Question

HELP!! do not understand how to do steps a-e

21. Here is another opportunity to balance a redox equation. In aqueous solution containing hydrochloric acid, methanol, CH,OH, reacts with potassium dichromate (dichromate Cr2O72) to produce formaldehyde, CH2O, and chromium(III) chloride. Note that both methanol and formaldehyde are covalent molecules and, thus, do not dissociate into ions. a. Write the unbalanced equation showing the appropriate chemical formulas and subscripts for the two reactants and two products ignoring the acid and water for now: Helaa + CH ₂ OH at KCR of CH 2 Oth b. Remove the spectator ions (potassium and chloride), ignore the acid for now, and re-write the ionic equation showing only those species undergoing oxidation and reduction (i.e., the skeleton equation): ㅓ ㅓ 에 내 니 애t 여 대가 e. Using the half reaction method, balance the redox equation, but save reinserting the spectator ions and hydrochloric acid for the last step: CH3OHA (3) croqr (crow- d. Reinsert the spectator ions and assume that hydrochloric acid is used to make the acidic solution. Show the balanced redox reaction in molecular form: e. Identify the two reactants as either the oxidizing or reducing agent. Just label and circle which is which

Answer 1

page-1 (a) en on (ad) + K, ch of (adh This is the unbalanced equation for and two products ignoring the acid 4 - CH₂O (98) + crely the two reactanto and water. two reactanto 1967 1 1 CH₂ OH (ad) +(2kt crot (ad) - CH₂O (96) + st y spectator crotiad) ions fice Jad o by removing spectator ions, and rewrite the sduation spectator in CH₂OH (ad) + crot (ad) - CH₂O lad) + cast (ado / 3 + 9 3X1 CH ₂ OH lad) - Cho (ad] + 2Httre ) half charge = 0 . Charge = 0 reaction Crot (ad) + 14 ht to - 2 Gr (ad) + 74₂0 2nd half charge = +6 charge =+6 heaction now no. of electrons in both half reaction should be equal, so multiply ed lip by 3 , and rewrite it in 3 (CH₃OH) (9) - 3 (Ch₂0) (ad) + out to go from both للمه - sides. Mon (ad) + (22 Oz t 14 h → 3(CH₂ of lad) + 6MT + 20 +7H2O equation (hi)

So, we no. of rewrite the can H equation lies after page-2 balancing the | 3 (CH₂ ou ) lad) + Cr2O Can t gutes 3+ 4 This is balanced 3 (Cų0) aq + 268 1967) + 7H20 (49) by half reaction chemical method. equation A d now, on reinserting the spectator ions. - 3(eu ou) (ad) + 2 kt ca o 1997 + 84t 1997 - 3 (CH2O) (ad) + 2003 to ce can +7420 - rewriting in molecular form, | 3 (ch, on) cad) + K, Co2 Oz (ad) + 8H (98) 3 (CH₂) (ad) + 2 cr clg (as) + 7H20 (and) molecular form → balanced redox reaction in CHyon (ad ] + KC & Of 1997 r CH₂O (ad) , reduced oxidised Hydrogen (is removed So it is reducing so it is oxidising agent agent So, finally I Choon k(22 Oq - Reducing agent oxidising agent