Respuesta :
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:
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:
[tex] Eº cell = Eº cathode - Eº anode [/tex]
[tex] Eº cell = EºAg⁺/Ag - Eº Mg²⁺/Mg [/tex]
[tex] Eº cell =0.80 V - (-2.37 V) [/tex]
Eº cell = 3.17 V
