Converting measured potentials to standard potentials: In investigating the properties of rare and expensive metal such as rhenium `(Z = 75)`, it is both impractical and uneconomical to prepare cells with standard concentrations. The find the standard potential of the `Re^(3+)(aq.)//Re(s)` electrode, the following cell is constructed:
`Pt(s) | Re(s)| Re^(3+)(aq., 0.018 M)|| Ag^(+)(aq., 0.010 M) | Ag(s)`
The potential of this cell is found to be `0.42 V` with the `Re` electrode as the node. Calculate the standard potential of the half reaction
`Re^(3+)(aq.)+3e^(-)hArrRe(s)`
Strategy: Table 3.1 lists the `E_(@)` for the `Ag^(+)(aq.) | Ag(s)` electrode as `0.80`. We can find the `E^(@)` of the other electrode by finding the `E^(@)` of the cell through the application of the Nernst equation.

Explanation

The cell reaction is `Re(s)+3Ag^(+)(aq.)hArr Re^(3+)(aq.)+3Ag(s)`
The two half-cell reaciton are
`{:(Re(s)hArrRe^(3+)(aq.)+3e^(-)),(Ag^(+)(aq.)+e^(-)hArr Ag(s)):}`
We obtain the overall reaction by multiplying the silver half-reaction by three. Thus, `n = 3`. The expression for `Q` is
`Q = (Re^(3+))/(C_(Ag^(+))^(3))`
Now we can us ethe Nernst equation for the standard potential of the cell:
`E_("cell") = E_("cell")^(@) - (0.0592 V)/(n) "log Q"`
`0.42 V = E_("cell")^(@) - (0.0592 V)/(3) "log"(0.0018)/(0.010)^(3)`
or ` E_("cell")^(@) = 0.48 V`
Since
`E_("cell")^(@) = E_("cathode")^(@) - E_("Anode")^(@)`
`0.48 V = (0.80 V) - E_("Anode")^(@)`
or `E_("Anode")^(@) = 0.32 V`
which is the standard reduction potential of the `Re^(3+)(aq.) | Re(s)` half-cell reaction.