First of all, we need tô calculate moles of KCl
next, mass=mol×molar mass
Answer:
7 methyl 3 octyne is the answer option D
Answer:
E° = 0.00 V
E = 0.079 V
Explanation:
We can identify both half-reactions occurring in a concentration cell.
Anode (oxidation): Al(s) → Al³⁺(1.0 × 10⁻⁵ M) + 3 e⁻ E°red = -1.66 V
Cathode (reduction): Al³⁺(0.100 M) + 3 e⁻ → Al(s) E°red = -1.66 V
The global reaction is:
Al(s) + Al³⁺(0.100 M) → Al³⁺(1.0 × 10⁻⁵ M) + Al(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 = -1.66 V - (-1.66 V) = 0.00 V
To calculate the cell potential (E) we have to use the Nernst equation.
E = E° - (0.05916/n) .log Q
where,
n: moles of electrons transferred
Q: reaction quotient
E = 0.00 V - (0.05916/3) .log (1.0 × 10⁻⁵/0.100)
E = 0.079 V
