Writing the cell reaction from the cell notation:
Given the following shorthand notation
`Pt(s) | Sn^(2+) (aq., 1 M), Sn^(4+)(aq., 1 M) || Fe^(3+)(aq., 1 M),Fe^(2+)(aq., 1 M) | Pt(s)`
Write a balanced equation for the cell reaction and give a brief description of the cell.
Strategy: We can write the overall cell reaction from the cell notation by first writing the approprite half-cell reactions. We can obtain the cell half-reactions simply by reading the short-hand notation. The shorthand notation specifies the anode (on the extreme left), the cathode (on the extreeme right), and the reactants in tghe half-cell compartments. To find the balanced equation for the cell reaction, we first write the approprite half-cell reactions. two half-reactions after multiplying each (if necessary) by an approprite factor so that the electrons will cancel. The result is the cell reaction.

Explanation

Because the anode always appears at the left in the shorthand notation, the oxidation half-reation that occurs in the anode half-cell is
`Sn^(2+)(aq.) rarr Sn^(4+)(aq.) + 2e^(-)`
The reducation half-reaction that occurs in the cathode half cell is
`Fe^(3+)(aq.) +e^(-) rarr Fe^(2+)(aq)`
We multiply the cathode half-reaction by a factor of `2` so that the electrons will cancel when we sum the two half-reactions to give the cell reaction.
`2 xx [Fe^(3+)(aq.) + e^(-) rarr Fe^(2+)(aq.)]`
Summing the half-cell reactions gives the overall cell reaction :
`Sn^(2+)(aq.) + 2Fe^(3+)(aq.) rarr Sn^(4+)(aq.) + 2Fe^(3+)(aq.)]`
The cell consists of a `Pt` wire anode dipping into an `Sn^(2+)` solution such as `Sn(NO_(3))_(2)(aq.)` and another `Pt` wire dipping into a `Fe^(3+)` solution such as `Fe(NO_(3))_(3)(aq.)`. As usual, the anode and cathode half cells must be conneted by a wire a salt bridge containing inert ions.