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Math  /  Algebra

Question5. Calculate the activity coefficients of sodium and chloride ions for a 0.04 M NaCl solution
6. What is an electrode of the first kind? Write a general Nernst equation for this indicator electrode.
7. A solution of 0.25MCr2O72−0.25 \mathrm{M} \mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-} and 0.35MCr3+0.35 \mathrm{M} \mathrm{Cr}^{3+} has a pH of 2.0 . What is the Ecell \mathrm{E}_{\text {cell }} of this half-cell?

Studdy Solution

STEP 1

1. We are given a solution of 0.25 M Cr2O72−0.25 \, \text{M} \, \text{Cr}_2\text{O}_7^{2-} and 0.35 M Cr3+0.35 \, \text{M} \, \text{Cr}^{3+}.
2. The solution has a pH of 2.0.
3. We need to calculate the EcellE_{\text{cell}} of this half-cell.
4. We assume standard conditions for temperature (25°C or 298 K).
5. The Nernst equation will be used to find the cell potential.

STEP 2

1. Write the balanced half-reaction.
2. Determine the standard reduction potential.
3. Use the Nernst equation to find the cell potential.

STEP 3

Write the balanced half-reaction for the reduction of Cr2O72−\text{Cr}_2\text{O}_7^{2-} to Cr3+\text{Cr}^{3+}.
The balanced half-reaction in acidic solution is:
Cr2O72−+14H++6e−→2Cr3++7H2O \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- \rightarrow 2\text{Cr}^{3+} + 7\text{H}_2\text{O}

STEP 4

Determine the standard reduction potential (E∘E^\circ) for the half-reaction.
The standard reduction potential for the reaction is typically found in a table of standard electrode potentials:
E∘=1.33 V E^\circ = 1.33 \, \text{V}

STEP 5

Use the Nernst equation to find the cell potential (EcellE_{\text{cell}}).
The Nernst equation is given by:
E=E∘−RTnFln⁡Q E = E^\circ - \frac{RT}{nF} \ln Q
Where: - EE is the cell potential. - E∘E^\circ is the standard reduction potential. - RR is the universal gas constant (8.314 J/mol K8.314 \, \text{J/mol K}). - TT is the temperature in Kelvin (298 K298 \, \text{K}). - nn is the number of moles of electrons transferred in the reaction (66 for this reaction). - FF is Faraday's constant (96485 C/mol96485 \, \text{C/mol}). - QQ is the reaction quotient.
Calculate QQ:
Q=[Cr3+]2[Cr2O72−][H+]14 Q = \frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}][\text{H}^+]^{14}}
Given: - [Cr3+]=0.35 M[\text{Cr}^{3+}] = 0.35 \, \text{M} - [Cr2O72−]=0.25 M[\text{Cr}_2\text{O}_7^{2-}] = 0.25 \, \text{M} - [H+]=10−2 M[\text{H}^+] = 10^{-2} \, \text{M} (from pH = 2)
Q=(0.35)20.25×(10−2)14 Q = \frac{(0.35)^2}{0.25 \times (10^{-2})^{14}}
Calculate EcellE_{\text{cell}}:
E=1.33−8.314×2986×96485ln⁡((0.35)20.25×(10−2)14) E = 1.33 - \frac{8.314 \times 298}{6 \times 96485} \ln \left( \frac{(0.35)^2}{0.25 \times (10^{-2})^{14}} \right)
E=1.33−0.02576ln⁡(0.12250.25×10−28) E = 1.33 - \frac{0.0257}{6} \ln \left( \frac{0.1225}{0.25 \times 10^{-28}} \right)
E=1.33−0.00428ln⁡(4.9×1026) E = 1.33 - 0.00428 \ln (4.9 \times 10^{26})
E=1.33−0.00428×61.6 E = 1.33 - 0.00428 \times 61.6
E=1.33−0.264 E = 1.33 - 0.264
E=1.066 V E = 1.066 \, \text{V}
The EcellE_{\text{cell}} of this half-cell is approximately 1.066 V1.066 \, \text{V}.

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