Question

Calculating the cell emf for Nonstandard condition : A galvanic cell is constructed at `298 K` as follows. One half-cell consists of a chlorine gas electrode (i.e. `Cl_(2)//Cl^(-)` couple) with the partial pressure of `Cl_(2) = 0.100` bar and `C_(CI-) = 0.100 M`. The other half-cell involves the `MnO_(4)^(-)//Mn^(2+)` couple in acidic solution with `C_(MnO_(4)^(-)) = 0.100 M, C_(Mn^(2+)) = 0.100 M`, and `C_(H+) = 1.00 M`. Apply the Nernst equation to datermine the cell potenital for this cell.
Strategy: First datermine the overall cell reaction and then calculate the standard cell potential `(E_("cell")^(@))` from the standard electrode potenital in Table 3.1. then use the Nernst equation to find the cell potential `(E)` under cited conditions.

Answer 1

The `MnO_(4)^(-)//Mn^(2+)` half-cell reaction has the more positive reduction potential `(+ 1.507 V)`, so we write it first. Then we write the `Cl_(2)//Cl^(-)` half-cell reaction as an oxidation, balance the electron transfer, and add the two half-reactions and their standard potentials to obtain the overall cell reaction and its `E_("cell")^(@)`
`{:(2[MnO_(4)^(2-)+8H^(+)+5e^(-)hArrMn^(2)+4H_(2)O]" "+1.507V),(5[2Cl^(-)hArrCl_(2)+2e^(-)]" "-1.360V),(bar(2MnO_(4)^(-)+16H^(+)+10Cl^(-)hArr 8H_(2)O "]"+5Cl_(2)E_("cell")^(@)=+0.147 V)):}`
In the overall reaction, `n = 10`, we then apply the Nernst equation to this overall reaction by subsitution suitable concentration and partial pressure values. Bacause `Cl_(2)` is a gaseous substance, its term in the Nernst equation involves its partial pressure, `P_(Cl_(2))`.
`E_("cell") = E_("cell")^(@)-(0.592 V)/(n) "log"(C_(Mn^(2+))^(2)P_(Cl_(2))^(5))/(C_(MnO_(4)^(-))^(2)C_(H+)^(16)C_(Cl^(-))^(10))`
` = 0.147 V - (0.0592 V)/(10) "log" ((0.100)^(2)(0.100)^(5))/((0.100)^(2)(0.100)^(16)(0.100)^(10))`
` = 0.147 V - (0.0592 V)/(10) "log" (1.00 xx 10^(21))`
` = 0.147 V - (0.0592 V)/(10) "log" (21.00)`
`= 0.017 V`