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प्रश्न
Write the Nernst equation and emf of the following cell at 298 K:
\[\ce{Fe_{(s)} | Fe^{2+} (0.001 M) || H^+ (1 M) | H2_{(g)} (1 bar) | Pt_{(s)}}\]
उत्तर
The cell reaction is as follows:
\[\ce{Fe_{(s)} + 2H^+ (1 M) -> Fe^{2+} (0.001 M) + H2 (1 bar)}\]
Hence, n = 2,
The Nernst equation for the emf of this cell will be as follows –
`"E"_"cell" = ("E"_("H"^+//1/2"H"_2)^Θ - "E"_("Fe"^(2+)//"Fe")^Θ) - 0.059/2 log_10 (["Fe"^(2+)] xx "pH"_2)/(["H"^+]^2)`
∴ `"E"_"cell" = [0 - (-0.44)] - 0.059/2 log_10 (0.001 xx 1)/(1)^2`
= `0.44 - 0.059/2 log_10 10^-3`
= `0.44 - 0.059/2 xx (-3)`
= 0.529 V
संबंधित प्रश्न
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