0.5L OF A 0.6M SnSO4 SOLUTION IS ELECTROLYSED FOR A PERIOD OF 30 MINUTES USING A CURRENT OF 4.6A . IF INERT ELECTRODES ARE USED WHAT IS THE FINAL CONCENTRATION OF Sn2+ REMAINING IN SOLUTION?

Question

Sunil Kumar FP · Answer

Faraday's laws can be summarized by

m=(Q/F)(M/Z)

where:

mis the mass of the substance liberated at an electrode in grams
Qis the total electric charge passed through the substance
F= 96485 C mol−1is theFaraday constant
Mis the molar mass of the substance
zis the valency number ofionsof the substance (electrons transferred per ion).

MOLE of SnSO4=(4.6*30*60)/(96500*2)
=.0429 mole
mole of SnSO4 taken=.5*.6=.3 mole
mole left=.3-.0429=.257 mole
concentration=.257/.5=.514 M

Rahul Kumar · Answer

Initial mole of Sn²+ = 0.5×0.6= 3 mole Weight of Sn²+ deposited = 119÷2 × 4.6 × 30 × 60/ 96500 = 5.10528 Mole of Sn²+ deposited = 5.10528 ÷ 119 = 0.0429 Mole of Sn²+ left = 0.3 - 0.0429 = 0.257 Concentration of Sn²+ = 0.257 ÷ 0.5 = 0.514 M

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0.5L OF A 0.6M SnSO4 SOLUTION IS ELECTROLYSED FOR A PERIOD OF 30 MINUTES USING A CURRENT OF 4.6A . IF INERT ELECTRODES ARE USED WHAT IS THE FINAL CONCENTRATION OF Sn2+ REMAINING IN SOLUTION?

0.5L OF A 0.6M SnSO4 SOLUTION IS ELECTROLYSED FOR A PERIOD OF 30 MINUTES USING A CURRENT OF 4.6A . IF INERT ELECTRODES ARE USED WHAT IS THE FINAL CONCENTRATION OF Sn2+ REMAINING IN SOLUTION?

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