Question

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.

Answer 1

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.