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प्रश्न
Write the Nernst equation and emf of the following cell at 298 K:
\[\ce{Pt_{(s)} | Br^- (0.010 M) | Br2_{(l)} || H^+ (0.030 M) | H2_{(g)} (1 bar) | Pt_{(s)}}\]
उत्तर
The cell reaction is as follows:
\[\ce{2Br^- (0.010 M) + 2H^+ (0.030 M) -> Br2_{(l)} + H2 (1 bar)}\]
Hence, n = 2,
According to the Nernst equation for the cell, the emf is given below:
`"E"_"cell" = ("E"_("H"^+//1/2"H"_2)^Θ - "E"_(1/2"Br"_2//"Br"^-)^Θ) - 0.059/2 log_10 ("pH"_2)/(["Br"^-]^2["H"^+]^2)`
= `[0 - (+ 1.08)] - 0.059/2 log_10 1/((0.010)^2 (0.030)^2)`
= `-1.08 - 0.059/2 log_10 (1.11 xx 10^7)`
= −1.08 − 0.208
= −1.288 V
संबंधित प्रश्न
Calculate the e.m.f. of the following cell at 298 K:
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[log 2 = 0.3010, log 3 = 0.4771, log 10 = 1]
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Given \[\ce{E^0_{cell}}\] = 0.44 V.
Calculate the emf of the following cell at 298 K.
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Write the Nernst equation and emf of the following cell at 298 K:
\[\ce{Fe_{(s)} | Fe^{2+} (0.001 M) || H^+ (1 M) | H2_{(g)} (1 bar) | Pt_{(s)}}\]
Write the Nernst equation and emf of the following cell at 298 K:
\[\ce{Sn_{(s)} | Sn^{2+} (0.050 M) || H^+ (0.020 M) | H2_{(g)} (1 bar) | Pt_{(s)}}\]