Question

Na2s2o3+i2=na2s4o6+nai balance the equation by ion electron method

Answer 1

Na2S2O3 +I2 = Na2S4O6+NaI balance the equation by ion electron method.

here we see iodine , I2 is reduced (as oxidation number changes from 0 to -1) and sulphur is oxidised. ( as oxidation number of s changes from +2 to +2.5).

half reduction reaction : $I_2\rightleftharpoons 2I^++2e^-$....(1)

half oxidation reaction : $2S_2O_3^{2-}+2e^-\rightleftharpoons S_4O_6^{2-}$

now adding $2Na^+$ ions both sides in equations (1) and (2)

then, $2Na^++I_2\rightleftharpoons 2NaI+2e^-$....(1)

and $2Na_2S_2O_3+2e^-\rightleftharpoons Na_2S_4O_6$....(2)

now adding equations (1) and (2) we get,

$2Na_2S_2O_3+I_2\rightleftharpoons N_2S_4O_6+2NaI$ This is our balanced chemical equation.

Answer 2

Na2S2O3 +I2 = Na2S4O6+NaI balance the equation by ion electron method.

here we see iodine , I2 is reduced (as oxidation number changes from 0 to -1) and sulphur is oxidised. ( as oxidation number of s changes from +2 to +2.5).

half reduction reaction : I_2\rightleftharpoons 2I^++2e^-I2⇌2I++2e− ....(1)

half oxidation reaction : 2S_2O_3^{2-}+2e^-\rightleftharpoons S_4O_6^{2-}2S2O32−+2e−⇌S4O62−

now adding 2Na^+2Na+ ions both sides in equations (1) and (2)

then, 2Na^++I_2\rightleftharpoons 2NaI+2e^-2Na++I2⇌2NaI+2e− ....(1)

and 2Na_2S_2O_3+2e^-\rightleftharpoons Na_2S_4O_62Na2S2O3+2e−⇌Na2S4O6 ....(2)

now adding equations (1) and (2) we get,

2Na_2S_2O_3+I_2\rightleftharpoons N_2S_4O_6+2NaI2Na2S2O3+I2⇌N2S4O6+2NaI This is our balanced chemical equation.