Same quantity of charge is being used to liberate iodine (at anode) and a metal `M` (at cathode). The mass of metal `M` liberated is `0.617 g` and the liberated iodine is completely reduced by `46.3 mL` of `0.124 M` sodium thio-sulphate. Calculate equivalent weight of metal. Also calculate the total time to bring this change if `10` ampere current passed through solution of metal iodide.

Same quantity of charge is being used to liberate iodine (at anode) an

1.	Same quantity of charge is being used to liberate iodine (at anode) and a metal `M` (at cathode). The mass of metal `M` liberated is `0.617 g` and the liberated iodine is completely reduced by `46.3 mL` of `0.124 M` sodium thio-sulphate. Calculate equivalent weight of metal. Also calculate the total time to bring this change if `10` ampere current passed through solution of metal iodide.
Answer» Correct Answer - Eq.wt `=107.468` `I_(2) +2Na_(2)S_(2)O_(3) rarr 2NaI +Na_(2)S_(4)O_(6)` Mili moles of `Na_(2)S_(2)O_(3) = 46.3 xx 0.124 = 5.7412` Mili mole of `I_(2) = 2.8706` `2I^(-) rarr I_(2) +2e^(-)` `n_(E) = 2n_(I_(2)) = 5.7412 xx 10^(-3)` `5.7412 xx 10^(-3)` moles `e^(-)` deposited `= 0.617g` So 1 mole `e^(-)` will deposite `= (0.617)/(5.7412 xx 10^(-3))` `= 107.468g` Eq.wt `=107.468g`

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