Could someone please explain to me how we get to this answer?
I know (Ecell = Ecathode − Eanode)
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Could someone please explain to me how we get to this answer? I know (Ecell =...
Can someone please help me? I keep getting stuck, I don’t know how to get log on the other side of the equation. I can easily follow another tutorial, but I can’t figure out the ending and what step I need to do to get it! This was previously answered by someone else and was wrong. I need to know HOW to work it out please! Given the measured cell potential, Ecel, is-0.3555 V at 25 C in the following...
An electrochemical cell based on the following reaction has astandard cell voltage (Eocell) of 0.48 V Sn (s) + Cu2+ (aq) ---> Sn2+ (aq) + Cu (s) What is the standard reduction potential of tin? Sn2+ (aq) +2e- ---> Sn(s) a. -0.14 V b. 0.14 V c. -0.82 V d. 0.82 V e. none of the above
For the cell shown, the measured cell potential, Ecell, is -0.3629 V at 25 °C. Pt(s) | H,(g, 0.707 atm) | H+ (aq, ? M) || Cd2+ (aq, 1.00 M)| Cd(s) The balanced reduction half-reactions for the cell, and their respective standard reduction potential values, Eº, are 2 H+ (aq) + 2e - H (9) E° = 0.00 V Cd2+ (aq) + 2e — Cd(s) E' = -0.403 V Calculate the H+ concentration. M = 0.031 Incorrect
For the cell shown, the measured cell potential, Ecell, is -0.3687 V at 25 °C. Pt(s) | H, (8,0.873 atm) | H+ (aq, ? M) || Cd2+ (aq, 1.00 M) | Cd(s) The balanced reduction half-reactions for the cell, and their respective standard reduction potential values, Eº, are 2 H+ (aq) + 2e → H, (g) E° = 0.00 v Cd2+ (aq) + 2e- Cd(s) E° = -0.403 V Calculate the H+ concentration. [H+) = .0588 M Incorrect
ALEKS data table may not include the value, if not please just include the E value and I will go through the table and leave the answer in the comments A certain half-reaction has a standard reduction potential Ered- provide at least 0.50 V of electrical power. The cell will operate under standard conditions. Note for advanced students: assume the engineer requires this half-reaction to happen at the anode of the cell 0.53 V. An engineer proposes using this half-reaction...
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...
QUESTION 4 Using the table below and the Nearnst Equation E-E° (0.0592/n)logQ Find the voltage of the cell: Cu2+(aq) + Ni(s) ? Cu(s) + Ni2+(aq) at 25° C if the concentrations of the soluble species are [Cu2+1-0.050 M and [Ni]- 1.40 M Reduction Reaction Sn2++ 2e-? Sn Cu2+ + 2e- ?Cu" Reduction Potential, Ecell 0.14 Volts 0.34 Volts 0.25 Volts +0.77 Volts Fe3+ + e-? Fe2+ 0.50 V 0.54 V 0.62 V 0.66 V
please solve A to d to get thumps up. Question 11 5 pts Consider a galvanic cell consisting of the following two redox couples: Sn2+ (aq, 0.010 M) + 2e --> Sn (s) E.-0.14 V Cst (aq, 0.010 M) + e --> Cs(s) E--2.92 V a. Write the equation for the half-reaction at the cathode Question 12 5 pts b. Write the equation for the half-reaction at the anode HTML Editor Question 13 6 pts c. Write a shorthand cell...
Hi can someone help me with these questions, Thank you: Question 8 (1 point) The standard potential of the cell Ni(s) Ni2+(aq) || Cl(aq) AgCl(s) Ag(s) is +0.45 V at 25°C. If the standard reduction potential of the AgCl|Ag|Ci couple is +0.22 V, calculate the standard reduction potential of the Ni2+INi couple. -0.45 V +0.23 V -0.67 v +0.67 v 0 -0.23 V Question 6 (1 point) The standard reduction potential of the Br2/Brand Sna/Sn couples are +1.07 and -0.14...
A) Use tabulated electrode potentials to calculate ΔG∘ for the reaction. 2K(s)+2H2O(l)→H2(g)+2OH−(aq)+2K+(aq) B) (Refer to the following standard reduction half-cell potentials at 25∘C: VO2+(aq)+Ni2+(aq)2H+(aq)++2e−e−→ →Ni(s)VO2+(aq) +H2O(l)E∘=−0.23V E∘=0.99V) An electrochemical cell is based on these two half-reactions: Oxidation:Reduction:Ni(s)VO2+(aq,0.024M)+2H+(aq,1.4M)+e−→→Ni2+(aq,1.8M)+2e−VO2+(aq,1.8M)+H2O(l) Calculate the cell potential under these nonstandard concentrations. C) Standard reduction half-cell potentials at 25∘C Half-reaction E∘ (V ) Half-reaction E∘ (V ) Au3+(aq)+3e−→Au(s) 1.50 Fe2+(aq)+2e−→Fe(s) − 0.45 Ag+(aq)+e−→Ag(s) 0.80 Cr3+(aq)+e−→Cr2+(aq) − 0.50 Fe3+(aq)+3e−→Fe2+(aq) 0.77 Cr3+(aq)+3e−→Cr(s) − 0.73 Cu+(aq)+e−→Cu(s) 0.52 Zn2+(aq)+2e−→Zn(s) − 0.76...