Question

If 0. 5371 g of zn was plated onto an electrode, how much charge (in coulombs) was used to reduce the zn2+ to zn(s)?

Answer 1

The amount of charge required to reduce Zn²⁺ to Zn is 1585.26 coulomb.

What is electrolysis?

The process that has been equipped with the use of electric current to break the chemical compound and deposited at the constituent electrodes is termed as electrolysis.

The reaction of reduction of Zn can be given as:

Zn²⁺ + 2e⁻ → Zn (s)

Molar mass of Zn = 65.39 g

The equivalent of Zn²⁺ has been:

Equivalent = atomic mass/equivalent weight

Equivalent Zn²⁺ = 65.39/32.69

Equivalent Zn²⁺ = 2 mol

1 equivalent = 1 F charge used

2 mol equivalent = 2 F charge used

Charge in coulomb can be given as:

1 Faraday = 96500 coulomb

2 F = 96500 * 2 coulomb

2 F = 1,93,000 coulomb

The moles of Zn to be deposited is:

Moles = weight /molar weight

Moles = 0.5371 g/ 65.39 g/mol

Moles = 0.00821 mol

Therefore, the charge used for the deposition of 0.00821 mol is:

Charge = 0.00821 * 1,93,000 coulomb

Charge = 1585.26 coulomb

1585.26 coulomb charge was used to reduce the Zn²⁺ to Zn.

Learn more about electrolysis, Here:

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