Given the Na+/Na and K+/K half-cells, determine the overall electrochemical reaction that proceeds spontaneously and the E0 value. Answer in units of V
Can you please explain the steps?
I don't understand the question. Is that between Na and K? Any water involved?
Yes, and there is no water involved.
To determine the overall electrochemical reaction and its standard potential (E0), you need to follow these steps:
Step 1: Identify the half-reactions
The Na+/Na half-cell represents the reduction of sodium ions (Na+) to sodium metal (Na):
Na+ + e- → Na
The K+/K half-cell represents the reduction of potassium ions (K+) to potassium metal (K):
K+ + e- → K
Step 2: Balance the half-reactions
To balance the half-reactions, ensure the number of electrons (e-) on the left side of the equation is equal to the number of electrons on the right side. In this case, both reactions already have one electron on each side, so no additional balancing is needed.
Step 3: Determine the overall reaction
Since both half-reactions involve reductions (gaining electrons), we can add them together to get the overall reaction. To do this, we multiply one or both of the half-reactions by a suitable number to cancel out the electrons:
2(Na+ + e- → Na)
2(K+ + e- → K)
The overall reaction is:
2Na+ + 2K+ → 2Na + 2K
Step 4: Determine the standard potential (E0)
The standard potential, denoted as E0, is the measure of how likely a half-reaction is to occur under standard conditions. It is given in volts (V). To determine the E0 of the overall reaction, you need to look up the standard reduction potentials of the half-reactions from a reference table, such as the NIST Standard Reference Database.
Na+ + e- → Na has a standard reduction potential of +2.71 V.
K+ + e- → K has a standard reduction potential of -2.92 V.
The overall E0 can be calculated by summing the reduction potentials of the half-reactions:
E0 = E0cathode - E0anode
E0 = (-2.92 V) - (+2.71 V)
E0 = -5.63 V
Therefore, the overall electrochemical reaction that proceeds spontaneously is:
2Na+ + 2K+ → 2Na + 2K
And the E0 value for this overall reaction is -5.63 V.
To determine the overall electrochemical reaction that proceeds spontaneously using the given half-cells, you need to first identify the half-reactions and their respective standard reduction potentials, E°.
1. Na⁺/Na half-cell: The half-reaction for the Na⁺/Na half-cell can be written as:
Na⁺(aq) + e⁻ → Na(s)
The standard reduction potential, E°, for this half-reaction is +2.71 V.
2. K⁺/K half-cell: The half-reaction for the K⁺/K half-cell can be written as:
K⁺(aq) + e⁻ → K(s)
The standard reduction potential, E°, for this half-reaction is -2.92 V.
To determine the overall electrochemical reaction, you need to combine the two half-reactions in a way that cancels out the electrons to balance the equation. Since the half-reactions are reduction reactions, you need to reverse the oxidation half-reaction (K⁺/K) and multiply it by the appropriate number of electrons to balance the charge.
By reversing the K⁺/K half-reaction and multiplying by the appropriate number of electrons, the overall balanced equation, when the two half-reactions are added together, becomes:
2Na⁺(aq) + 2K(s) → 2Na(s) + 2K⁺(aq)
The E° value for the overall reaction is the sum of the standard reduction potentials of the two half-reactions. Therefore, the E° value for the overall electrochemical reaction is:
E° = (+2.71 V) + (-2.92 V)
E° = -0.21 V
Thus, the overall electrochemical reaction that proceeds spontaneously is:
2Na⁺(aq) + 2K(s) → 2Na(s) + 2K⁺(aq)
with an E° value of -0.21 V.