A $$60$$ ml sample solution of KI was electrolysed for $$579$$ $$sec$$ using a $$5$$ amp constant current. The produced, completely reacted with $$0.20$$ M sodium thiosulphate solution and unreacted KI solution required $$30$$ ml of $$0.04$$ M acidic solution of $$KMnO4$$. If the original molarity of KI solution can be expressed in scientific notation as Y × $$10-1$$. Calculate the value of Y.

Question

A $$60$$ ml sample solution of KI was electrolysed for $$579$$ $$sec$$ using a $$5$$ amp constant current. The   produced, completely reacted with $$0.20$$ M sodium thiosulphate solution and unreacted KI solution required $$30$$ ml of $$0.04$$ M acidic solution of $$KMnO4$$. If the original molarity of KI solution can be expressed in scientific notation as Y × $$10-1$$. Calculate the value of Y.

Infinity Learn · Accepted Answer

$$nF=579$$×$$596500=0.032I$$−→$$I2+2e$$−$$nF=2nI2nI2=00.015mol$$⇒$$nKI=0.03=30$$×$$10$$−$$3eI$$−$$=eKMNO4nI$$−$$=301000$$×$$.04$$×$$5=6$$×$$10$$−$$3$$[KI]$$=36$$×$$10$$−$$3$$×$$10360=0.6$$[KI]$$=6$$×$$10$$−$$1M$$

Electrochemical cells

Unlock the full solution & master the concept.

Ready to Test Your Skills?

free study material

Electrochemical cells

Unlock the full solution & master the concept.

nF=579×596500=0.032I−→I2+2e−nF=2nI2nI2=00.015mol⇒nKI=0.03=30×10−3eI−=eKMNO4nI−=301000×.04×5=6×10−3[KI]=36×10−3×10360=0.6[KI]=6×10−1M

Ready to Test Your Skills?

free study material