Advertisements
Advertisements
प्रश्न
Complete the following statement by selecting the correct alternative from the choices given:
For a spontaneous reaction ΔG° and E° cell will be respectively:
विकल्प
-ve and +ve
+ve and -ve
+ve and +ve
-ve and -ve
उत्तर
-ve and +ve
APPEARS IN
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]
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]
Calculate the value of Ecell at 298 K for the following cell:
`(Al)/(Al^(3+)) (0.01M) || Sn^(2+) ((0.015 M))/(Sn)`
`E° _(Al^(3+))/(AI)= -1.66 " Volt and " E° _(Sn^(2+)) /(Sn) = -0.14` volt
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.
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 value of \[\ce{E^\circ}\]cell, E cell and ΔG that can be obtained from the following cell at 298 K.
\[\ce{Al/Al^3+ _{(0.01 M)} // Sn^{2+} _{(0.015 M)}/Sn}\]
Given: \[\ce{E^\circ Al^3+/Al = -1.66 V; E^\circ\phantom{.}Sn^2+/Sn = -0.14 V}\]
Write the Nernst equation and emf of the following cell at 298 K:
\[\ce{Fe_{(s)} | Fe^{2+} (0.001 M) || H^+ (1 M) | H2_{(g)} (1 bar) | Pt_{(s)}}\]
Write the Nernst equation and emf of the following cell at 298 K:
\[\ce{Sn_{(s)} | Sn^{2+} (0.050 M) || H^+ (0.020 M) | H2_{(g)} (1 bar) | Pt_{(s)}}\]
Write the Nernst equation and emf of the following cell at 298 K:
\[\ce{Pt_{(s)} | Br^- (0.010 M) | Br2_{(l)} || H^+ (0.030 M) | H2_{(g)} (1 bar) | Pt_{(s)}}\]
