0.617g metal is produced at cathode during electrolysis of a metal iodide . 46.3 mL of 0.124 ( M ) `Na_(2)S0_(4)` solution is required to reduce the `I_(2)` produced at anode completely . Find the equivalent mass of the metal . [ I = 127 ]

Explanation

`"1000 mL 1 (M) "Na_(2)SO_(3)=(1)/(2)" mol "I_(2)-="1 mol I"`
`therefore" 46.3 mL of 0.124 (M) "Na_(2)S_(2)O_(3)-=5.7412xx10^(-3)"mol I"`
`I^(-)rarrI+e" So, I mol I "-=96500C`
`5.7412xx10^(-3)"mol I"-="554.02C again, 554.02C "-="0.617g metal"`
`therefore 96500C-=107.47g" metal"`
`therefore" Equivalent mass of the metal "=107.47.`