Question

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?

Answer 1

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

Answer 2

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.

Answer 3

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.