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.

Explanation

Eq. of metal = Eq. of iodine = Eq. of hypo `= (i.t)/(96500)`
`:. (0.617)/(E) = (46.3)/(1000) xx 0.124 xx 1`
[For hypo : `2SO_(2)O_(3)^(2-) rarr S_(4)O_(6)^(2-)+2e, E = (M)/(1)`]
`:. E = 107.47`
Also, `(10 xx t)/(96500) = (46.3)/(1000) xx 0.124 xx 1`
`t = 55.4` second