Calculate the current density of an electrochemical cell where a nickel electrode is oxidized to form nickel (II) ions, while a copper electrode is reduced by the nickel (II) ions. The cell has a voltage of 0.65 volts and a resistance of 50 ohms. The surface area of the nickel electrode is 10 cm² while that of the copper electrode is 15 cm². The current density of the electrochemical cell is expressed in A/cm².

Question

Calculate the current density of an electrochemical cell where a nickel electrode is oxidized to form nickel (II) ions, while a copper electrode is reduced by the nickel (II) ions. The cell has a voltage of 0.65 volts and a resistance of 50 ohms. The surface area of the nickel electrode is 10 cm² while tha

Answer 1

To calculate the current density of the electrochemical cell, we first need to find the current  I  flowing through the cell. We can use Ohm's law to do this:Ohm's law: V = I * RWhere V is the voltage, I is the current, and R is the resistance.Rearranging the formula to solve for I:I = V / RNow, we can plug in the given values:I = 0.65 volts / 50 ohmsI = 0.013 A  amperes Next, we need to find the current density  J . Current density is defined as the current per unit area. Since the nickel electrode is being oxidized, we will use the surface area of the nickel electrode to calculate the current density:J = I / AWhere J is the current density, I is the current, and A is the surface area of the nickel electrode.Plugging in the values:J = 0.013 A / 10 cmJ = 0.0013 A/cmSo, the current density of the electrochemical cell is 0.0013 A/cm.