A student wants to calculate the efficiency of an electrochemical cell with a standard potential of -0.74 volts. The cell is set up with a copper electrode immersed in a solution of copper(II) sulfate and a silver electrode immersed in a solution of silver nitrate, connected by a salt bridge. The student measures a current of 0.5 amperes flowing through the cell for a duration of 60 minutes. Calculate the efficiency of the electrochemical cell in terms of the amount of reactants consumed and the amount of product formed.

Question

A student wants to calculate the efficiency of an electrochemical cell with a standard potential of -0.74 volts. The cell is set up with a copper electrode immersed in a solution of copper(II) sulfate and a silver electrode immersed in a solution of silver nitrate, connected by a salt bridge. The student measures a current of 0.5 amperes flowing through the cel

Answer 1

To calculate the efficiency of the electrochemical cell, we need to determine the amount of reactants consumed and the amount of product formed. We can do this by using Faraday's laws of electrolysis.First, let's convert the duration of the experiment to seconds:60 minutes * 60 seconds/minute = 3600 secondsNow, we can calculate the total charge passed through the cell:Charge  Q  = Current  I   Time  t Q = 0.5 A  3600 s = 1800 CoulombsNext, we need to determine the number of moles of electrons transferred using Faraday's constant  F = 96485 C/mol :Moles of electrons  n  = Charge  Q  / Faraday's constant  F n = 1800 C / 96485 C/mol = 0.01865 molNow, let's write the half-reactions for the electrochemical cell:Copper half-reaction: Cu + 2e  CuSilver half-reaction: Ag + e  AgSince the copper half-reaction requires 2 moles of electrons for every mole of copper, we can calculate the moles of copper formed:Moles of Cu = Moles of electrons / 2Moles of Cu = 0.01865 mol / 2 = 0.009325 molFor the silver half-reaction, 1 mole of electrons is required for every mole of silver, so the moles of silver formed are equal to the moles of electrons:Moles of Ag = 0.01865 molNow, we can calculate the mass of copper and silver formed using their molar masses  Cu = 63.55 g/mol, Ag = 107.87 g/mol :Mass of Cu = Moles of Cu  Molar mass of CuMass of Cu = 0.009325 mol  63.55 g/mol = 0.5926 gMass of Ag = Moles of Ag  Molar mass of AgMass of Ag = 0.01865 mol  107.87 g/mol = 2.0106 gThe efficiency of the electrochemical cell can be expressed as the ratio of the mass of products formed to the mass of reactants consumed. Since the mass of reactants consumed is equal to the mass of products formed in this case, the efficiency is 100%.However, it's important to note that the standard potential of the cell is -0.74 V, which indicates that the cell is not spontaneous and requires an external voltage to drive the reaction. The efficiency calculated here is based on the assumption that the external voltage is sufficient to drive the reaction and that there are no side reactions or losses occurring in the cell.