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प्रश्न
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.
उत्तर
Fe(s) | Fe2+ (0.01 M) | | H+ (1 M) | H2(g) (1 bar) Pt(s)
\[\ce{E^0_{cell}}\] = 0.44 V
Reaction at anode: \[\ce{Fe -> Fe^{2+} + 2e^-}\]
Reaction at cathode: \[\ce{2H^+ + 2e^- -> H2}\]
Overall: \[\ce{Fe + 2H^+ -> Fe^{2+} + H2}\]
According to the Nernst equation,
E = `"E"_"cell"^0 - 0.0591/"n" log (["Fe"^(2+)])/(["H"^+]^2)`
As n = 2
E = `0.44 - 0.0591/2 log 0.01/1`
= `0.44 - 0.0591/2 (-2 log 10)`
= `0.44 + 2 xx 0.0591/2`
= 0.4991 V
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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:
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[Given Eocell = +2.71 V, 1 F = 96500 C mol–1]
Calculate the value of Ecell at 298 K for the following cell:
`(Al)/(Al^(3+)) (0.01M) || Sn^(2+) ((0.015 M))/(Sn)`
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\[\ce{{a}A + {b}B ⇔ {c}C + {d}D}\]
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Calculate the value of \[\ce{E^\circ}\]cell, E cell and ΔG that can be obtained from the following cell at 298 K.
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