ Cancel out species that are on both sides (electrons, H+, H2O) 3ClO- +2CrO2- + H2O => 3Cl- +2CrO42- + 2H+   (note 8H+ - 6H+ means 2H+ on the right; 4H2O-3H2O means 1 H2O onleft, and 6e- - 6e- = 0)

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