hce_nthu
113年
化學與物理
第 6 題
Given the following standard reduction potentials at $25^{\circ}\text{C}$:
$\text{Cu}^{2+}\text{(aq)} + 2e^- \rightarrow \text{Cu(s)} \quad E^{\circ} = 0.34 \text{ V}$
$\text{Zn}^{2+}\text{(aq)} + 2e^- \rightarrow \text{Zn(s)} \quad E^{\circ} = -0.76 \text{ V}$
In a galvanic cell, the copper compartment contains a copper electrode and $[\text{Cu}^{2+}] = 1.00\text{ M}$. The zinc compartment contains a zinc electrode and $\text{Zn}^{2+}$ at an unknown concentration. The compartment containing the zinc ($1.00\text{ L}$ of solution) was titrated with $0.600\text{ M H}_2\text{EDTA}^{2-}$, resulting in the reaction:
$\text{H}_2\text{EDTA}^{2-} + \text{Zn}^{2+} \rightleftharpoons \text{Zn(EDTA)}^{2-} + 2\text{ H}^+$
The potential of the cell was monitored to determine the stoichiometric point for the process, which occurred when $400.0\text{ mL}$ of $\text{H}_2\text{EDTA}^{2-}$ solution was added. What is $E_{\text{cell}}$ at the halfway point of titration?
$\text{Cu}^{2+}\text{(aq)} + 2e^- \rightarrow \text{Cu(s)} \quad E^{\circ} = 0.34 \text{ V}$
$\text{Zn}^{2+}\text{(aq)} + 2e^- \rightarrow \text{Zn(s)} \quad E^{\circ} = -0.76 \text{ V}$
In a galvanic cell, the copper compartment contains a copper electrode and $[\text{Cu}^{2+}] = 1.00\text{ M}$. The zinc compartment contains a zinc electrode and $\text{Zn}^{2+}$ at an unknown concentration. The compartment containing the zinc ($1.00\text{ L}$ of solution) was titrated with $0.600\text{ M H}_2\text{EDTA}^{2-}$, resulting in the reaction:
$\text{H}_2\text{EDTA}^{2-} + \text{Zn}^{2+} \rightleftharpoons \text{Zn(EDTA)}^{2-} + 2\text{ H}^+$
The potential of the cell was monitored to determine the stoichiometric point for the process, which occurred when $400.0\text{ mL}$ of $\text{H}_2\text{EDTA}^{2-}$ solution was added. What is $E_{\text{cell}}$ at the halfway point of titration?
- A $1.04 \text{ V}$
- B $1.07 \text{ V}$
- C $1.10 \text{ V}$
- D $1.13 \text{ V}$
- E $1.16 \text{ V}$
思路引導 VIP
如果在實驗中，我們已知反應達到「半當量點」，而這時除了溶質的莫耳數減半之外，溶液的總體積也隨之改變了，這會如何影響能斯特方程式中反應商 $Q$ 的數值呢？
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AI 詳解
AI 專屬家教
非常好！你能精準算出 $1.13 \text{ V}$，代表你對電位計滴定與能斯特方程式 (Nernst equation) 的結合應用掌握得十分紮實。
化學計量與濃度運算
首先，我們透過滴定終點（當量點）得知 $\text{Zn}^{2+}$ 的初始量：滴定耗費 $0.400 \text{ L} \times 0.600 \text{ M} = 0.240 \text{ mol}$ 的 $\text{H}_2\text{EDTA}^{2-}$。當反應進行到半當量點時，剛好有一半的鋅離子被絡合，此時溶液中剩餘 $0.120 \text{ mol}$ 的 $\text{Zn}^{2+}$。關鍵在於體積的變化：此時總體積為 $1.00 \text{ L} + 0.200 \text{ L} = 1.20 \text{ L}$，因此鋅離子濃度變為 $[\text{Zn}^{2+}] = 0.120 / 1.20 = 0.100 \text{ M}$。
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