The total reduction potential of a cell can be calculated by subtracting the standard reduction potential of the oxidation half-reaction from that of the reduction half-reaction. For potassium (K) being reduced, the standard reduction potential is approximately -2.93 V, while for copper (Cu) being oxidized, its reduction potential is +0.34 V. Thus, the total reduction potential of the cell is calculated as: E_cell = E_reduction (Cu) - E_reduction (K) = 0.34 V - (-2.93 V) = 3.27 V. This positive value indicates that the cell reaction is spontaneous.
-1.95
-1.95V
silver reduced , gold oxidized
Silver (Ag) reduced, gold (Au) oxidized
In the reaction between potassium (K) and oxygen (O₂), potassium is oxidized and oxygen is reduced. Potassium loses electrons to form potassium ions (K⁺), while oxygen gains electrons to form oxide ions (O²⁻). This transfer of electrons defines the oxidation and reduction process, where oxidation refers to the loss of electrons and reduction refers to the gain of electrons. Thus, K is the reducing agent, and O₂ is the oxidizing agent in this reaction.
-3.27V
The total reduction potential of a cell where potassium is reduced and copper is oxidized can be calculated by finding the difference in the standard reduction potentials of the two half-reactions. The reduction potential for potassium reduction (K⁺ + e⁻ → K) is -2.92 V, and the oxidation potential for copper oxidation (Cu → Cu²⁺ + 2e⁻) is 0.34 V. Therefore, the total reduction potential of the cell is -2.92 V - 0.34 V = -3.26 V.
-1.95
-1.95
Any pair of half-reactions where the reduction potential of the half-reaction being oxidized is greater than the reduction potential of the half-reaction being reduced will have a negative total reduction potential. This results in a thermodynamically unfavorable reaction.
-1.95V
silver reduced , gold oxidized
The total reduction potential of the cell can be calculated by finding the difference between the reduction potentials of the two half-reactions at standard conditions. The reduction potential for K reduction is -2.92 V and for Cu oxidation is 0.34 V. So, the total reduction potential for the cell would be (-2.92 V) - 0.34 V = -3.26 V.
The reduction potential of Na is -2.71 V and the reduction potential of Zn is -0.76 V. When Na is reduced, it gains electrons, so its reduction potential is written as a positive value (+2.71 V). When Zn is oxidized, it loses electrons, so its oxidation potential is -0.76 V. Therefore, the total reduction potential of the cell is +2.71 V - (-0.76 V) = +3.47 V.
Silver (Ag) reduced, gold (Au) oxidized
In the reaction between potassium (K) and oxygen (O₂), potassium is oxidized and oxygen is reduced. Potassium loses electrons to form potassium ions (K⁺), while oxygen gains electrons to form oxide ions (O²⁻). This transfer of electrons defines the oxidation and reduction process, where oxidation refers to the loss of electrons and reduction refers to the gain of electrons. Thus, K is the reducing agent, and O₂ is the oxidizing agent in this reaction.
-1.95