Answer :
To determine which electrode will be positive and what voltage will be shown on the voltmeter, we need to apply the Nernst equation for the given galvanic cell setup.
1. Nernst Equation:
[tex]\[ E = \frac{RT}{nF} \ln \left(\frac{C_2}{C_1}\right) \][/tex]
where:
- [tex]\( E \)[/tex] is the cell potential (voltage),
- [tex]\( R \)[/tex] is the universal gas constant (8.314 J/(mol·K)),
- [tex]\( T \)[/tex] is the temperature in Kelvin,
- [tex]\( n \)[/tex] is the number of moles of electrons transferred,
- [tex]\( F \)[/tex] is Faraday's constant (96485 C/mol),
- [tex]\( C_2 \)[/tex] is the concentration of the right half cell,
- [tex]\( C_1 \)[/tex] is the concentration of the left half cell.
2. Given Data:
- Left half cell concentration ([tex]\( C_1 \)[/tex]): 20.0 mM = 0.020 M,
- Right half cell concentration ([tex]\( C_2 \)[/tex]): 4.00 M,
- Temperature ([tex]\( T \)[/tex]): 30.0°C = 303.15 K (after converting from Celsius to Kelvin).
3. Constants:
- [tex]\( R = 8.314\ \text{J/(mol·K)} \)[/tex],
- [tex]\( F = 96485\ \text{C/mol} \)[/tex],
- [tex]\( n = 2 \)[/tex] (since the metal [tex]\( M \)[/tex] typically forms [tex]\( M^{2+} \)[/tex] ions in solution).
4. Calculating the Voltage (E):
[tex]\[ E = \frac{(8.314\ \text{J/(mol·K)} \times 303.15\ \text{K})}{(2 \times 96485\ \text{C/mol})} \ln \left(\frac{4.00}{0.020}\right) \][/tex]
[tex]\[ E = \frac{(8.314 \times 303.15)}{(2 \times 96485)} \ln (200) \][/tex]
[tex]\[ E \approx 0.07\ \text{V} \][/tex]
5. Determining the Positive Electrode:
To identify which electrode will be positive, observe the sign of [tex]\( E \)[/tex]:
- If [tex]\( E \)[/tex] is positive, the right electrode is positive because [tex]\( \frac{C_2}{C_1} > 1 \)[/tex], which implies a spontaneous reaction in that direction.
Given our calculation shows [tex]\( E \approx 0.07\ \text{V} \)[/tex], which is positive, the right electrode is the positive electrode.
6. Final Answer:
- Positive Electrode: Right
- Voltage: [tex]\( 0.07\ \text{V} \)[/tex]
So, filling out the table:
[tex]\[ \begin{array}{|c|c|} \hline \text{Which electrode will be positive?} & \text{Right} \\ \hline \begin{array}{l} \text{What voltage will the voltmeter show? Assume its positive lead is connected to the} \\ \text{positive electrode.} \\ \text{Be sure your answer has a unit symbol, if necessary, and round it to 2 significant} \\ \text{digits.} \end{array} & 0.07\ \text{V} \\ \hline \end{array} \][/tex]
1. Nernst Equation:
[tex]\[ E = \frac{RT}{nF} \ln \left(\frac{C_2}{C_1}\right) \][/tex]
where:
- [tex]\( E \)[/tex] is the cell potential (voltage),
- [tex]\( R \)[/tex] is the universal gas constant (8.314 J/(mol·K)),
- [tex]\( T \)[/tex] is the temperature in Kelvin,
- [tex]\( n \)[/tex] is the number of moles of electrons transferred,
- [tex]\( F \)[/tex] is Faraday's constant (96485 C/mol),
- [tex]\( C_2 \)[/tex] is the concentration of the right half cell,
- [tex]\( C_1 \)[/tex] is the concentration of the left half cell.
2. Given Data:
- Left half cell concentration ([tex]\( C_1 \)[/tex]): 20.0 mM = 0.020 M,
- Right half cell concentration ([tex]\( C_2 \)[/tex]): 4.00 M,
- Temperature ([tex]\( T \)[/tex]): 30.0°C = 303.15 K (after converting from Celsius to Kelvin).
3. Constants:
- [tex]\( R = 8.314\ \text{J/(mol·K)} \)[/tex],
- [tex]\( F = 96485\ \text{C/mol} \)[/tex],
- [tex]\( n = 2 \)[/tex] (since the metal [tex]\( M \)[/tex] typically forms [tex]\( M^{2+} \)[/tex] ions in solution).
4. Calculating the Voltage (E):
[tex]\[ E = \frac{(8.314\ \text{J/(mol·K)} \times 303.15\ \text{K})}{(2 \times 96485\ \text{C/mol})} \ln \left(\frac{4.00}{0.020}\right) \][/tex]
[tex]\[ E = \frac{(8.314 \times 303.15)}{(2 \times 96485)} \ln (200) \][/tex]
[tex]\[ E \approx 0.07\ \text{V} \][/tex]
5. Determining the Positive Electrode:
To identify which electrode will be positive, observe the sign of [tex]\( E \)[/tex]:
- If [tex]\( E \)[/tex] is positive, the right electrode is positive because [tex]\( \frac{C_2}{C_1} > 1 \)[/tex], which implies a spontaneous reaction in that direction.
Given our calculation shows [tex]\( E \approx 0.07\ \text{V} \)[/tex], which is positive, the right electrode is the positive electrode.
6. Final Answer:
- Positive Electrode: Right
- Voltage: [tex]\( 0.07\ \text{V} \)[/tex]
So, filling out the table:
[tex]\[ \begin{array}{|c|c|} \hline \text{Which electrode will be positive?} & \text{Right} \\ \hline \begin{array}{l} \text{What voltage will the voltmeter show? Assume its positive lead is connected to the} \\ \text{positive electrode.} \\ \text{Be sure your answer has a unit symbol, if necessary, and round it to 2 significant} \\ \text{digits.} \end{array} & 0.07\ \text{V} \\ \hline \end{array} \][/tex]
