Question

A galvanic (voltaic) cell consists of an electrode composed of magnesium in a 1.0 M magnesium ion solution and another electrode composed of silver in a 1.0 M silver ion solution, connected by a salt bridge. Calculate the standard potential for this cell at 25 °C.

Answer 1

Answer:

3.17 V

Explanation:

The cell is operating under standard conditions. These standard conditions include; that the reaction takes place at 298 Kelvin (room temperature), the pressure of the system is 1 atmosphere (standard pressure), and the solutions have a Molarity of 1.0 M for both the anode and cathode solutions. All these conditions are satisfied in the cell under review in the question.

Hence;

E°anode (magnesium)= -2.37 V

E°cathode (silver) = 0.80 V

E°cell= E°cathode -E°anode

E°cell= 0.80-(-2.37)

E°cell= 0.80 + 2.37

E°cell= 3.17 V

Hence the standard cell potential of this cell at 25°C is 3.17 V

Answer 2

Answer:

The standard potential at 25ºC is 3.17 V.

Explanation:

The anode in  a galvanic cell is the electrode at which oxidation occurs and the cathode is the electrode  at which reduction occurs.

The overall cell reaction  will be the sum of two half-cell reactions. The standard reduction potentials are:

Mg²⁺ (1.0 M) + 2e⁻  →  Mg (s)          Eº = -2.37

Ag⁺ (1.0 M) + e⁻ → Ag (s)                 Eº= +0.80

Since the reactants are in their standard states (1.0 M) and at 25ºC we can write the half-cell reactions as follows:

Anode (oxidation):                        Mg (s) → Mg²⁺ (1.0 M) + 2e⁻

Cathode (reduction):     2Ag⁺ (1.0 M) + 2e⁻ → 2Ag (s)

Overall:            Mg (s) +2 Ag⁺ (1.0 M) + 2e⁻ → 2Ag (s) + Mg²⁺ (1.0 M) + 2e⁻

In order to balance the overall equation we multiply the reduction of  Ag⁺ by 2. We can do so because, as an intensive property, E° is not affected by  this procedure.

The standard emf of the cell, E°cell , which is composed of a contribution  from the anode and a contribution from the cathode, is given by:

$Eº cell = Eº cathode - Eº anode$

$Eº cell = EºAg⁺/Ag - Eº Mg²⁺/Mg$

$Eº cell =0.80 V - (-2.37 V)$

Eº cell = 3.17 V