Question

What is E∘ (cell) for the cell Mn∣ ∣​M²⁺(aq)∣| ∣ Ag⁺(aq)∣ Ag ?

Answer 1

The standard cell potential (E°(cell)) for a galvanic cell composed of Mn/Mn²⁺ and Ag⁺/Ag half-cells is +1.98 V, calculated using the standard reduction potential for each half-cell.

Step-by-step explanation:

The question is asking for the standard cell potential (E°₂₇₈₃₆₀₇₈₃(cell)) for a galvanic cell composed of manganese (Mn/Mn²⁺) and silver (Ag⁺/Ag) half-cells. To calculate this, we need to know the standard reduction potentials for both half-reactions. From standard reduction potential tables, we can obtain E°₂₇₈₃₆₀₇₈₃ for the reduction of Ag⁺ to Ag(s) and the oxidation of Mn(s) to Mn²⁺. The cell potential is calculated by subtracting the anode potential (which is the reverse of the oxidation potential) from the cathode potential:

E°₂₇₈₃₆₀₇₈₃(cell) = E°₂₇₈₃₆₀₇₈₃(cathode) - E°₂₇₈₃₆₀₇₈₃(anode)

The standard electrode potentials are as follows:

• Mn²⁺/Mn = -1.18 V (E°₂₇₈₃₆₀₇₈₃(oxidation) which gets reversed for the cell calculation)
• Ag⁺/Ag = +0.80 V (E°₂₇₈₃₆₀₇₈₃(reduction))

Therefore, the E°₂₇₈₃₆₀₇₈₃(cell) for the Mn| Mn²⁺(aq)|| Ag⁺(aq)| Ag cell is:

E°₂₇₈₃₆₀₇₈₃(cell) = +0.80 V (cathode) - (-1.18 V) (anode)

E°₂₇₈₃₆₀₇₈₃(cell) = +1.98 V

Answer 2

The standard cell potential, E∘(cell), for the given cell is -0.38 V.

Step-by-step explanation:

The question asks for the standard cell potential, E∘(cell), for the cell Mn| Mn²⁺(aq)|| Ag⁺(aq)| Ag. The standard cell potential, E∘(cell), is a measure of the electrical potential difference between the two half-cells of a redox reaction under standard conditions. It is calculated by subtracting the reduction potential of the anode from the reduction potential of the cathode. In this case, the anode is Mn| Mn²⁺(aq) and the cathode is Ag⁺(aq)| Ag.

The standard reduction potential of Mn²⁺(aq) is +1.18 V, and the standard reduction potential of Ag⁺(aq) is +0.80 V. So, the standard cell potential (E∘(cell)) can be calculated as follows;

E∘(cell) = E∘(cathode) - E∘(anode) = (+0.80 V) - (+1.18 V) = -0.38 V