JEE Advanced 2013 Paper 2 Chemistry Q19

Chemistry · Electrochemistry

Problem

The standard reduction potential data at

25\,^{\circ}\mathrm C

are

E^\circ(\ce{Fe^3+,Fe^2+})=+0.77\,\mathrm V

E^\circ(\ce{Fe^2+,Fe})=-0.44\,\mathrm V

E^\circ(\ce{Cu^2+,Cu})=+0.34\,\mathrm V

E^\circ(\ce{Cu+,Cu})=+0.52\,\mathrm V

E^\circ[\ce{O2(g)+4H+ +4e- -> 2H2O}]=+1.23\,\mathrm V

E^\circ[\ce{O2(g)+2H2O+4e- -> 4OH-}]=+0.40\,\mathrm V

E^\circ(\ce{Cr^3+,Cr})=-0.74\,\mathrm V

and

E^\circ(\ce{Cr^2+,Cr})=-0.91\,\mathrm V

. Match List I with List II. Column I — P:

E^\circ(\ce{Fe^3+,Fe})

; Q:

E^\circ(\ce{4H2O <=> 4H+ +4OH-})

; R:

E^\circ(\ce{Cu^2+ + Cu -> 2Cu+})

; S:

E^\circ(\ce{Cr^3+,Cr^2+})

Column Ii — 1:

-0.18\,\mathrm V

; 2:

-0.40\,\mathrm V

; 3:

-0.04\,\mathrm V

; 4:

-0.83\,\mathrm V