Question

Calculate the cell potential for the reaction as written at 25.00 °C, given that [Mg2 Use the standard reduction potentials in this table 0801 M and [Fe+] 0.0120 M. Mg(s) + Fe2+(aq) Mg2+(aq) + Fe(s)

Answer 1

The given cell reaction is: Mg2 (aq)Fe(s) Mg (s)+Fe (aq) E° Mg/Mg -2.38 V and E *,=-0.41V The standard half-cell potentials values are The half-cell Mg Mg 2+ Fe acts as anode and the Fe acts as cathode The standard cell potential, E is calculated by the following formula: Е — Е°, - Eo anode cetl cathode -E°, = E° FeFe --0.41V -(2.38V) =1.97 V The cell potential for the given reaction is calculated by using the Nernst equation: RT - In Q Е. — Е _ el nF Here, R is the gas constant, T is the temperature, in reaction, F is the Faraday constant, and Q is reaction quotient n is the number of electrons transferred For temperature 25°C the Nernst equation becomes 0.0592 -log Q Е.e — Езан - catl n Mg as 1.97V, and the reaction quotient, Q 2, E Substitute the value of n as n as Fe2 the above equation Mg E1.97V- 0.0592 log 2 cell

2+ Substitute the value of as 0.801 M and as 0.0120 M in the above equation shown below as (0.801 M -log '(0.0120 M 2 E1.97V-0.0592. call =1.92V