The measured Eocell=+0.583 V. Using EoCu2+/Cu= +0.337 V, calculate Eo Ni2+/Ni in volts Ni (s) | Ni2+ (aq) ||Cu2+ (aq) | Cu (s) no correct answer +0.920 -0.256 -0.246 +0.246
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The measured Eocell=+0.583 V. Using EoCu2+/Cu= +0.337 V, calculate Eo Ni2+/Ni in volts Ni (s) | Ni2+ (aq) ||Cu2+ (aq) | Cu (s)
no correct answer
+0.920
-0.256
-0.246
+0.246
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- Estimate Ecell in volts. Cu(s)|Cu2+(3.00x10-4)||Ag+(0.15M)|Ag(s) Eocell=+0.620 0.5 0.8 0.7 0.600 0.9The E°cell = 0.135 V for the reaction %3D 312(s) + 5Cr20,2(aq) + 34H*(aq) → 6103 (aq) + 10Cr³+(aq) + 17H2O What is Ecell if [Cr2072] = 0.044 M, [H*] = 0.16 M, [IO3] = 0.00016 M, and [Cr3*] = 0.0041 M? Ecell = i Vline notation Pt(s) | X+(aq), X2+(aq) || Y3+(aq) | Y(s) X2+ + e- X+ Eo = 0.34 V Y3+ + 3 e- Y Eo = -2.2 V Pt4+ + 4 e- Pt Eo = 1.20 V Calculate the EMF.
- XE"(aq) II Y3+ ELECTROCHEMISTRY: *(aq) I Y(s) Given the line notation: Pt(s) IX* (aq) where: x2+ + e - x+ E° = 1.3 V y3+ + 3 e - Y E° = -2.15 V Pt4+ + 4 e E° = 1.20 V Pt Calculate the EMF. Input values only with 2 decimal places. Do not include the unit. If positive value: DO NOT include the positive (+) sign If negative value: Please include the negative (-) signCalculate Ksp for Hg.Cl, from std potential data: Hg,Cl2(s) > Hg, + + 201 Mercury E°,V +0.788 Hg3* + 2e=2Hg(1) 2H9²+ + 2e¯= Hg3* Hg2+ + 2e-= Hg(l) Hg,Cl2(s)+ 2e¯ =2Hg(l) + 2Cl- Hg,SO4(s)+ 2e-=2Hg(1) + SO? +0.920 +0.854 +0.268 +0.615The described electrochemical cell (concentration cell) was set up: Cu|CuSO4 (c1) // CuSO4 (c2)|Cu Determine which electrode is the positive terminal (+) of the cell and calculate the cell potential E (electromotive force, EMF) for the given concentrations of the electrolyte CuSO4 in the cell (25oC): c1 = 0.01 mol dm-3, c2 = 1×10-4 mol dm-3
- Pt(s) IX*(aq), X2+, ELECTROCHEMISTRY: (aq) I| Y3+ *(aq) | Y(s) Given the line notation: where: x2+ + e - x+ E° = 1.17 V y3+ + 3 e - Y E° = -2.16 V Pt4+ + 4 e → Pt E° = 1.20 V Calculate the EMF. Input values only with 2 decimal places. Do not include the unit. If positive value: DO NOT include the positive (+) sign If negative value: Please include the negative (-) signA chloride ion selective electrode (ISE) calibration curve was measured to be E(mV, vs ref)=-109.8 + 54.5 x PCI; Concentration of CI is in mol/L If a chloride solution was placed in a beaker of sodium acetate and the potential vs the same reference electrode was E=72.47 mV. What is the concentration of chloride in mmol/L (mM)? Report your answer to 3 decimal places.Pt(s) |X* (aq), X2* (ag) || Y3+ Pt(s) |X' (aq). |x* Given the line notation: (aq) | Y(s) where: x2+ . x* E° = 0.84 V + e - y3+ + 3 e → Y E° = -1.56 V Pt4+ + 4 e → Pt E° = 1.20 V Calculate the EMF.
- ELECTROCHEMISTRY: x2+ (aq) || Y3+ Given the line notation: Pt(s) |X*(ag). (aq) IY(s) where: x2+ + e - x* E° = 0.47 V y3+ + 3 e - Y E° = -2.65 V Pt4+ + 4 e E° = 1.20 V - Pt %3D Calculate the EMF. Input values only with 2 decimal places. Do not include the unit. If positive value: DO NOT include the positive (+) sign If negative value: Please include the negative (-) sign2VO2"(aq) + Zn(e) + 4H*(ag) → 2V02*(ag) + Zn²*(aq) + 2H2OM a. How many moles of electrons are transferred overall? b. If E°cell = +1.76 V and Ecell = +1.25 V and [VO2*] and [VO2*] are both 1.0 M, and the pH is 5.00, what is the [Zn*]?Oxidizing Reducing - Eo Electrode E°cathode anode = E°cell Agent Agent Potential (V) Au3*(aq) + 3 e Au(s) +1.50 Br2(t) + 2 e 2 Br (aq) +1.08 Cathode Anode Hg2*(aq) + 2 e Hg(1) +0.855 Ag*(aq) + e Ag(s) +0.80 Hg22*(aq) + 2 e 2 Hg(t) +0.789 Cu2* (aq) + 2 e Cu(s) +0.337 Sn*(aq) + 2 e Sn2*(s) +0.15 2 H30*(aq) + 2 e H2(g) + 2 H20(s) 0.00 Sn2* (aq) + 2 e Ni2*(aq) + 2 e cd2*(aq) + 2 e Sn(s) -0.14 Ni(s) -0.25 Cd(s) -0.40 AI3* (ag) + 3 e Al(s) -1.66 Mg2+(aq) + 2 e Li*(aq) + e Mg(s) -2.37 Li(s) -3.045 Click on an Oxidizing Agent and a Reducing Agent. Select Ag+ as the oxidizing agent. What is the value of E°cathode? X V Energy Free Gibbs 1 1 1 1 1