
a. Fe'is titrated with Cell. What is the potential at the indicator electrode 2/5 of the...
If a platinum indicator electrode is used to measure [Cu^22^++]
(standard reduction potential of Cu^22^++ is 0.337 V), and
[Cu^22^++] = 0.0053 M, what is the cell potential (V) if the
reference electrode is SCE and the measurement is taken at 298
K?
Indicator Electrode--Quantification (Homework) Homework Unanswered If a platinum indicator electrode is used to measure [Cu2+] (standard reduction potential of Cu +is 0.337 V), and [Cu2+ o.0053 M, what is the cell potential (V) if the reference electrode...
A solution of K:Cr04 was titrated with AgNO, using a silver indicator electrode and a s.??. reference electrode. The product, Ag2Cr04, has a Ksp = 1.1 x 10-12. Eo-0.558 Then potential at the equivalence point would be 0.328 V Hint: 1. Use Ksp to calculate [Ag 2. Use Nernst equation 3. Decide which form is more available based on the pH 4. Do not forget Nernst equation
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A cell composed of a platinum indicator electrode and a silver-silver chloride reference electrode in a solution containing both Fe2* and Fe3+ has a cell potential of 0.699 V. If the silver-silver chloride electrode is replaced with a saturated calomel electrode (SCE), what is the new cell potential? Number 0.675
In a 3-electrode cell filled with a solution of dopamine, the potential of the working electrode is scanned from +0.2 V to +0.7 V vs. a Ag/AgCl reference electrode. The current rises exponentially and then peaks at +0.48 V followed by a decay (gradual decrease) in current. What causes the decrease in current? The concentration of dopamine near the electrode decreases. All of the dopamine in the solution is consumed. A second reaction begins to interfere at such positive potentials....
Calculate the potential of the indicator electrode when 40% of Fe2+ has been titrated with Ce(IV). Fe2+ +Ce4+ <--> Fe3+ + Ce3+ (Fe3+/Fe2+): E=0.771 V (Ce4+/Ce3+): E=1.44 V
A 100.00mL solution of 0.100 M NaCl was titrated with 0.100 M AgNO3. A SCE indicator electrode (E+=0.241V) was the anode and an Ag wire was the cathode for this precipitation titration. Calculate the voltage after the addition of 65 mL of AgNO3 given that Ksp(AgCl)=1.8x10^(-10). Answer=0.081 V. Please show steps, thank you! :)
Questions Consider an electrochemical cell consisting of an Fe(s) electrode, Fe(NO3)2 electrolyte connected through a salt bridge to a Ag wire coated in AgCl(s) immersed in an aqueous KCl solution. Is the standard cell and balanced galvanic cell reaction (If needed, refer to Table 17-1.) 0.669 V, Fe(8) + Agci(9) - Fel*(0) + Ag(s) + Cl(aq) 0.669 V, Fe(s) + 2ACH() - Tel(aq) + 2A(8) + 2Cl(9) -0.669 V, Fe(s) + 2Ags) - Fed(0) + 3A(s) + 3Cl(aq) -0.225 V,...
The standard reduction potential of the Ag Ag electrode is +0.80 V and the standard potential of the cell Fe(s) Fe3+(aq) || Ag" (aq) Ag(s) is +0.84 V. What is the standard reduction potential of the Fe3+1Fe electrode? +1.64 V +0.04 V O-1.64v -0.04 v -0.12 V Question 10 (1 point) In the following cell, A is a standard Co2+1Co electrode connected to a standard hydrogen electrode. If the voltmeter reading is -0.28 V, which half-reaction occurs in the left-hand...
What is the standard cell potential of a cell created using a cadmium electrode in 0.45 M Cd(NO3)2(aq) and a graphite electrode in an aqueous solution that is 0.45 M in both Fe2+ and Fe3+? 0.38 V 1.95 V 1.94 V 1.17 V 1.18 V What is the standard cell potential of a cell which utilizes the following reaction: Zn (s) + Ni2+ (aq) → Ni (s) + Zn2+ (aq) All solutions are 1.0 M and the reaction occurs at...
A 100.0 mL solution of 0.0200 M Fe 3 + in 1 M HClO 4 is titrated with 0.100 M Cu + , resulting in the formation of Fe 2 + and Cu 2 + . A Pt indicator electrode and a saturated Ag ∣ ∣ AgCl electrode are used to monitor the titration. Write the balanced titration reaction. titration reaction: -> ⟶ Complete the two half‑reactions that occur at the Pt indicator electrode. Write the half‑reactions as reductions. half‑reaction:...