A student wants to determine the power output of a galvanic cell containing a zinc electrode and a copper electrode. The zinc electrode has a surface area of 20 cm² and is coated with a layer of zinc which weighs 0.5 grams. The copper electrode has a surface area of 15 cm² and is coated with a layer of copper that weighs 0.4 grams. The cell operates at standard conditions (25°C and 1 atm), and the electrolyte is a solution of 1 M copper sulfate and 1 M zinc sulfate. Calculate the power output of the electrochemical cell.

Question

A student wants to determine the power output of a galvanic cell containing a zinc electrode and a copper electrode. The zinc electrode has a surface area of 20 cm² and is coated with a layer of zinc which weighs 0.5 grams. The copper electrode has a surface area of 15 cm² and is coated with a layer of copper that weighs 0.4 grams. The cell operates at standard conditions (25°C

Answer 1

To calculate the power output of the electrochemical cell, we need to determine the cell potential and the current flowing through the cell.First, let's find the cell potential  E_cell  using the standard reduction potentials of the half-reactions:Zn + 2e  Zn  E = -0.76 V Cu + 2e  Cu  E = +0.34 V The overall cell reaction is:Zn + Cu  Zn + CuThe cell potential  E_cell  is the difference between the reduction potentials of the two half-reactions:E_cell = E Cu/Cu  - E Zn/Zn  = 0.34 V -  -0.76 V  = 1.10 VNext, we need to find the current  I  flowing through the cell. To do this, we need to determine the rate of the reaction. Since the reaction is limited by the mass of the zinc and copper, we'll use the mass of the limiting reactant to find the moles of electrons transferred.The molar mass of Zn is 65.38 g/mol, and the molar mass of Cu is 63.55 g/mol.Moles of Zn:  0.5 g  /  65.38 g/mol  = 0.00765 molMoles of Cu:  0.4 g  /  63.55 g/mol  = 0.00630 molSince the moles of Cu are less than the moles of Zn, Cu is the limiting reactant. The reaction involves the transfer of 2 moles of electrons per mole of Cu:Moles of electrons: 2 * 0.00630 mol = 0.0126 molNow, let's assume that the reaction takes place over a period of 1 hour  3600 seconds . We can calculate the average current  I  in Amperes:I =  moles of electrons * Faraday's constant  / timeI =  0.0126 mol * 96485 C/mol  / 3600 s  0.339 AFinally, we can calculate the power output  P  of the electrochemical cell using the formula P = E_cell * I:P =  1.10 V  *  0.339 A   0.373 WThe power output of the electrochemical cell is approximately 0.373 watts.