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Sure, I'll walk you through the step-by-step solution for calculating the reduction potential [tex]\( E_{\text{red}} \)[/tex] for the given half-cell reaction involving [tex]\( \text{Cr}_2\text{O}_7^{2-} \)[/tex] and [tex]\( \text{Cr}^{3+} \)[/tex].
### Step-by-Step Solution:
1. Identify Given Data:
- Standard electrode potential, [tex]\( E^\circ = 1.5 \)[/tex] V
- Concentration of [tex]\( \text{Cr}_2\text{O}_7^{2-} \)[/tex] is [tex]\( 4 \times 10^{-4} \)[/tex] M
- Concentration of [tex]\( \text{Cr}^{3+} \)[/tex] is [tex]\( 2 \times 10^{-2} \)[/tex] M
- pH of the solution is 1
2. Determine the number of electrons transferred ([tex]\( n \)[/tex]):
The balanced reaction is:
[tex]\[ \text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ + 6 \text{e}^- \rightarrow 2 \text{Cr}^{3+} + 7 \text{H}_2\text{O} \][/tex]
From this equation, we see that 6 electrons ([tex]\( n = 6 \)[/tex]) are transferred in the reaction.
3. Calculate the concentration of [tex]\(\text{H}^+\)[/tex] from pH:
The pH is given as 1. Using the definition of pH:
[tex]\[ \text{pH} = -\log[\text{H}^+] \][/tex]
Thus,
[tex]\[ [\text{H}^+] = 10^{-1} = 0.1 \text{ M} \][/tex]
4. Calculate the reaction quotient ([tex]\( Q \)[/tex]):
The reaction quotient [tex]\( Q \)[/tex] is given by:
[tex]\[ Q = \frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}] \cdot [\text{H}^+]^{14}} \][/tex]
Plugging in the given concentrations:
[tex]\[ [\text{Cr}_2\text{O}_7^{2-}] = 4 \times 10^{-4} \text{ M} \][/tex]
[tex]\[ [\text{Cr}^{3+}] = 2 \times 10^{-2} \text{ M} \][/tex]
[tex]\[ [\text{H}^+] = 0.1 \text{ M} \][/tex]
Thus,
[tex]\[ Q = \frac{(2 \times 10^{-2})^2}{(4 \times 10^{-4}) \cdot (0.1)^{14}} \][/tex]
Simplifying:
[tex]\[ Q = \frac{4 \times 10^{-4}}{4 \times 10^{-4} \cdot 10^{-14}} = \frac{4 \times 10^{-4}}{4 \times 10^{-18}} = 10^{14} = 99999999999999.92 \][/tex]
5. Use the Nernst Equation to find [tex]\( E_{\text{red}} \)[/tex]:
The Nernst equation is:
[tex]\[ E = E^\circ - \frac{0.0591}{n} \log_{10}(Q) \][/tex]
Substituting the known values:
[tex]\[ E = 1.5 - \frac{0.0591}{6} \log_{10}(99999999999999.92) \][/tex]
Evaluate the term:
[tex]\[ \log_{10}(99999999999999.92) \approx 14 \][/tex]
So,
[tex]\[ E = 1.5 - \frac{0.0591}{6} \times 14 = 1.5 - 0.1379 = 1.3621 \text{ V} \][/tex]
### Final Answer:
The reduction potential [tex]\( E_{\text{red}} \)[/tex] for the given half-cell reaction is [tex]\( 1.3621 \text{ V} \)[/tex].
### Step-by-Step Solution:
1. Identify Given Data:
- Standard electrode potential, [tex]\( E^\circ = 1.5 \)[/tex] V
- Concentration of [tex]\( \text{Cr}_2\text{O}_7^{2-} \)[/tex] is [tex]\( 4 \times 10^{-4} \)[/tex] M
- Concentration of [tex]\( \text{Cr}^{3+} \)[/tex] is [tex]\( 2 \times 10^{-2} \)[/tex] M
- pH of the solution is 1
2. Determine the number of electrons transferred ([tex]\( n \)[/tex]):
The balanced reaction is:
[tex]\[ \text{Cr}_2\text{O}_7^{2-} + 14 \text{H}^+ + 6 \text{e}^- \rightarrow 2 \text{Cr}^{3+} + 7 \text{H}_2\text{O} \][/tex]
From this equation, we see that 6 electrons ([tex]\( n = 6 \)[/tex]) are transferred in the reaction.
3. Calculate the concentration of [tex]\(\text{H}^+\)[/tex] from pH:
The pH is given as 1. Using the definition of pH:
[tex]\[ \text{pH} = -\log[\text{H}^+] \][/tex]
Thus,
[tex]\[ [\text{H}^+] = 10^{-1} = 0.1 \text{ M} \][/tex]
4. Calculate the reaction quotient ([tex]\( Q \)[/tex]):
The reaction quotient [tex]\( Q \)[/tex] is given by:
[tex]\[ Q = \frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}] \cdot [\text{H}^+]^{14}} \][/tex]
Plugging in the given concentrations:
[tex]\[ [\text{Cr}_2\text{O}_7^{2-}] = 4 \times 10^{-4} \text{ M} \][/tex]
[tex]\[ [\text{Cr}^{3+}] = 2 \times 10^{-2} \text{ M} \][/tex]
[tex]\[ [\text{H}^+] = 0.1 \text{ M} \][/tex]
Thus,
[tex]\[ Q = \frac{(2 \times 10^{-2})^2}{(4 \times 10^{-4}) \cdot (0.1)^{14}} \][/tex]
Simplifying:
[tex]\[ Q = \frac{4 \times 10^{-4}}{4 \times 10^{-4} \cdot 10^{-14}} = \frac{4 \times 10^{-4}}{4 \times 10^{-18}} = 10^{14} = 99999999999999.92 \][/tex]
5. Use the Nernst Equation to find [tex]\( E_{\text{red}} \)[/tex]:
The Nernst equation is:
[tex]\[ E = E^\circ - \frac{0.0591}{n} \log_{10}(Q) \][/tex]
Substituting the known values:
[tex]\[ E = 1.5 - \frac{0.0591}{6} \log_{10}(99999999999999.92) \][/tex]
Evaluate the term:
[tex]\[ \log_{10}(99999999999999.92) \approx 14 \][/tex]
So,
[tex]\[ E = 1.5 - \frac{0.0591}{6} \times 14 = 1.5 - 0.1379 = 1.3621 \text{ V} \][/tex]
### Final Answer:
The reduction potential [tex]\( E_{\text{red}} \)[/tex] for the given half-cell reaction is [tex]\( 1.3621 \text{ V} \)[/tex].
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