Consider the following cell reaction
`Zn +2Ag^(o+)rarr Zn ^(2+)+2Ag.`
Given that
`E^(c-)._(Zn^(2+)(aq)Zn(s))=-0.76V`
`E^(c-)._(Ag^(o+)(aq)Ag(s))=0.80V`
`a.` Calculate the standard `EMF` fo the cell.
`b.` Which ion is more powerful oxidizing agent ?
`c.` Which metal is more powerful reducing agent ?

Explanation

`a.` The cell reaction is written ini two half reactions are as`:`
`Zn(s) rarr Zn^(2+)(aq)+2e^(-) (` oxidatoin half reaction `)`
`2Ag^(o+)(aq)+2e^(-) rarr 2Ag (` reduction half reaction `)`
Here `E^(c-)` is given in standard reduction potential.
`E^(c-)._(Ag^(o+)|Ag)gtE^(c-)._(Zn^(2+)|Zn).`
Therefore, `Ag^(o+)|Ag` will undergo reduction and will act as cathode and `Zn|Zn^(2+)` will undergo oxidation and will act as anode.
`:. E^(c-)._(cell)=(E^(c-)._(reduction))_(cathode)-(E^(c-)._(reduction))_(anode)`
`=0.80-(-0.76)=1.56V`
Alternatively
Since standard reduction potential of `Ag^(o+)|Ag` is greater than standard reduction of `Zn^(2+)|Zn,` thus, `Zn|Zn^(2+)` will undergo oxidation and `Ag^(o+)|Ag` will undergo reduction .
`E^(c-)._(o x i d ation Zn|Zn^(2+))=-E^(c-)._(reduction(Zn^(2+)|Zn))`
`=-(0.76)=0.76V`
`E^(c-)._(cell)=E^(c-)._(o x i dation)+E^(C-)._(reduction)`
`=0.76+0.80=1.56V`
`b.` Cell rection `:`
`2Ag^(o+)(1M)+Zn(s) rarr Zn^(2+)(1M)+2A(s)`
According to the cell reaction, `Ag^(o+)` ion is reduction to `Ag` and hence `Ag^(o+)` ion is a stronger oxidizing agent.
`c.` According to the cell reaction, `Zn(s)` is oxidized to `Zn^(2+)` ions and hence `Zn` metal is a stronger reducing agent.