Consider these generic half-reactions. Half-reaction E° (V) X+(aq)+e−⟶X(s) 1.52 Y2+(aq)+2e−⟶Y(s) −1.17 Z3+(aq)+3e−⟶Z(s) 0.84 Identify the strongest oxidizing agent. X+ Y2+ X Z Y Z3+ Identify the weakest oxidizing agent. X Z3+ Y Y2+ Z X+ Identify the strongest reducing agent. Z3+ X+ Y2+ Z X Y Identify the weakest reducing agent. Y Z X+ X Y2+ Z3+ Which substances can oxidize Z ?

     strongest oxidizing agent:

     weakest oxidizing agent:

     strongest reducing agent: Y

     weakest reducing agent: X

      will oxidize Z

Explanation:

The higher the reduction potential of a species, higher will be the tendency to consume electrons from another species. Hence higher will be the oxidizing power of it's oxidized form and lower will be the reducing power of it's reduced form.

Alternatively, higher reduction potential value suggests that the oxidized form of the species acts as a stronger oxidizing agent and the reduced form of the species acts as a weaker reducing agent.

Order of reduction potential:

So, strongest oxidizing agent:

     weakest oxidizing agent:

     strongest reducing agent: Y

     weakest reducing agent: X

As reduction potential of the half cell is higher than the reduction potential of the half cell therefore will oxidize Z into and itself gets converted into X.

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml