A student wants to find out the cell potential of a voltaic cell made up of a nickel electrode and an iron electrode. The nickel electrode is in a 0.1 M nickel (II) sulfate solution, and the iron electrode is in a 0.1 M iron (II) sulfate solution. What is the cell potential of this voltaic cell at standard conditions (25°C and 1 atm of pressure) given the standard reduction potentials for Ni2+/Ni and Fe2+/Fe are -0.25 V and -0.44 V, respectively?

Question

A student wants to find out the cell potential of a voltaic cell made up of a nickel electrode and an iron electrode. The nickel electrode is in a 0.1 M nickel (II) sulfate solution, and the iron electrode is in a 0.1 M iron (II) sulfate solution. What is the cell potential of this voltaic cell at standard conditions (25°C and 1 atm of pre

Answer 1

To find the cell potential of the voltaic cell, we need to determine which electrode will act as the anode  oxidation  and which will act as the cathode  reduction . The electrode with the lower reduction potential will be the anode, and the electrode with the higher reduction potential will be the cathode.In this case, the reduction potentials are:Ni2+/Ni: -0.25 VFe2+/Fe: -0.44 VSince -0.44 V is lower than -0.25 V, the iron electrode will act as the anode, and the nickel electrode will act as the cathode.Now we can calculate the cell potential using the Nernst equation:E_cell = E_cathode - E_anodeE_cell =  -0.25 V  -  -0.44 V E_cell = 0.19 VSo, the cell potential of this voltaic cell at standard conditions  25C and 1 atm of pressure  is 0.19 V.