Advertisements
Advertisements
प्रश्न
For the cell \[\ce{Mg_{(s)}|Mg^{2+}_{( aq)}||Ag^+_{( aq)}|Ag_{(s)}}\], calculate the equilibrium constant at 25°C and maximum work that can be obtained during operation of cell.
Given: \[\ce{E^0_{{Mg^{2+}|Mg}}}\] = −2.37 V and \[\ce{E^0_{{Ag^{+}|Ag}}}\] = 0.80 V
Advertisements
उत्तर
oxidation at anode
\[\ce{Mg -> Mg^{2+} + 2e^-}\]..........(1) `("E"_"ox"^0)` = 2.37 V
Reduction at cathode
\[\ce{Ag^+ + e^- -> Ag}\] ......(2) `("E"_"red"^0)` = 0.80 V
∴ `"E"_"cell"^0 = ("E"_"ox"^0)_"anode" + ("E"_"red"^0)_"cathode"`
= 2.37 + 0.80
= 3.17 V
Overall reaction
Equation (1) + 2 × equation (2)
\[\ce{Mg + 2Ag^{2+} -> Mg^{2+} + 2Ag}\]
∆G° = –nFE°
= –2 × 96500 × 3.17
= –611810 J
∆G° = –6.12 × 105 J
W = 6.12 × 105 J
∆G° = –2.303 RT log Kc
log Kc = `(6.12 xx 10^5)/(2.303 xx 8.314 xx 298)`
Kc = Antilog of (107.2)
APPEARS IN
संबंधित प्रश्न
Define the following term:
Fuel cell
Define cathode
Use the data given in below find out the most stable ion in its reduced form.
`"E"_("Cr"_2"O"_7^(2-)//"Cr"^(3+))^⊖`= 1.33 V `"E"_("Cl"_2//"Cl"^-)^⊖` = 1.36 V
`"E"_("MnO"_4^-//"Mn"^(2+))^⊖` = 1.51 V `"E"_("Cr"^(3+)//"Cr")^⊖` = - 0.74 V
For the given cell, \[\ce{Mg | Mg^{2+} || Cu^{2+} | Cu}\]
(i) \[\ce{Mg}\] is cathode
(ii) \[\ce{Cu}\] is cathode
(iii) The cell reaction is \[\ce{Mg^+ Cu^{2+} -> Mg^{2+} + Cu}\]
(iv) \[\ce{Cu}\] is the oxidising agent
Consider the following diagram in which an electrochemical cell is coupled to an electrolytic cell. What will be the polarity of electrodes ‘A’ and ‘B’ in the electrolytic cell?
Which of the following statements about galvanic cell is incorrect
Cell reaction is spontaneous when
If the half-cell reaction A + e– → A– has a large negative reduction potential, it follow that:-
Which of the following is incorrect?
What is an electrochemical cell? What does it consist of?
