Answer:
Option A:
Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)
Explanation:
The half reactions given are:
Zn(s) → Zn^(2+)(aq) + 2e^(-)
Cu^(2+) (aq) + 2e^(-) → Cu(s)
From the given half reactions, we can see that in the first one, Zn undergoes oxidation to produce Zn^(2+).
While in the second half reaction, Cu^(2+) is reduced to Cu.
Thus, for the overall reaction, we will add both half reactions to get;
Zn(s) + Cu^(2+) (aq) + 2e^(-) → Cu(s) + Zn^(2+)(aq) + 2e^(-)
2e^(-) will cancel out to give us;
Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)
Answer:
a chart
Explanation:
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The pH of a solution at 25. 0 °C that contains 2. 95 × 10^-12 m hydronium ions is 13.5.
<h3>What is pH? </h3>
pH is defined as the concentration of the hydrogen bond which is released or gained by the species in the solution which depicts the acidity and basicity of the solution.
<h3>What is pOH? </h3>
pOH is defined as the concentration of the hydronium ion present in solution.
pOH value is inversely proportional to the value of pH.
pH value increases, pOH value decreases and vice versa.
Given,
Total H+ ions = 2.95 ×10^(-12)M
<h3>Calculation of pH</h3>
pH = -log[H+]
By substituting the value of H+ ion in given equation
= log(2.95× 10^(-12) )
= 13.5
Thus we find that the pH of a solution at 25. 0 °C that contains 2. 95 × 10^-12 m hydronium ions is 13.5.
learn more about pH:
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