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Q.

Consider the following cells in which all the electrodes are immersed in aqueous solutions of their electrolyte:

            Cell number                                                               Cell

                   1                                                          Pt |H2|HA || Ag+| 0.1MAg 

                   2                                                         Pt |H2|HB || Ag+| 0.1MAg

                   3                                                         Pt |H2|HC || Ag+| 0.1MAg

                   4                                                         Pt |H2|HD || Ag+| 0.1MAg

                   5                                                         Pt |H2|HE || Ag+| 0.1MAg

                   6                                                         Pt |H2|HF || Ag+| 0.1MAg

                   7                                                         Pt |H2|HG || Ag+| 0.1MAg

 The freezing point of 0.1M aq acids are: (assume molarity is equal to molality) 

          Acid (0.1M) HA  HB  HC  HD  HE  HF  HG  FP. (K)272.5272271.8271.5271272.10272.80              

[Assume all acids are weak acids and pressure of H2 in each cell is 1 bar]
If the cell number (represented in table) of cell which has minimum e.m.f. at 298 K is x, then x2 is

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a

35

b

25

c

15

d

45

answer is Y.

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Detailed Solution

Given that,  The freezing point of 0.1M aq acids are: (assume molarity is equal to molality) 

          Acid (0.1M) HA  HB  HC  HD  HE  HF  HG  FP. (K)272.5272271.8271.5271272.10272.80              

So, The value of H+is maximum, e.m.f, will be minimum. 

H+=Cα

 If α,i,ΔTf, freezing point decreases so e.m.f decreases. 

 Hence acid HE have maximum [H'], so x=5

Then x2 is = 25

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