Question

Can someone explain how to get these answers based on the activity series?

36) Of the reactions below, only is not spontaneous. A) Mg (s)+2HCI (aq)MgCl,(aq) +H,(g) B) 2Ag (s)+2HNO ,(aq) ->2A^NO (aq) +H,(g) C) 2Ni (s)H2SO,(aq) -» Ni,SO4(aq) + H2 (g) D) 2AI (s) 6HB' (aq) ->2AIBT,(aq) +3H2(g) E) Zn (s)+2HI (aq) -»Znl,(aq) +H,(g) Answer: B Diff: 4 Page Ref: Sec.4.4 37) Based on the activity series, which one of the reactions below will occur? A) Zn (s)+Mnl2(aq) ->Znl,(aq) +Mn (s) B) SnCl, (aq) Cu (s)-» Sn (s)+CuCl,(aq) C)2AgNO, (aq) +Pb (s)->2Ag (s) +Pb(NO3)2(aq) D) 3Hg )+2Cr(NO3)3(aq) ->3Hg(NO3)2+2Cr (s) E)3FeBr (aq)2Au (s)->3Fe (s)+2AuBr3 (aq) Answer: C Diff: 4 Page Ref: Sec.4.4

Answer 1

Oxidation potential = -(Reduction potential)

When a reaction occurs we have to see which species is gettig oxidized and which is getting reduced. This is checked by looking at their oxidation states. If oxidarion states is increased in the reaction then we say that species is oxidised. On the other hand if oxidation state reduces in a reaction then that species is reduced.

So oxidation potential of oxidized species plus reduction potential of reduced species gives us overall cell potential . And for a spontaneous reaction this total cell potentail shpulf be positive.

Q1 Based on electrochemical senes, we see whether the reaction ocas om not for a spontaneous reaction t of the cell should be positive. Ecece (Esothote + Exi Jovente Reduction For eg: Agt té - Ag E= 0.80V a Ht té 1H,

. Reduction and Hy so as to of Ag" should occur. should oxidise to Hi get total E positive But in oxidiced Reaction to Agt ® Ag is getting (Eni - O Bov). & Ht te → IH EO Total E: Reaction -0.80V is not spontaneous > Take another Mg + g HQ - exemple. Mall, + H₂ +2 Ploxidebar) + Reducten Englingar 2.37 tilH = 0 Total = (+ve) 2.37V Spontaneous

6 2Ni + H₂SO4 → Nizsoy + H₂ +2 Enitet = 0.26 V (This is oxidation Potental) (This is Reduction Potential) H/H, 50 Total E= positive ① 201 + 6HAN - ZA1Br + 3H₂ oxidation +3 Eoxidation (Allnear) = 1.66V E Reduction (H/F) = 0 Total E Eoxidation et tReductni 1.66U

+2 ② zu + ZHI - Zn + H₂ Cantzitoride 0.76V. ENTH = 0 Total positive E Q2: Similarly Q2 can be attempted

Stronger Oxiding Weaker reducing 2.87 1.78 1.51 1.36 1 33 1.23 1.09 O 80 F:6) +20 2 F (a) H.O.(aq) + 2 H+ (aq) + 2 - 2 H.O() MnO,-) + 8H') + 50 - Mn(q) + 4 HOU! CH ) + 2 - 2C ) Croa ) + 14 Haq) + 6 2 Craq) + 7 HOUD O.() + 4 H aq) + 4 2 H.00) Br(!) + 2e- - 2 Bria) Ag() + Ages) Fo (4) + -Fea) 0.) + 2 H aq) + 2 HO) 1(s) + 2 - 2 Flag) O.(s) + 2 H OM) + 4 4OH ) Cu(a) +20 Cus) Sn(m) + 2 - Snap 2 H() +2 H ) Pb (aq) + 2e- bis) Ni (aq) 2 NI) Cd (aq) + 2 → Cd() Fe + 2 Fels) Zn ) + 2 Zn() 2 H.O(I) + 2 - H() + 2OH(aq) Alaq) + 3e Als) Mg() + 2e Mg(s) Na ) e - Nas) Litlag) + c - Lis) 0.77 0.70 0.14 0.40 034 0.15 -0.13 -0.26 -0.40 -0.45 -0.76 -0.83 -1.66 -2.37 -2.71 -104 Weaker Oxidizing asent Stru reducing

Last image is downloaded from web, gives the series of electrochemical potential values . Note that all these values given are reduction potential values. Oxidation potential values are equal in magnitude but opposite in sign.

If you find any problem in solving Q2,feel free to ask.