For each of the following pairs of half-cells,
determine the overall electrochemical reaction
that proceeds spontaneously.
Na+ | Na, Ni2+ | Ni
What are your reduction potentials? My tables probably don't agree with your tables.
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To determine the overall spontaneous electrochemical reaction for each pair of half-cells (Na+ | Na, Ni2+ | Ni), we need to compare the reduction potentials of each half-cell.
First, let's write down the half-cell reactions:
1. NA+ | NA:
Na+ + e- -> Na
2. Ni2+ | Ni:
Ni2+ + 2e- -> Ni
Now, let's look up the reduction potentials (E°) for each half-cell reaction. The reduction potentials are typically provided in tables. Assuming standard conditions (1 M concentration and 25°C), the values are:
E°(Na+ | Na) = -2.71 V
E°(Ni2+ | Ni) = -0.26 V
To determine the overall electrochemical reaction, we need to compare the reduction potentials and consider the electrons transferred in the half-cell reactions.
Since the more positive reduction potential belongs to the Ni2+ | Ni half-cell, it means that Ni2+ is more easily reduced compared to Na+. Hence, the reduction of Ni2+ to Ni will occur spontaneously.
To balance the electrons and make both half-cell reactions compatible, we need to multiply the half-reactions by appropriate stoichiometric coefficients and combine them:
2(Na+ + e- -> Na) + Ni2+ + 2e- -> Ni
2Na+ + Ni2+ -> 2Na + Ni
Therefore, the overall electrochemical reaction that proceeds spontaneously for the given pair of half-cells is:
2Na+ + Ni2+ -> 2Na + Ni
To determine the overall electrochemical reaction for each pair of half-cells, we need to compare the standard reduction potentials (E°) of the half-reactions. The half-reactions with higher reduction potentials will be reduced, while those with lower reduction potentials will be oxidized.
Let's start with the first pair of half-cells:
Na+ | Na (sodium half-cell)
Ni2+ | Ni (nickel half-cell)
To find the reduction potentials for each half-reaction, we can refer to a standard reduction potential table. Let's assume the values of E° are as follows:
Na+ + e- → Na (E° = -2.71 V)
Ni2+ + 2e- → Ni (E° = -0.25 V)
The greater the value of E°, the stronger the reduction potential. In this case, since the reduction potential of Na+ | Na is more negative (-2.71 V) than that of Ni2+ | Ni (-0.25 V), the half-reaction with Ni2+ will be reduced and the half-reaction with Na+ will be oxidized.
Now, we can write the overall electrochemical reaction by multiplying the half-reactions by appropriate coefficients to balance the electrons. In this case, the Ni2+ half-reaction needs to be multiplied by 2 to balance the electrons:
2Na+ + 2e- → 2Na (oxidation)
Ni2+ + 2e- → Ni (reduction)
Now, we can add these two equations together to get the overall electrochemical reaction:
2Na+ + Ni2+ → 2Na + Ni
Therefore, the overall electrochemical reaction that proceeds spontaneously for the given pair of half-cells is:
2Na+ + Ni2+ → 2Na + Ni