Question

Balance the following reaction Partial credit will be given for each properly balanced half reaction Balance in acid solution: The reducing agent is The oxidation state of tungsten in Wosis Bow Br Wos. (Brz sertBe) 6X (3Hot W W O2 + 6H't be 3 +2 6B 18 Br 718E glots W 303 +18Htle O +6br7 738-312Br230,77 Balance in basie solution: The oxidizing agent is The oxidation state of carbon in HCHO is - CHOH + cio" Hсно + Clot (CH3OH HC HO +26) 1+2 (12+10:30 02 +40)2 +2 CHOH -> Hichote 2 34 2010 2005 12/10 HC110H FC)0, BACHO

please solve both with the correct answers

Answer 1

In acidic medi Brez tw Br + woz 1. Assign the oxidation numbers for all atoms involved in the equations and find the species undergoing oxidation and reduction. Reduction gain of ke- +6 -2 Brytw 3x=-! -Y3 Br- & Wor - 1 X-Y3 oxidation (logs of te). 2. half reactions are Brza Br- (Reduction half reaction) w - wo₂ (oxidations half reaction] 3. balance the half realtion other than h&o. Brz - 3Br & w 2003 wt 204 two

to Balance Oxygen atom (in oxidation half Reaction). (1) Balance odrom by adding sufficient number of water molecules on the deficient side in oxygenaton-heed of 3 tho molecule. W + 3H20 Woz (1) Balance H-atom by adding Htion. There is need of 6Ht W+3tho - Wo₃ + 6H+. (111) Balance charge in oxidation half reaction by é are added on right side. & There is need of be as oxidation no. charge o toto W+3140 - wo₂+ 6 H + +60 # To Balance Reduction halb Reaction, 383 3 Br- (Reduction half Rebection) There is no need to balalance o and H-atom. Balancanga ge electrons in reduction half Reaction by addition of é to left sich. - Braté 3 Br 4. Now to equalize the number of electrons in both the half reach To balance the number of e- in both half leul reactions multiply the Reduction halt reaction by 6. W +3th wo₃ +6 H + +6e- (Brith 3 Br ] X6 6 BritsetW+370 wozt 6H+ +60 41835 acidic medium. this is the final balanced equation in 6 Br+w+840 18 Br + wozt 6H+

The reducing agent The oxidation state w of tungust en in wog is +6 2. Balance in Basic solution. The oxidizing agent is cloz The oxidation state of carbon in HCHO loss of e- (oxidation) is o I 13 HCHO + c 2+1 -2+1 di 0712 +4 (rea Gain of te CH₂OH oxidized toucho. lattia) (Reduction) 2. ts +4 CHOH HCHO oxidation helf reaction). (103 (102 (Reduction half reaction) HoH is oxidased to HICHO coxidation number increased from 2 to tos & coz is reduced to cl0₂ (oxidation number decreased from ts totu). 3. Balancing oxidation half reaction. CHLOH HCHO noe is no need to Boulance atoms other than t fo. Also no need. to balance o.atom. Balance H by adding sufficient number of the molecull tor! each atoms needed on the side deficient in Hydrogenaten. Now add an equal no. of oH l'ons to opposite side eth OH HCHO+2ho motho molecule as ar atomis 1 ress. AZ OH+2014 HCHO+2140 (as tho molekule added 2 or added on oprosiou side Balance charge by addition of eronotia addition of e to oxidation half reaction on Right side and Reduri -on half reaction by on left side.

CHLOH + 204 HCHO +240 the This is last thoH(-23 to MCHoltes oxidation haff realtion (1) Balancing Reduction halt realtion. clo; coz Bij et atoms are balanced (11) Balance oxygen atom by oH (103 C10₂ + OH (to balancero-aton) (11) Balance Hydrogen atoms by to und equal no. of ot on opposite Sicle cloz + tho (102+ oH + oH coz tho - 10 + 204 (iv) Add of te addition of 1 e culto (103 + tho (102+204 (coz € +5 to ClO2 (+4) tler This balanced reduction half reaction

4. Multiply the Reduction half reaction by 2 and then add the CHHOH +20H HCHO +240 +2e- [eloz t tot le 202 + 200] X2 2010 +24otzé (10₂+ you CHOH + 204 +24103 +2440 HCHO+240 + ClO2 + you balance equation : HOH +21lon - HCHO+ cozt 2000