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2 years ago

What current (in a) is required to plate out 2. 96 g of nickel from a solution of ni2 in 27. 12 minutes?

Chemistry

2 answers:

amid [387]2 years ago

6 0

The current required to plate out 2.96 g of nickel from the solution of Ni²⁺ in 27.12 minutes is 5.95 A

<h3>Balanced equation </h3>

We'll begin by writing the balanced equation showing the number of faraday required to plate nickel. This is given below

Ni²⁺ + 2e —> Ni

Molar mass of Ni = 59 g/mol

Mass of Ni from the balanced equation = 1 × 59 = 59 g

Number of faraday = 2 F

1 faraday = 96500 C

2 faraday = 2 × 96500 = 193000 C

SUMMARY

From the balanced equation above,

59 g of Nickel was deposited by 193000 C of electricity

<h3>How to determine the quantity of electricity </h3>

From the balanced equation above,

59 g of Nickel was deposited by 193000 C of electricity

Therefore,

2.96 g of Nickel will be deposited by = (2.96 × 193000) / 59 = 9682.71 C of electricity

<h3>How to determine the current </h3>

Quantity of electricity (Q) = 9682.71 C
Time (t) = 27.12 mins = 27.12 × 60 = 1627.2 s
Current (I) =?

I = Q / t

I = 9682.71 / 1627.2

I = 5.95 A

Learn more about Faraday's law:

brainly.com/question/24795850

Margaret [11]2 years ago

4 0

The current required to plate out 2. 96 g of nickel from a solution of ni2 in 27. 12 minutes is 5.95 ampere.

<h3>What is current?</h3>

Current is a charged particle moving through an electrical conductor.

The equation is $\rm Ni^2^+ + 2e = Ni$

step 1: calculate the mass of Ni

Molar mass of Ni is 59 g/mol

59 × 1 = 59 g

Step2: calculate the number of Faraday

no. of Faraday is 2F

1 Faraday = 96500 C

then, 2 Faraday is equal to 2 × 96500 = 193000 C

Step 3: calculate the quantity of electricity

59 g of Nickel is deposited by, 193000 C of electricity

So, for 2.96 g of nickel

$\dfrac{2. 96\;g \times 193000}{59} = 9682.71 \;C$

Step4: calculate the current:

Electricity Q = 9682.71 C

Time = $27.12 \times 60 = 1627.2 \;s$

To find current

$I = \dfrac{Q}{t} \\\\I = \dfrac{9682.71 }{1627.2}\\I = 5.95 A$

Thus, the current required is 5.95 ampere.

Learn more about current

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