Question

16.52 Balance each of the following unbalanced equations for cell reactions in acidic conditions, then calculate the standard cell emf, and decide whether the equation is written in the direction of spontaneous reaction. (a) Sn2+ (aq) + Ag(s) > Sn(s) + Ag+ (aq) (b) Al(s) + Sn4+ (aq) → Sn²+ (aq) + AP+ (aq) (C) CIO3+ (aq) + Ce3+ (aq) —Cl(aq) + Ce++ (aq) (d) Cu(s) + NO3 (aq) —> Cu2+ (aq) + NO(g) 16 .1

Answer 1

Steps to find the orication numbers of alom. each and finel the orication and suelenchon half coll meretion. . 9. For each haff cell reachon, balance baloree the atom other than or H. 3. for acidic meelium . for every on o nom add 420 on. The deficient oxygen side. - and then for balancing H atoms 247 I for each molecule of water are added to the opposite srce. and tor alkaline / basic meelium for each o-com add one molecule of H₂o on same side and 204 on the opposste side. 4. Bd then balomee at the charge by adding eo to the side deficient in negative by alang Charge ho then to equalise the number of ef of the two half peaction, multiply both with suitable integer and then add the bith equation.

so we will follow the above steps to : hal once the following equation. (as Smp empot Agis) - Snes) + Agt.. oxidation! Ag Agt C Ağ - Agt té - / atoms an sime on both side, no preseneca 1. of oxygen crtom sme to so only we. have to blo balance the charge by led to the product side) Reduction : snit (aa) snes) is same no numbers of entom out both siche so no need of beilonce . balance the charge. smetyt ze oso(s) —© £!: -0.17 so. to camel caneel the nee electron in . The final net equation we must make. surre both peactant annel papo clued side have same maumbers of eo swi? 0 x 2 + 2 x Ag Agt té 5m2++20 sn (s) sm?* cag] + AGCE) — 2A9Sm(s)

w for The half now from stined and cell pth Ag (5). the table tind the weelnetron potential neaction - Agtapt = Food - 0.9994. Foxis-0.9994. and som? (aq) +24" => smesa de pe = -0-14. El . = Eveel + I oxid LA N fecell = coul4 -0.7994 = 10-9394: co E eu for the above equation is-0-9394.) is negative so AG = net can yo so.. not spontaneous ass auto non spontaneous O AGLo spontaneous . and also remember En is an intensive prooperty so it does not change if we muit multiply with any integer so at Ag Ered is not multiplied by 2 (1) Alcg+ Sn 4* caq) =$? tem + Al 3+ fag) Oxidation Al(s): - Al Pfaglit 3e? .no speel 2 :-1066 foox = 1066. Reduction sn24 - snut +20 0.15 v E predl. a

117 2x iquanton 0 + 24 (alcs) - 3n equation west conj) * 30-) 2 Alis) + 3 594+ (aq) 12a + 2 50 ²t (aq) full = 0.15 + 1066. z counso che ) a co 181 V so spontaneouse I so I o dos coq) + 10 3t cag) or cash + Colt (aq) oxidation ce 3t caq) . cest caqja (83+ (aq) = crying to co - 0 £ °ox = 1061 vi e spade pred uchon ( Audie meedrum Qo, cad) - (og) a add uo - dozcaa) - u Too) + 3H₂O ald Ht on opposite side of the added. + ñ do₂ +64 + + 3 H₂O how balance the charge. Ered 4105 +66+ 6érou + 340 - 0 6 x equation & equation 4 6x (cest caq) cut (ag) te 40,- (aq) + 6H7 +5l 4 +3420

AL om - ceston & 40₃ can) + 6 Heag) 6clean & Yapt 3420 (aq) is the balanced equation. Fan = F 8x + o reed for 403.caq) + 6H*can) + Ge → 059)+3420 Errored = 1047 (Clz (g) + 24 24+ caq)). Fordel. 5136. co (aa) + 6Hf of. be su cag) + 3 H20 Fºz 1047 +1:36 = 2083, So po 24-uu. so . E cell à 2 lofl +2083 so so spontanious to all yo ; AG Lo (d) cues) + NO3-cap) : ?teaq) a Nolg) oxidation cu (3) co the change the balance : culs) - cuit (aq) + 20 charge - cuit (aq) +20 Ereds 0.337 & Fox = -0.337 Reduction parte balance No caq) - NO (9) Nadatom is balanced so now оя 24м Рож

Nos can) - Nocg) No Tan) + H₂O no lg) + 3H2o. Nos cagle - Nolght 2tho add HT Noz(aq) + 4H+ - NOCg) + 2H20 so Nos + 449%+46 NO(g) + 24120:-* .. 0x2 + ① a la rue (s) - curtean + 22-). Nos - +44* Hye - Nol g) + 21,0 ) zu (s) + Wozcan) + 4H plag) —2curtcaq) No 69 ) + 2 H20 - equation ... is the balanced fel. Elle + femeli for 0.334. Epped in 0.96. So flees 2 -0.337 +0.96 = 6.623 fruito so AGLO so spontaneous,