selectivity coefficient Kot immersed in 1.00 mM NaCl at pH 8.00, a potential of -38 mV (versus S.C.E.) was recorded. (a) Neglecting activity coefficients, calculate the potential with Equation 14-10 if the electrode were immersed in 5.00 mM NaCl at pH 8.00. (b) What would the potential be for 1.00 mM NaCl at pH 3.87? You can see that pH is a critical variable for the sodium electrode 36. When this electrode was H*

selectivity coefficient Kot immersed in 1.00 mM NaCl at pH 8.00, a potential of -38 mV (versus S.C.E.) was recorded. (a) Neglecting activity coefficients, calculate the potential with Equation 14-10 if the electrode were immersed in 5.00 mM NaCl at pH 8.00. (b) What would the potential be for 1.00 mM NaCl at pH 3.87? You can see that pH is a critical variable for the sodium electrode 36. When this electrode was H*

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Indicate the option that is true:(1). An indicator electrode has an upper limit of measurement; That is, it does not serve as a precise analytical instrument when the concentrations of the analyte are very high.(2). Some electrodes take a long time to respond, but that is usually not an insurmountable drawback.(3). The representation of the potential of an indicator electrode against the analyte concentration is a straight line for analyte concentrations below a certain value.(4). For activities lower than the lower detection limit, the electrode always measures the same potential.
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For the following electrode, obtain the equation that relates the potential it develops as a function of pY. How is that electrode classified? Pb / PbY2-(0.15M), Y4-(0.12M)
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e) Using the half reaction below sHE Pou3- HPO,2 + 2H* + 2e HPO+H;O E--0,234 V and acid dissociation constants for H3PO4 (K=7.11 × 10°; K2 = 6.32 × 10*, K7.1 x 1014), calculate standard reduction potential for the half reaction he Oy H2POª + H* + 2e HPO + H;O
Calculate potentials after the addition of 10.00, 25.00, 50.00, and 50.10 mL of the reagent. Where necessary, assume that [H+ ] = 1.00 throughout. (a) 50.00 mL of 0.1000 M V2+ with 0.05000 M Sn4+
6. f. Consider the following coupled redox reaction. O₂ + 2NADH + 2H* = 2H₂O + 2NAD+ Calculate AG" (at 310 K) if the standard reduction potentials for O2 and NAD+ are 0.815V and -0.315V respectively. g. If you wanted to shift the above reaction in f. to the left, how would you adjust the pH? Explain briefly. Lab-related questions. a. In lab, you made a 5 M NaCl stock solution. How would you make a 10- mL 0.5 M NaCl solution from your stock solution? i. 9 mL 5 M NaCl and 1 mL water ii. 5 mL 5 M NaCl and 5 mL water iii. 1 mL 5 M NaCl and 9 mL water iv. 0.1 mL 5 M NaCl and 9.9 mL water b. In the first week of lab, you made a pH 7.4 phosphate buffer by mixing some H₂PO4 with HPO4². The pK₂ of H₂PO4 is 7.2. Which of the following is true? i. ii. iii. iv. 7 More H₂PO4* needs to be added because the pH > pKa More HPO4²- needs to be added because the pH > pKa More H₂PO4 needs to be added because more acid is needed to maintain a pH of 7.4. More HPO4²- needs to be added because it is more…
The two solutions are contained (no dilution) in a single beaker with a saturated calomel electrode and an inert electrode (e.g., Pt) immersed in the solution. Relevant information: Cr0,/Cr*: Fe*/Fe*: For SCE: E° = +1.33 V E° = +0,77 V Erer +0.244 V Cr,O, + 6Fe" +14H" 2Cr + 6Fe* + 7H;0 The reaction quickly reaches equilibrium. What is the potential of this cell?
help please answer in text form with proper workings and explanation for each and every part and steps with concept and introduction no AI no copy paste remember answer must be in proper format with all working
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a- For electro chemical cell an Zn1 Zn*ce) || Cu co-sm)| Cur at constant tempecature reducing the foncentraldiens of of celt potentiat Value of 2+ both the Solutions, then value increaser then what would be the