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 0.25MCr2O720.25 \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
Use the Nernst equation to find the cell potential (EcellE_{\text{cell}}).
The Nernst equation is given by:
E=ERTnFlnQ E = E^\circ - \frac{RT}{nF} \ln Q
Where: - EE is the cell potential. - EE^\circ is the standard reduction potential. - RR is the universal gas constant (8.314J/mol K8.314 \, \text{J/mol K}). - TT is the temperature in Kelvin (298K298 \, \text{K}). - nn is the number of moles of electrons transferred in the reaction (66 for this reaction). - FF is Faraday's constant (96485C/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+]=0.35M[\text{Cr}^{3+}] = 0.35 \, \text{M} - [Cr2O72]=0.25M[\text{Cr}_2\text{O}_7^{2-}] = 0.25 \, \text{M} - [H+]=102M[\text{H}^+] = 10^{-2} \, \text{M} (from pH = 2)
Q=(0.35)20.25×(102)14 Q = \frac{(0.35)^2}{0.25 \times (10^{-2})^{14}}
Calculate EcellE_{\text{cell}}:
E=1.338.314×2986×96485ln((0.35)20.25×(102)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=1.330.02576ln(0.12250.25×1028) E = 1.33 - \frac{0.0257}{6} \ln \left( \frac{0.1225}{0.25 \times 10^{-28}} \right)
E=1.330.00428ln(4.9×1026) E = 1.33 - 0.00428 \ln (4.9 \times 10^{26})
E=1.330.00428×61.6 E = 1.33 - 0.00428 \times 61.6
E=1.330.264 E = 1.33 - 0.264
E=1.066V E = 1.066 \, \text{V}
The EcellE_{\text{cell}} of this half-cell is approximately 1.066V1.066 \, \text{V}.

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