Electrochemistry. Flashcards
(12 cards)
Can absolute electrode potential of an electrode be measured?
No, the absolute potential of an electrode can not be measured because the half-cell containing a single electrode cannot work on its own. It can work only in combination with another half cell.
Can E0cell or ΔrG0 for a cell reaction ever be equal to zero?
No, it cannot be equal to zero for a cell reaction proceeding in a particular direction (forward or backward direction)
Under what conditions is E0cell = 0 and ΔrG0 = 0 ?
At the condition of equilibrium, E0cell = 0 and ΔrG0 = 0
What does the negative sign in the expression E0Zn2+ /Zn = – 0.76 V means?
It implies that Zn is more reactive than hydrogen or it is a stronger reducing agent. In a cell containing zinc electrode and standard hydrogen electrode present in two half cells, zinc will be oxidized to Zn2+ ions while H+ ions will get reduced to hydrogen.
Aqueous copper sulphate solution and aqueous silver nitrate solution are electrolysed by 1 ampere current for 10 minutes in separate electrolytic cells. Will the mass of copper and silver deposited on the cathode be same or different ? Explain your answer.
The mass of copper and silver deposited on the cathode in the two electrolytic cells will not be the same.
According to second law of Electrolysis.
(Mass of Cu deposited /mass of Ag deposited) = (equivalent mass of Cu/equivalent mass of Ag)
= 31.75/ 108.
Depict the galvanic cell in which the cell reaction is : Cu + 2Ag+ → 2Ag + Cu2+.
In the cell reaction: Cu + 2Ag+ → 2Ag + Cu2+
At anode : Cu → Cu2+ + 2e
At Cathode : 2 Ag+ + 2e → 2Ag
The representation of cell = Cu(s) | Cu2+(aq) || Ag+(aq) | Ag (s)
Value of standard electrode potential for the oxidation of Cl– ions is more positive than that of water, even then in the electrolysis of aqueous sodium chloride, why are Cl– ions oxidised at anode instead of water ?
he oxidation reactions taking place at anode is:
2Cl– (aq) → Cl2 (g) + 2e– E0(oxid) = -1.36 V
2H2O (l) → O2 (g) + 4H+ (aq) + 4e– E0(oxid) = -1.23 V
Due to the overvoltage of oxygen (O₂), its liberation is kinetically slower than that of CI ions. Therefore, Cl ions are oxidised to Cl₂ gas.
What is electrode potential ?
Electrode potential appears at the interface between an electrode and electrolyte due to the transfer of charged species across the interface.
Why is alternating current used for measuring resistance of an electrolytic solution?
Alternate current is used for measuring the resistance of an electrolytic solution because direct current will cause the electrolysis of the electrolyte. As a result Concentration of the ions in the solution will change.
Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The Λm of ‘B’ increases 1.5 times while that of A increases 25 times. Which of the two is a strong electrolyte? Justify your answer.
Dilution of the electrolyte mainly decreases interionic attractive force, therefore Λm increases. In case of strong electrolyte it dissociated completely in solution so on dilution Λm increases to a small extent. On the other hand, the electrolyte weak is ionised only to a small extent. Upon dilution, its ionisation or dissociation increases to a large extent. That is why, Λm for weak electrolyte is more higher on dilution.
From the available information A is a weak electrolytes and B is a strong electrolyte.
How will the pH of brine (NaCl solution) be affected when it is electrolysed?
NaCl is a salt of strong acid and strong base. It is neutral in nature. pH of NaCl solution is 7. However, upon electrolysis of the aqueous solution, Na+ and OH– ions are formed. The solution becomes basic and its pH rises.
Write the Nernst equation for the cell reaction in the Daniell cell. How will the Ecell be affected when concentration of Zn2+ ions is increased ?
aniell cell involves the redox reaction :
Zn(s) + CuSO4(aq) → ZnSO4 (aq) + Cu(s)
Thus in the cell Zn(s) is oxidised to Zn²+ (aq) ions in the oxidation half cell while Cu²+ (aq) ions are reduced to Cu(s) in the reduction half cell.
According to Nernst equation,
Ecell = E0cell – (0.059/2) {log[Zn2+(aq)] / [Cu2+(aq)]
Hence Ecell decreases as the [Zn2+(aq)] increases.
