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प्रश्न
Write the Nernst equation and emf of the following cell at 298 K:
\[\ce{Sn_{(s)} | Sn^{2+} (0.050 M) || H^+ (0.020 M) | H2_{(g)} (1 bar) | Pt_{(s)}}\]
उत्तर
The cell reaction is as follows:
\[\ce{Sn_{(s)} + 2H^+ (0.020 M) -> Sn^{2+} (0.050 M) + H2 (1 bar)}\]
Hence n = 2,
According to this, the Nernst equation will be as follows:
`"E"_"cell" = ("E"_("H"^+//1/2"H"_2)^Θ - "E"_("Sn"^(2+)//"Sn")^Θ) - 0.059/2 log_10 (["Sn"^(2+)] xx "pH"_2)/(["H"^+]^2)`
∴ `"E"_"cell" = [0 - (-0.14)] - 0.059/2 log_10 (0.050 xx 1)/(0.020)^2`
= `0.14 - 0.059/2 log_10 125`
= 0.078 V
संबंधित प्रश्न
Calculate the e.m.f. of the following cell at 298 K:
Fe(s) | Fe2+ (0.001 M) | | H+ (0.01 M) | H2(g) (1 bar) | Pt(s)
Given that \[\ce{E^0_{cell}}\] = 0.44 V
[log 2 = 0.3010, log 3 = 0.4771, log 10 = 1]
Write the Nemst equation and explain the terms involved.
Calculate emf of the following cell at 298 K:
Mg(s) | Mg2+(0.1 M) || Cu2+ (0.01) | Cu(s)
[Given Eocell = +2.71 V, 1 F = 96500 C mol–1]
For a general electrochemical reaction of the type:
\[\ce{{a}A + {b}B ⇔ {c}C + {d}D}\]
Nernst equation can be written as:
Calculate ΔrG0 and log Kc for the following cell:
\[\ce{Ni(s) + 2Ag^+(aq) -> Ni^{2+}(aq) + 2Ag(s)}\]
Given that \[\ce{E^0_{cell}}\] = 1.05 V, 1F = 96,500 C mol–1.
The cell potential for the given cell at 298 K Pt | H2 (g, 1 bar) | H+ (aq) | | Cu2+ (aq) | Cu(s) is 0.31 V. The pH of the acidic solution is found to be 3, whereas the concentration of Cu2+ is 10-x M. The value of x is ______.
[Given: (\[\ce{E_{Cu^{2+}/Cu}}\]) = 0.34 V and `(2.303 " RT")/"F"` = 0.06 V]
Calculate the emf of the following cell at 298 K:
Fe(s) | Fe2+ (0.01 M) | | H+ (1 M) | H2(g) (1 bar) Pt(s)
Given \[\ce{E^0_{cell}}\] = 0.44 V.
Calculate the emf of the following cell at 298 K.
\[\ce{Cu/Cu^{2+}_{(0.025 M)}//Ag^+_{(0.005 M)}/Ag}\]
Given `"E"_("Cu"^(2+)//"Cu")^circ` = 0.34 V, `"E"_("Ag"^+//"Ag")^circ` = 0.80 V
1 Faraday = 96500 C mol -1
Calculate the value of ΔG that can be obtained from the following cell at 298K.
`(Al)/(Al3+) (0.01M) || (Sn^(2+) (0.015M))/(Sn)`
`E^\circ (Al^(3+))/(Al) = -1.66 V; E^\circ (Sn^(2+))/(Sn) = -0.14 V`
Write the Nernst equation and emf of the following cell at 298 K:
\[\ce{Fe_{(s)} | Fe^{2+} (0.001 M) || H^+ (1 M) | H2_{(g)} (1 bar) | Pt_{(s)}}\]
