Question

Balance the following redox reaction in basic solution. You will need to show all work including half-reactions, balancing of O, H, electrons, and the final balanced reaction.

Cr(OH)₃(s) + ClO₃⁻(aq)  ®  CrO₄²⁻(aq) + Cl⁻(aq)   

please show work

Answer 1

Redox Reaction given crion), (s) + coz (99) – crop eaq) + crear Step O: Assign oxidation numbers for each atom in equation. croH; & coz crop- & at oxidation no. of 8 -2 ON of H = t1 O.N. of 12-1 O.N. Of or in co(oh), => crt 3(-2) + 3(+1)=0 3) cr - 6 + 320 2 cra +3 O.N of er in croge - art 4(-2) = -2 cr- 8 = -2 cr= +6 O.N. Of cl in clos - I + 3(-2) 2-1 a-62-1 c=15 (13) (+5) cr(on), + clos (16) crog- & & (-1) Step U identify Reduction and oxidation reaction Reduction - Decrease in O.N. Oxidation » increase in O.N. (+5) Reduction : 103 ci- (ON. Of u decreases) oxidation: Er(OH)2 CroqP-(O.N. Of cn increases) +1) (13) (16)

Step ( Balance all other atoms except 'H' and O' atom Oxidation Reduction - Cr(OH)3 (103) Croqp c (cr balanced) (cl balanced) Step ® Balance the 'o' atom by adding Ho molecule Oxidation » cr(си), * 4,0 Croq?- 16 atom balanced) Reduction clos (+ 34,0 lo atom balanced) Step Balance the '' atom by adding ni ion Crop + 54+ (H' atom balanced) croH)₃ + H₂O coz + 6Ht R: 1+ 3H0 ( atom balanced.) Step for reaction in a basic medium, add one on on to each side for every Ht ſon present in the equation. CreOH)₂ + H₂O + 50H cro- +5H + +504 OR oxidation : cr(on)z + H20 + 501 - Croqr +5H0 R: 105 + 601 +6Htc + 604 + 3H₂O OR Reduction do + 6H₂ c+600 + 340

Step VID Balance the charge : for this, add e to the more positive side to equal the less positive side of reaction. O colon), + 40 + 504- crog + 5H₂O + 3 é R: cO₂ + 6H₂O + bé - ct + 3H0+ 604 Step VID Add both half reaction after multiplying each reaction coith appropriate number of electron. so as to cancel out electrons. croq? +54,0 + 36 ) x 2 O: (creon) + H₂O +504 - R (do + 6H₂O + bé or + 3H₂O + 604) x 1 0: 2 CH(OH)2 + 24,0 + 100M - zero, e + 10 Horse a + 340 + 604 R: co₂ + 6H₂O + bé + 200(0m),+ C103 + 10 OH + 84,0 2 croq? - +ci+604- + 1340 OR 2 cr(on) + 410; + 1001 -601 — peroq? 74 +134,0-840 OR ( Cr(on)C10, + 404 2000 - + + 5H₂O (Balanced Redox reaction)