Question

For each reaction below, identify the atom oxidized, the atom reduced, the oxidizing agent, the reducing agent, the oxidation half reaction, the reduction half reaction, and then balance the equation by the method of oxidation-reduction showing all electrons transfers.

KMnO4 NaClH2SO4 à Cl2 K2SO4 MnSO4 H20 Na2SO4 5. Cr2O3 6. K2CR2O7 H20 + S à SO2 + KOH KCIO3 C12H22011 à KCI + H20 + CO2 7 H2C204+K2MnO4 à CO2 K20 Mn203 H20 8. 9. Mn(NO3) 2 NaBiO3 HNO3 à HMnO4 Bi(NO3)3+ NaNO3 H20 H2C204+ KMnO à CO2 + K20 Mn2O3+ H20 10

Answer 1

+I +3-2 +6-2 +4-2 + 1 +1-2 +3 -2 K2 Mady t-2 + + Mn3 + H2O Oxidation Half 202 + 2e Rection Reduction Half 2 K2 Mnoy + be reaction Atom Oxidizied Atom Reduced Mn Reducing agent H2C204 Oxidizing agent K2Mnoy H2 C204 20, +2e +2H+ y3 2K2 Mnoy t 6e 6Ht Mn203 t 2K20 3H20 X1 Знас204 +2K2MmDy + 6e + 6Ht Mm t be +2K20 + 6H + зHо Bularced Equation ia 3 H2C204 + 2K2ny 6 Mm203 2K 3H,0

BALANCING OXIDATION-REDUCTION REACTIONS BY OXIDATION NUMBER METHOD In a balanced chemical reaction, total increase in oxidation number must be equal to the total decrease in oxidation number. This equivalence provides the basis for balancing redox reactions The general procedure involves the following steps Assign oxidation number to the atoms in the equation and find out which atoms are undergoing oxidation and reduction. Write separate equations for both. Find the change in oxidation number in each equation. Make the change equal in both the equations by multiplying with suitable integers. Add both the equations. Complete the balancing by inspection In ionic equations the net charges on both sides of the equation must be exactly the same. Use H+ ions in acidic reactions and OH- ions in basic reactions to balance the charge and number of hydrogen and oxygen atoms Following these steps we can balance the given equation

5J + tb-2 + + t7-2 + t6 -2 +H2504 +2 tb -2 + t6-2 +-) Nace t-2 KMnoy + Na304 Oxidation Half 2Nace Rection Reduction Half + Se Mmsoy KMnoy reaction Atom Reduced Atom Oxidizied Mn Oxidizing agent KMnoy Reducing agent Nacl Jx5 Naz soy t 2e t 2H 2 Nau +HSOy u, + 2KMMO t 3H2SO4 t1oe 1OHt 2 Mnsoy K2s04 0 xI + 2KMmoy + 8H,SOy + 10e-+ 1OHT つ 502 + 2Mnsoy + 5 Nas04 tK2SOy t 10 8H,O10H Balanud Equaabn ia + 8 H2SO4 502 t K2s0y 2 Mmioy + 2 KMnOy + 1o Nace 5 Na2SOy 8 H0

t1 t6 -2 +-2 O + 2 t KOH t 3 +4-2 3-2 Oxidation Half S So2 +4e Rection Reduction Half K2 cr20 be Cr203 reaction Atom Oxidizied Atom Reduced Reducing agent K2C20 Oxidizing agent St 2H,0 SO, H t uH J3 x 2 +6e + 6H r,03 t 2KOH 2H20 +2K Cr204 + 6H20 +12e 12 Ht 3S 3S02 + 2r203 12e +4KOH t 12Ht Чhо Balanud Equation wa +2K2C207 t 220 2r +4KO 3502 3S

+4-2 +1 -2 Ot 2 + C12H22011 +- +15-2 + H0 +t ке 2 Oxidation Half 12 CO 48e C12H2201 Rection Reduction Half KCe Kuo3 t be reaction Atom Reduced Atom Oxidizied C12H2201 Oxidizing agent Reducing agent коз + 13 H20 C12M220 12C2 t48e +48Ht +3H20 KD be +6H+ Kce C12H2201 8 KCD, 13 H0+ 48e 48 H 12C0, + &Kce 48e + 24 H20 +L48 H Balaned equanon wa 8K.ce +lHO + 8киbg 12 0 C12H22011 +