Balance the following equation in basic conditions. Phases are optional.
CoCl2+Na2O2 = Co(OH)3+Cl- +Na+
(redox and oxidation reactions)
CoCl2 --> Co(OH)3 & 1 e-
cobalt loses 1 e- going from Co+2 to Co+3
H2O2 & 2e- --> Co(OH)3
while O2)-2 take a total of 2 e- going from O-1 each to O-2
each
so we have to double the Cobalt to balance the electrons taken and
lost:
2CoCl2 --> 2 Co(OH)3 & 2 e-
H2O2 & 2e- --> Co(OH)3
combining these:
2CoCl2 (s) + 1Na2O2 (ag) ----> 2Co(OH)3 (ag)+Cl- (ag) + Na+
(ag)
let's balance the Na's:
2CoCl2 (s) + 1Na2O2 (ag) ----> 2Co(OH)3 (ag)+Cl- (ag) + 2Na+
(ag)
let's balance the Cl's:
2CoCl2 (s) + 1Na2O2 (ag) ----> 2Co(OH)3 (ag)+ 4Cl- (ag) + 2Na+
(ag)
let's balance the net -2 on the right by adding 2 OH- to the
left:
2CoCl2 & 1Na2O2 & 2 OH- ----> 2Co(OH)3 & 4Cl- &
2Na+
we need 4 more H's on the left to balance the 6 H's on the right,
so add neutral water:
2CoCl2 & 1Na2O2 & 2 OH- & 2 H2O ----> 2Co(OH)3 &
4Cl- & 2Na+
it is done, as a check on the work , we check oxygens which should
already balance, 6 oxygens --> 6 oxygens, its done
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I am having trouble balancing the following problem. I am following the steps for a basic solution, but I am coming up with the wrong answer. There is an answer on Chegg, but I am having difficulty following that response. The problems is: Balance the following equation in basic conditions. CoCl2 + Na2O2 --> Co(OH)3 + Cl- + Na+
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