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Sagot :
The current required to accumulate the 1.22 grams of nickel in 0.5 hours is 2.23 A.
What is current?
The current is given as the product of the charge with time. In the electrochemical analysis of the nickel, there will be a reduction of the nickel ion to nickel. The formation is given as:
[tex]\rm Ni^{2+}+2e^-\;\rightleftharpoons Ni[/tex]
There is the deposition of 1 mole of Ni with 2 electrons transfer. The transfer of charge for 1 mole that is 58.7 grams Nickel is:
[tex]\rm 58.7\;g=2\;\times\;96487\;C\\58.7\;g=192974\;C[/tex]
The mass of Ni to be deposited is 1.22 grams. The charge required is given as:
[tex]\rm 58.7\;grams\;Ni=192974\;C\\\\1.22\;grams\;Ni=\dfrac{192974}{58.7}\;\times\;1.22\;C\\\\1.22\;grams\;Ni=4010.7\;C[/tex]
The current required to transfer 4010.7 C of charge in 1800 seconds is given as:
[tex]\rm Charge=Current\;\times\;Time\\4010.7\;C=Current\;\times\;1800\;sec\\Current=2.23\;A[/tex]
Thus, the current required to accumulate the 1.22 grams of nickel in 0.5 hours is 2.23 A.
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