Question

how many gram of calcium will be formed at the cathode by the passage of 96500 coulomb of charge through the molten cacl2 solution

Answer 1

Explanation:

Weight of calcium deposited on cathode is W=Z*Charge(or)Q

Weight = $\frac{equivalent weight of calcium }{96500}$*96500= 20g of calcium is deposited

as equivalent weight of calcium is 20grams

Answer 2

Answer:

20 gram of calcium will be formed at the cathode by the passage of 96500 coulomb of charge through the molten CaCl₂ solution.

Explanation:

Half cell reaction of calcium deposition at the cathode:-

Ca²⁺(aq)   +  2e⁻ $\longrightarrow$  Ca(s)

From the above chemical equation we can say that for deposition of  one mole of calcium ions. We need two mole of electrons.

The one mole of electrons = 1 Faraday = 96500C

Molar mass of Calcium = 40g/mol

Therefore, 2 moles of electrons deposited calcium = 40g

2× 96500C charge passed to deposited Calcium = 40g

96500C charge will deposit Calcium on cathode $=\frac{40}{2\times 96500}\times 96500= 20g$

Therefore, 20g of calcium will be formed at the cathode by the passage of 96500 coulomb charge.

                                                                                   

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