Question

Balance each of the following equations according to the half-reaction

method:
1) CN-​ ​(aq) + ClO​2(​ aq) —> CNO-​ ​(aq) + Cl-​ ​(aq) (in acid)

2) MnO​4-​ ​(aq) +NO​2-​ ​(aq) → MnO​2(​ s) + NO3​ -​ ​(aq) (in base)

3) In which species does nitrogen have the highest oxidation number? a) NaNO​3

b) HNO​2 c) NO​2-​ d) NH​3 e) N​2

Answer 1

cont clg (No tell H₂H+ oto CN I el ICNO I al *=-1 X+2(-2)0 x+(-2)27 x=+4 xat ! oxidation half reaction | Reductim half geaction CN (NO eloch balance oxygen atoms by adding to CNT HOCNO co ci't sho balance hydrogen atoms | by adding th CN +40 (NO + 2H+ do + 4H+ _></'+ LHO balance charge by adding electrons CN THO JENOT 24h tzel (10 + 4 H t t se cilt to multiply with suitable integer to equalise number of electrons | JX2 Ox5 5cN'T SHO)5CN5+10 47/2c/o 48h + 100lmy załth to toe Linn add both SCM + so 5 CNO + oHt+ 10 et 2dlo + BAT + 1g - 2ci+ 440 5cNlt Hot Lalo - 5CNO + 2H² + zcil s

ning + Ng" ning & Not AHO o- oH - x+41-2)=- x=+7 +2(-2)=-1| X + 2(-2) 20 x+3 (-2) = x=+3 x=+4 Txas oxidation half reaction | Reduction half reaction NO, NO mno mng balance oxygen atoms by adding oft ima balance hydrogen atomo by adding to Repeat this up to balanceing of all oxygen and hydrogen atoms . No + 2 ot Ng + Ho no + 2H20 -) ming +4 0h balance charge by adding elections No + 2 077 __ NO + Hotel / mno + 2H0+ 3€/-) ming +404' multiply with suitable integer to equalise number of electrons. DX2 ox3 3ng '+ Gott - 3 No, +350 (2 mno + 4H0+ bê - aming + P04-1 add both 300 + son ~ 399 + 3thot Gel zmng'+ ' 40 + 68 – 2 mnout 8 OH" zmnga'+3w¢.'+ 40 => 2m19_43N9+'+ 204"}

HNO, NANO ++&+31-2)=0 Itx-620 x-520 x=+5 1+ X+2(-2)=0 1+X-420 2-320 x=+3 NH3 X+3(+1) 20 x=-3 NO- X+2(-2) = -1 x-4201 x=.4-1=+3 2x=0 o in N NaN : have highest oxidation state