What is the mass of silver deposited when 24,125 C of electricity is passed through a solution or silver salt. [Ag = 108, IF = 96,500 C ]

Answer Details

The mass of silver deposited can be calculated using the equation: mass = (current x time x atomic weight of silver) / (ionic charge x Faraday constant) In this case, the current is not given, but we are given the total charge passed through the solution, which is 24,125 C. We can assume that this charge was passed at a constant current over a certain amount of time. Therefore, we can rearrange the equation to solve for the current: current = (ionic charge x Faraday constant x mass) / (time x atomic weight of silver) The ionic charge for silver is +1, the Faraday constant is 96,500 C, and the atomic weight of silver is 108 g/mol. Substituting these values into the equation, we get: current = (1 x 96,500 x mass) / (time x 108) We don't know the time, so we can't solve for the current. However, we can use the fact that the current is constant to say that the total charge passed is equal to the current multiplied by the time: charge = current x time Substituting the expression for current into this equation, we get: charge = ((1 x 96,500 x mass) / (time x 108)) x time Simplifying, we get: charge = (96,500 x mass) / 108 Multiplying both sides by 108 and dividing by 96,500, we get: mass = (charge x atomic weight of silver) / 96,500 Substituting the values given in the question, we get: mass = (24,125 x 108) / 96,500 mass = 27 g Therefore, the mass of silver deposited is 27 g