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QUESTION 4

Learners are given the following two unknown half-cells:

- Half-cell 1: \( Q^{2+}(aq) \mid Q(s) \)
- Half-cell 2: Pt \(\left| R_2(g) \right| R(aq) \)

During an investigation to identify the two half-cells, the learners connect each half-cell alternately to a \( Cd^{2+}(aq) \mid Cd(s) \) half-cell under standard conditions. For each combination of two half-cells, they write down the net cell reaction and measure the cell potential. The results obtained for the two half-cell combinations are given in the table below.

[tex]\[
\begin{tabular}{|c|c|c|}
\hline
COMBINATION & NET CELL REACTION & CELL POTENTIAL \\
\hline
I & \( Q^{2+}(aq) + Cd(s) \rightarrow Cd^{2+}(aq) + Q(s) \) & 0.13 V \\
\hline
II & \( R_2(g) + Cd(s) \rightarrow Cd^{2+}(aq) + 2R(aq) \) & 1.76 V \\
\hline
\end{tabular}
\][/tex]

4.1 Write down three conditions needed for these cells to function as standard cells.

4.2 For Combination I, identify:

4.2.1 The anode of the cell (3)

4.2.2 \( Q \) by using a calculation (1)

(5)

Sagot :

GinnyAnswer
Let's solve the given questions step by step.

### 4.1 Write down THREE conditions needed for these cells to function as standard cells.
For a cell to function as a standard cell, the following conditions need to be met:

1. Temperature: The temperature should be 25 degrees Celsius (298 K).
2. Concentration: The concentration of all aqueous solutions should be 1 Molar (1 M).
3. Pressure: The gas pressure should be 1 atmosphere (atm).

### 4.2 For Combination I, identify:

#### 4.2.1 The anode of the cell:
The anode is the electrode where oxidation occurs. In the net cell reaction for Combination I:
[tex]\[ Q^{2+}(aq) + Cd(s) \rightarrow Cd^{2+}(aq) + Q(s) \][/tex]
Cadmium (Cd) is being oxidized to \( Cd^{2+} \). Oxidation occurs at the anode, so the anode of the cell in Combination I is:
[tex]\[ \text{Anode: } Cd(s) \][/tex]

#### 4.2.2 Q by using a calculation:
To identify Q, we need to determine the reduction potential of the half-cell \( Q^{2+}/Q \).

The standard cell potential (E° cell) is given by the difference between the standard reduction potentials of the cathode and the anode:
[tex]\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \][/tex]

For Combination I, the measured cell potential is 0.13 V. The net cell reaction is:
[tex]\[ Q^{2+}(aq) + Cd(s) \rightarrow Cd^{2+}(aq) + Q(s) \][/tex]

Given:
- The standard reduction potential for \( Cd^{2+}/Cd \) is -0.40 V.
- The cell potential \( E^\circ_{\text{cell}} \) is 0.13 V.

Let \( E^\circ(Q^{2+}/Q) = x \). The cathode in this combination would be \( Q^{2+}/Q \), and the anode would be \( Cd^{2+}/Cd \).

Using the formula for the cell potential:
[tex]\[ 0.13 \, V = x - (-0.40 \, V) \][/tex]
[tex]\[ 0.13 \, V = x + 0.40 \, V \][/tex]
[tex]\[ x = 0.13 \, V - 0.40 \, V \][/tex]
[tex]\[ x = -0.27 \, V \][/tex]

Therefore, the reduction potential of \( Q^{2+}/Q \) is:
[tex]\[ E^\circ(Q^{2+}/Q) = -0.27 \, V \][/tex]

#### Final results:
1. Standard Conditions:
- Temperature: 25 degrees Celsius (298 K)
- Concentration: 1 Molar
- Pressure: 1 atmosphere (atm)

2. Anode in Combination I:
- Cd(s)

3. Reduction Potential of Q:
- [tex]\( -0.27 \, V \)[/tex]
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