20 mL of 0.1 M HCl is divided into two equal parts and kept in two separate beakers. To one beaker 10 mL of 0.06 M NaOH is added and to the other 10 mL of 0.02 M NaOH is added. Two hydrogen electrodes are placed in the two solution which are linked through a salt bridge. what will be the emf of the cell formed?

20 mL of 0.1 M HCl is divided into two equal parts and kept in two sep

1.	20 mL of 0.1 M HCl is divided into two equal parts and kept in two separate beakers. To one beaker 10 mL of 0.06 M NaOH is added and to the other 10 mL of 0.02 M NaOH is added. Two hydrogen electrodes are placed in the two solution which are linked through a salt bridge. what will be the emf of the cell formed?
Answer» Solution :Each beaker initially contains 10ML of 0.1 M HCl 10 mL of 0.1 M HCl contains HCl=`10xx0.1`Millimole=1 millimole 10 mL of 0.06 M NAOH contains NaOH=`10xx0.06=0.6` millimole HCl left unneutralized in beaker1=1-0.6=0.4 millimole TOTAL volume=20 mL `THEREFORE`In beaker 1, [HCl] or `[H^(+)]=(0.4)/(20)=0.02M` 10 mL of 0.02 M NaOH contains NaOH=`10xx0.02=0.2` millimole `thereforeHCl` left unneutralized in beaker`2=1-0.20` millimole=0.80 millimole total volume=20mL `therefore`In beaker 2, [HCl] or `[H^(+)]=(0.80)/(2)=0.04M` Thus, we have a concentration cell for which `E_(cell)=(0.0591)/(1)"log"(0.04)/(0.02)=0.0178V=17.8mV`