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Q1. For the electrochemical cell, M | M^+ || X^- | X,E^° (M^+ | M)=0.44 V and E^° (X | X^- )=0.33 V. From this data one can deduce that
•  E_cell=0.77 V
•  -0.77 V
•  M+X→M^++ X^- is the spontaneous reaction
•  M^++ X^- →M+X is the spontaneous reaction
Solution

Q2.The relationship between Gibbs’ free energy change (Î”G) and emf (E) of a reversible electrochemical cell is given by
•  Î”G= nFE
•  Î”G =-nFE
•  Î”G= E/nF
•  Î”G = nF/E
Solution
The Gibb’s free energy change △G and emf (E^°) of a reversible electrochemical cell are related by the following expression. △G = -nFE_cell^° or = - nFE

Q3.  Standard reduction potential of an element is equal to :
•   + 1 × its reduction potential
•  + 1 × its standard oxidation potential
•  0.00 V
•  - 1 × its standard oxidation potential
Solution

Q4. If a strip of copper metal is placed in a solution of ferrous sulphate :
•  Copper will precipitate out
•  Iron will precipitate out
•  Both copper and iron will be dissolved
•  No reaction will take place
Solution
Cu is below Fe in electrochemical series.

Q5.For the cell Zn|Zn^(2+) || Cu^(2+) |Cu, if the concentration of Zn^(2+) and Cu^(2+) ions is doubled, the e.m.f. of the cell :
•  Remains same
•  Becomes zero
•  Reduces to half
•  Doubles
Solution

.

Q6. Also the [H^+] for problem 9 using the same data is :
•  0.133 M
•  0.00133 M
• 0.0133 M
•  None of these
Solution
[H^+ ]=c.Î±=0.0133 × 0.1 = 0.00133 M.

Q7.〖10〗^(-2) g atom of Ag can be oxidised to Ag^+ during the electrolysis of AgNO_3 solution using silver electrode by :
•  96.500 coulomb
•  9650 coulomb
•  96500 coulomb
•  965 coulomb
Solution

Q8.An alloy of Pb-Ag weighing 1.08 g was dissolved in dilute HNO_3 and the volume made to 100 mL. A silver electrode was dipped in the solution and the emf of the cell set up Pt(s),H_2 (g) | H^+ (1 M) || Ag^+ (aq) | Ag (s) Was 0.62 V. If E_cell^° = 0.80 V, what is the percentage of Ag in the alloy? [At 25℃ , RT /F = 0.06]
•  25
•  50
•  2.5
•  10
Solution

Q9.At 25°C, the standard e.m.f. of cell having reactions involving a two electron change is found to be 0.295 V. The equilibrium constant of the reaction is :
•  29.5 ×〖10〗^(-2)
•  10
•  29.5 ×〖10〗^10
•  〖10〗^10
Solution

Q10. A conductivity cell has two platinum electrodes of 1.2 cm^2area, separated by a distance of 0.8 cm. The cell constant is :
•  1.5 cm^(-1)
•  0.66 cm^(-1)
•  0.96 cm^(-1)
• 0.66 cm
Solution
Cell constant =1/a=0.8/1.2=0.66 cm^(-1).

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