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Posted by on Thursday, May 8, 2014 at 6:01pm.

A current of 3.40 A is passed through a Ni(NO3)2 solution for 1.80 hours. How much nickel is plated out of the solution?

  • Chemistry 104 - , Thursday, May 8, 2014 at 6:51pm

    Current=charge/time (s)

    Where

    current=3.40A

    Convert hr to seconds (s):

    1.80hrs*(60 min/1 hr)*(1 min/60s)= time in s

    Solve for charge:

    seconds*current=charge (C)

    1 mole of e^-s=9.65 x 10 4 C

    Solve for moles:

    C*(1 mole/9.65 x 10 4 C)=moles of Ni(NO3)2

    Solve for mass:

    moles of Ni(NO3)2*(182.703 g/mole)= mass of Ni(NO3)2

  • oops--Chemistry 104 - , Thursday, May 8, 2014 at 7:08pm

    Note that there are two mols of electrons and you must take that into account. A minor problem is that you will not plate out Ni(NO3)2 but Ni metal.

  • Chemistry 104 --Opps - , Thursday, May 8, 2014 at 7:30pm

    Dr. Bob222 caught that for me. Look for **** for corrections.

    Current=charge/time (s)

    Where

    current=3.40A

    Convert hr to seconds (s):

    1.80hrs*(60 min/1 hr)*(1 min/60s)= time in s

    Solve for charge:

    seconds*current=charge (C)

    1 mole of e^-s=9.65 x 10^4 C

    Solve for moles:

    **** C*(1 mole/9.65 x 10^4 C)=moles of e^-s

    **** The half reaction is the following:

    Ni2+ + 2e ---> Ni

    So, 2 moles of e^- is needed for 1 mole of Ni:

    Solve for moles of Ni:

    **** moles of e^-s*(1 mole of Ni/2 mole of e^-)= moles of Ni

    Solve for mass:

    **** moles of Ni*( 58.69 g/mole)= mass of Ni

    Dr. Bob222 is correct: the question states Ni not Ni(NO3)2.

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