**Voltaic Cell Calculation**A voltaic cell using Pb²⁺/Pb and Ni²⁺/Ni half-cells is set up at standard conditions, and each compartment has a volume of 355 mL. What is the cell potential, in volts (V), after it has delivered 0.150 amperes (A) for 41.0 hours at 25°C? (The standard electrode potential, E°, for Pb²⁺/Pb is -0.130 V and for Ni²⁺/Ni is -0.250 V.)**Cell Potential Calculation Interface:**- Numerical input interface for entering the calculated cell potential in volts.- A variety of numerical keys are available, including basic digits (0-9), a decimal point, and a sign change button (+/-).- Additional options include a backspace key (indicated by an arrow) and a clear key (C).- A function to multiply inputs by 10 is also present.Users can enter their answer in the designated space above the keypad.

Here we have to determine the Ecell of the following given standard galvanic cell after passing 0.15…

A voltaic cell using Pb²+/Pb and Ni²+/Ni half-cells is set up at standard conditions, and each compartment has a volume of 355 mL. What is the cell potential, in V, after it has delivered 0.150 A for 41.0 hours at 25°C? (E° for Pb²+/Pb = -0.130 V and Eº for Ni²+/Ni = -0.250 V.)

A voltaic cell using Pb²+/Pb and Ni²+/Ni half-cells is set up at standard conditions, and each compartment has a volume of 355 mL. What is the cell potential, in V, after it has delivered 0.150 A for 41.0 hours at 25°C? (E° for Pb²+/Pb = -0.130 V and Eº for Ni²+/Ni = -0.250 V.)

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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Question

**Voltaic Cell Calculation**

A voltaic cell using Pb²⁺/Pb and Ni²⁺/Ni half-cells is set up at standard conditions, and each compartment has a volume of 355 mL. What is the cell potential, in volts (V), after it has delivered 0.150 amperes (A) for 41.0 hours at 25°C? (The standard electrode potential, E°, for Pb²⁺/Pb is -0.130 V and for Ni²⁺/Ni is -0.250 V.)

**Cell Potential Calculation Interface:**

- Numerical input interface for entering the calculated cell potential in volts.
- A variety of numerical keys are available, including basic digits (0-9), a decimal point, and a sign change button (+/-).
- Additional options include a backspace key (indicated by an arrow) and a clear key (C).
- A function to multiply inputs by 10 is also present.

Users can enter their answer in the designated space above the keypad.

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2. (a) A voltaic cell is constructed as follows: Zn | Zn²+ | H*|H₂ The half-cell reactions are: Zn →Zn²+ze E = 0.76 V If the voltage of the cell is 0.45V at 25°C when [Zn²] = 1.0 M and PH₂ = 1.0 atm, what is [H+]? 2H+ + 2e- → H₂ EⓇ = 0.0V (b) For the voltaic cell described in part (a), find the pH of the cathodic solution when pH2 = 1.0 atm and [Zn²+]= 0.1 M and the E*cell = 0.542 V. (c) What amount of time is needed to plate 2.08 g of copper at a constant current flow of 1.26 A from a 0.1 M CUSO4 solution? Cu²+ +ze--- Cu(s)
Use the Nernst equation to determine the cell potential, Ecell, of the galvanic cell consisting of the following half-cell systems: Ag+(0.010M) + e- --> Ag Eo= +0.80 V Cr3+(O.010M) + 3e- --> Cr Eo= -0.74 V
4. A cell uses the following reaction: a. Zn (s) + 2 H* (aq) → Zn²* (aq) + H2 (g) (a) Determine the emf of the cell under standard conditions. (b) When PH,=1.0 atm and [Zn²*] = 1.0 M, the cell potential is 0.560 V. Determine the concentration of H* in the cathode compartment. Zn2“(aq) + 2e¯ → Zn(s) -0.763 2H*(aq) + 2e¯ –→ H2(g) b.
(ii) State which species is the oxidising agent.
A galvanic cell is prepared according to the redox reaction shown below. A voltmeter connected to this galvanic cell under standard conditions reads 0.45 V, and AG, xn = -79 kJ/mol. If the concentration of Fe2+ is increased to 2.0 M, then Ecell will and AGrxn will be than AGxn- Fe(s) + 2H*(aq) Fe2* + H2(g) decrease, less negative O decrease, more negative O increase, more negative O increase, less negative
Please don't provide handwritten solution...
The cell potential of the following cell at 25°C is 0.475 V: Zn|Zn?t (1.0 M) || H* (test solution) H2 (1.0 atm)|Pt What is the pH of the test solution? The standard reduction potential for Zn+ is –0.76 V. (Enter your answer to two significant figures.) pH =
A chemist designs a galvanic cell that uses these two half-reactions: half-reaction standard reduction potential 2+ Zn (aq)+2e - Zn(s) :-0.763 V (red |Cro (aq)+4 H2O(1)+3e Cr(OH)3(s)+5OH (aq) :-0.13 V = - (red Answer the following questions about this cell. Write a balanced equation for the half-reaction that happens at the cathode. x10 Write a balanced equation for the half-reaction that happens at the anode. Write a balanced equation for the overall reaction that powers the cell. Be sure the reaction is spontaneous as written.
(c) A galvanic cell is constructed using the following two half-cells at 25°C and 1 atm. The concentration of each species is given in the half-cell equation. Fe* (aq, 1.10 mol dm³) +3e → Fe2+ (aq, 0.86 mol dm³) E° = +0.77 V Ag* (aq, 1.35 mol dm³) + e → Ag(s) E° = +0.88 V
This is the full question. Please help explain.
A voltaic cell utilizes the following reaction:2 Fe3+(aq) + H2(g)---->2 Fe2+(aq) + 2 H+(aq) (a) What is the emf of this cell under standard conditions?(b) What is the emf for this cell when [Fe3+] = 3.50 M, PH2=0.95 atm, [Fe2+] = 0.0010 M, and the pH in both half-cellsis 4.00?