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

Eq.wt `=107.468`

Explanation

`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`