To solve the problem of determining which electrode is positive and the voltage the voltmeter will show, we'll use the Nernst equation which is commonly applied to electrochemical cells.
### Step-by-Step Solution:
1. Identify the Key Information:
- Electrolyte Concentration in Left Half Cell: 30.0 mM ([tex]\(3.00 \times 10^{-2} M\)[/tex])
- Electrolyte Concentration in Right Half Cell: 3.00 M
- Temperature: [tex]\(30.0^{\circ}C\)[/tex]
2. Convert Temperature to Kelvin:
- Temperature in Kelvin ([tex]\(T\)[/tex]) can be calculated using the formula:
[tex]\[
T (K) = T (^{\circ}C) + 273.15
\][/tex]
Therefore,
[tex]\[
T = 30.0 + 273.15 = 303.15 \,K
\][/tex]
3. Calculate the Reaction Quotient (Q):
- The reaction quotient [tex]\(Q\)[/tex] is the ratio of the ion concentrations in the two half-cells. Since the left half-cell has a concentration of 30.0 mM and the right half-cell has a concentration of 3.00 M, we get:
[tex]\[
Q = \frac{[\text{electrolyte}_{\text{left}}]}{[\text{electrolyte}_{\text{right}}]} = \frac{30.0 \times 10^{-3} \,M}{3.00 \,M} = 0.01
\][/tex]
4. Determine the Positive Electrode:
- Generally, in a galvanic cell, the cell with the higher concentration is the positive electrode. Thus, the right half-cell (3.00 M) will be the positive electrode in this case.
5. Use the Nernst Equation to Calculate Voltage:
- The Nernst equation at temperature [tex]\(T\)[/tex] with gas constant [tex]\(R\)[/tex] and Faraday's constant [tex]\(F\)[/tex] is given by:
[tex]\[
E = \frac{RT}{nF} \ln{\left(\frac{1}{Q}\right)}
\][/tex]
Here, [tex]\(R = 8.314 \, \text{J/mol} \cdot K\)[/tex], [tex]\(F = 96485 \, \text{C/mol}\)[/tex], and [tex]\(n = 2\)[/tex] for the sulfate ion (assuming it loses or gains two electrons during the reduction/oxidation).
Inserting the values:
[tex]\[
E = \frac{8.314 \, \text{J/mol} \cdot K \times 303.15 \,K}{96485 \, \text{C/mol}} \ln{\left(\frac{1}{0.01}\right)}
= \left(\frac{8.314 \times 303.15}{96485}\right) \ln{(100)}
= \left(\frac{2519.6511}{96485}\right) \times 4.60517
= 0.0261 \times 4.60517 \approx 0.1203 \,V
\][/tex]
6. Final Answer:
- The positive electrode will be the right electrode.
- The voltmeter will show a voltage of approximately [tex]\(0.12 \,V\)[/tex].
Thus, summarizing the answers:
- Which electrode will be positive? The right electrode.
- What voltage will the voltmeter show? [tex]\(\boxed{0.12 \,V}\)[/tex]
