They tell you to have caution.
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Answer:</h3>
A. 1.4 V
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Explanation:</h3>
We are given the half reactions;
Ni²⁺(aq) + 2e → Ni(s)
Al(s) → Al³⁺(aq) + 3e
We are required to determine the cell potential of an electrochemical cell with the above half-reactions.
E°cell = E(red) - E(ox)
From the above reaction;
Ni²⁺ underwent reduction(gain of electrons) to form Ni
Al on the other hand underwent oxidation (loss of electrons) to form Al³⁺
The E.m.f of Ni/Ni²⁺ is -0.25 V and that of Al/Al³⁺ is -1.66 V
Therefore;
E°cell = -0.25 V - (-1.66 V)
= -0.250V + 1.66 V
= + 1.41 V
= + 1.4 V
Therefore, the cell potential will be +1.4 V
C) Noble gases
The six noble gases are helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn). Their atomic numbers are, respectively, 2, 10, 18, 36, 54, and 86.
Answer:
E° = 1.24 V
Explanation:
Let's consider the following galvanic cell: Fe(s) | Fe²⁺(aq) || Ag⁺(aq) | Ag(s)
According to this notation, Fe is in the anode (where oxidation occurs) and Ag is in the cathode (where reduction occurs). The corresponding half-reactions are:
Anode: Fe(s) ⇒ Fe²⁺(aq) + 2 e⁻
Cathode: Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an
E° = 0.80 V - (-0.44 V) = 1.24 V
Answer:
212°F , 100°C , 373.15°K
They are all the same, in different units.
