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For the cell, Pt| Cl(2)(g, 0.4 bar) | Cl...

For the cell, Pt| Cl_(2)(g, 0.4 bar) | Cl^(-)(aq, 0.1 M) || Cl^(_1)(aq, 0.01 M)| Cl_(2) (g, 0.2 bar) | Pt`
The measured potential at 298 K is ___________ V. [use `(2.303RT)/(F) = 0.0591`]

Text Solution

Verified by Experts

The correct Answer is:
`-0.05`
`E=E^(0)-(0.0591)/(2)"log"((0.1)^(2)(0.2))/((0.4)(0.01)^(2))`, where `E^(0)=0`
solving `E=-0.05`
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Knowledge Check

  • Mark the correct Nernst equation for the given cell. F_((s))|Fe^(2+)(0.001 M) ||H^(+)(1M) | H_(2(g)) ( 1 bar ) | Pt_((s)) is

    A
    `E_"cell"=E_"cell"^@-"0.591"/2 "log" ([Fe^(2+)][H^+]^2)/([Fe][H_2])`
    B
    `E_"cell"=E_"cell"^@-"0.591"/2 "log" ([Fe][H^+]^2)/([Fe^(2+)][H_2])`
    C
    `E_"cell"=E_"cell"^@-"0.0591"/2 "log" ([Fe^(2+)][H_2])/([Fe][H^+]^2)`
    D
    `E_"cell"=E_"cell"^@-"0.0591"/2 "log" ([Fe][H_2])/([Fe^(2+)][H^+]^2)`