Advertisements
Advertisements
Question
Arrange the following metals in the order in which they displace each other from the solution of their salts.
Al, Cu, Fe, Mg and Zn.
Advertisements
Solution
Mg, Al, Zn, Fe, Cu
APPEARS IN
RELATED QUESTIONS
What is the SI unit tor electrochemical equivalent?
If one mole electrons is passed through the solutions of CrCl3, AgNO3 and NiSO4, in what ratio Cr, Ag and Ni will be deposited at the electrodes?
Zinc can be coated on iron to produce galvanized iron but the reverse is not possible. It is because ____________.
Cell equation: \[\ce{A + 2B^- -> A^{2+} + 2B}\]
\[\ce{A^{2+} + 2e^- -> A}\] E0 = +0.34 V and log10 k = 15.6 at 300 K for cell reactions find E0 for \[\ce{B^+ + e^- -> B}\]
Describe the construction of Daniel cell. Write the cell reaction.
Two metals M1 and M2 have reduction potential values of −xV and +yV respectively. Which will liberate H2 and H2SO4.
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
Write a note on sacrificial protection.
Use the data given in below find out which option the order of reducing power is correct.
`"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
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
Use the data given in below find out the most stable oxidised species.
`E^0 (Cr_2O_1^(2-))/(Cr_(3+))` = 1.33 V `E^0 (Cl_2)/(Cl^-)` = 1.36 V
`E^0 (MnO_4^-)/(MN^(2+))` = 1.51 V `E^0 (Cr^(3+))/(Cr)` = – 0.74 V
The quantity of charge required to obtain one mole of aluminium from Al2O3 is ______.
A galvanic cell has electrical potential of 1.1V. If an opposing potential of 1.1V is applied to this cell, what will happen to the cell reaction and current flowing through the cell?
Given the data at 25°C
\[\ce{Ag + I- -> AgI + e-}\]; E° = – 0.152 V
\[\ce{Ag -> Ag+ + e-}\]; E° = – 0.800 V
The value of log Ksp for AgI is ______.
Cell reaction is spontaneous when
If the half-cell reaction A + e– → A– has a large negative reduction potential, it follow that:-
If the value of Ksp for Hg2Cl2 (s) is X then the value of X will be ____ where pX = - log X.
Given:
\[\ce{Hg2Cl2 + 2e- -> 2Hg(l) + 2Cl-}\], E° = 0.27 V
\[\ce{Hg+2 + 2e- -> 2Hg(l)}\] E° = 0.81 V
Which of the following is incorrect?
What should be the signs (positive/negative) for \[\ce{E^0_{cell}}\] and ΔG0 for a spontaneous redox reaction occurring under standard conditions?
What are electrochemical reactions?
