Question

Reaction 1 and 2: Oxidizing agents. N20 > Au+ ; reducing agents: Au > N2 Reaction 1 and 3: Oxidizing agents: Au+ A > Cr3+ 4; reducing agents: Cr > Au Reaction 2 and 3: Oxidizing agents: N20 4 > Cr3+ ; reducing agents : Cr > N2 Overall ranking: Oxidizing agents : [Select] > [Select] Select] ; reducing agents : [Select] [Select] > [Select]

i just need the last thing using what I've already put in. Here was the question attached to it:

Answer 1

Er represents reductional potential - Ex represents oxidation potential o oxidizing Agents CoA) are themselves highly reducible .., stronger oA will have higher EUR values o Reducing Agents (RA) are themselves wighly oxidisable s stronges RA will have higher Eox values Reaction @ and o Au* laq) + en - Au (s) Ee of Aut = 1.69 V Aucs) ; Aut (aq) te toe of Au = - 169 Y NO (g) at 2147 lap) +26 Nalg) + Hole) EUR of Nho = 1770 Nz (g) + H2011 - N2OCg) +243 Coq) +2= Eox of N2 = -1.777 TO.A. → Neo > Aut IR.A. Au > Na t o Reaction ♡ & € Aut loq) te ~ An (s) ER of Aut = 1.69 v Au Cs) — Aut caq) +e- for of Au = -1.65 V cm 3+ (aq) + Be + Cris) Ee of Co2+ = -0.74 V Cr(s) — (x3+ (aq) + 3e- fox of (r = 0.74 V Io.A. - Aut> (+34 | R.A. ~ cr > Au Reaction @ x ② fe of N20 = 1.774 ER = -0.74 v (of <p3+) Box of Ny = -1.774 Eux of co = 0.74 V

0. A. → R.A. - Ngo > Co > Cy3+ N₂ Combining 0.② 8 ② , we have over all order 0.A. - N2O > Aut > (7 37 R.A. y cr> Au JN