10800C ofelectricity through the electrolyte deposited 2. 977 gram of metal with atomic mass of 106.4 gram per mole find the charge on the metal cations

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Solution

Answer : Charge on metal cation = +4 Let the charge on the metal ion be n+. The reduction half-reaction would be M(n+) +n(e-) ---> M(s) (1 mol = 106.4 g) Quantity of electricity required for depositing 106.4 g of metal = n x 96500 C Quantity of electricity required for depositing 1 g of metal = n x 96500/106.4 C Quantity of electricity required for depositing 2.977 g of metal = n * 96500 * 2.977/106.4 = n2700 C Quantity of electricity actually passed = 10800 C Quantity of electricity required = Quantity of electricity passed Now, 10800 = n2700 n = 10800/2700 n = 4

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