The spontaneous redox reaction`//s` among the follow is`//` are
`(a)` `2Fe^(3+)+Fe rarr 3Fe^(++)`
`(b)Hg_(2)^(++) rarr Hg^(++)+Hg`
`(c )` `3AgCl+NO+2H_(2)O rarr 3Ag+3Cl^(-)+NO_(3)^(-)+4H^(+)`
Given that
`E_(Fe(+++)//Fe^(++))^(.)=0.77V" "E_(Fe^(++)//Fe)^(.)=-0.44V`
`E_(Hg^(+-)//Hg)^(.)=0.85V" "E_(Hg^(++)//Hg_(2))^(.)=0.92V`
`E_(AgCl//Ag)^(.)=0.22V " "E_(NO_(3)//NO)^(.)=0.96V`

Options

A. `a`
B. `a,b,c`
C. `a,b`
D. `a,c`

Answer

1

Explanation

Only for this reaction `E^(0)` will come out to be positive, calculate using relation
`DeltaG^(@)=DeltaG_(1)^(@)+DeltaG_(2)^(@)" "`and `" "DeltaG^(@)=-nFE_(cell)^(@)`