Short Answer Type
Long Answer TypeShort Answer TypeCan a nickel spatula be used to stir the solution of CuSO4? Support your answer with reason. (E°Ni2+/Ni = – 0.25 V, E°Cu2+/Cu = + 0.34 V)
Long Answer TypeCalculate the cell emf at 25° C for the following cell:
Ni(s) | Ni2+ (0.01 M) || Cu2+ (0.1 M) | Cu(s)
[Given E°Ni2+/Ni = – 0 25 V, E° Cu = + 0.34 V, 1 F = 96500]
Calculate the maximum work that can be accomplished by operation of this cell.
The emf of a cell corresponding to the reaction
Zn(s) + 2H+ (aq) → Zn2+ (0.1 M) + H2(g) (1 atm) is 0.28 V at 15° C.Write the half cell reactions and calculate the pH of the solution at the hydrogen electrode.
From the table, standard electrode potents at 298 k are:
(E°Fe3+/VF = 0.77 V, E°I2/I– = 0.54 V)
(E°Ag+/Ag = E°Cu2+/Cu = 0.34)
(E°Fe3+/Fe= 0.77 V, E°Br2/Br- = 1.08 V)
(E°Ag+/Ag= 0.8 V E°Fe3+/Fe2+ = 0.77 V)
(E°Fe3+/Fe2+= 0.77 V,E°Br2/Br- = 1.08 V)
(a) 
In this reaction, Fe3+ is reduced to Fe2+ and I– is oxidised to I2. The cell giving above reaction will be
As E0 is positive, the reaction between Fe3+ (aq) and I– (aq) occurs as indicated by possible reaction given above.
(b) 
Here, in this reaction, Ag+ is reduced to Ag (i.e., it should be cathode) and Cu(s) is oxidised to Cu2+(aq) (i.e., it should be anode).
The cell can be represented as
As E°cell is positive, the reaction between (Ag+ (aq) and Cu(s) occurs as indicated by possible reaction given above.
(c) 
In this reaction Fe3+ is reduced to Fe2+ (i.e., Fe3/Fe2+ electrode should be cathode) and Br is oxidised to Br2 (i.e., Br2/Br– electrode should be anode.
The cell can be represented as:
As E°cell is negative, no reaction will occur between Fe3+ (aq) and Br–(aq).
(d) 
Two half-cell reactions can be expressed as:
As E°cell is negative, no reaction occurs between Fe3+(aq) and Ag(s).
(e) 
The two half-cell reactions are
As E°cell is positive, the reaction is feasible, i.e., reaction between Br2(aq) and Fe2+ (aq) occurs as indicated by possible reaction given above.
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