Calculate the potential for each cell as written at 25 °C. Indicate whether the metal electrode in the half-reaction opposite the standard hydrogen electrode, S.H.E., would be the anode or the cathode if the cell was shorted and electrons were allowed to flow freely. S.H.E. Zn2+ (aq, 0.0550 M) | Zn(s) = -0.762 V Zn2+ /Zn The Zn electrode is the 0.762 Ecell = V O cathode. Incorrect O anode. Pt(s) | V3+(aq, 0.430 M), V2+(aq, 0.00780 M) || S.H.E. = -0.255 V The Pt electrode is the -0.1713 Ecell = V Incorrect: cathode. O anode.

Calculate the potential for each cell as written at 25 °C. Indicate whether the metal electrode in the half-reaction opposite the standard hydrogen electrode, S.H.E., would be the anode or the cathode if the cell was shorted and electrons were allowed to flow freely. S.H.E. Zn2+ (aq, 0.0550 M) | Zn(s) = -0.762 V Zn2+ /Zn The Zn electrode is the 0.762 Ecell = V O cathode. Incorrect O anode. Pt(s) | V3+(aq, 0.430 M), V2+(aq, 0.00780 M) || S.H.E. = -0.255 V The Pt electrode is the -0.1713 Ecell = V Incorrect: cathode. O anode.

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...

See similar textbooks

Similar questions

A certain metal M forms a soluble sulfate salt MSO4. Suppose the left half cell of a galvanic cell apparatus is filled with a 2.50 M solution of MSO, and the right half cell with a 25.0 mM solution of the same substance. Electrodes made of M are dipped into both solutions and a voltmeter is connected between them. The temperature of the apparatus is held constant at 40.0 °C. left Which electrode will be positive? x10 O right What voltage will the voltmeter show? Assume its positive lead is connected to the positive electrode. necessary, and round it to 2 significant Be sure your answer has a unit symbol, digits.
The following half-cells are on the left and coupled with the standard hydrogen electrode on the right to form a galvanic Calculate the cell potential. If the cell were shorted, indicate whether the electrodes shown would act as an anode or a cathode. a. Cu Cu²+ (0.0567 M), (Cu²+/Cu = 0.337 V) V Potential = b. Cu | Cul (sat'd), KI (0.0939 M), (ECul/Cu = -0.185 V) Potential = V 3+ c. Pt | Fe³+ (0.160 M), Fe²+ (0.08340 M), (E Potential = V Fe³+/Fe²+ = = 0.771 V)
Cu = reduction potential of +0.52. Mg is in pic. Thanks
The redox reaction 3 Cl, (g) + 2 Al(s) → 2 Al³ † (aq) + 6 CI"(aq) consists of the two half-reactions shown. Cl, (g) + 2 e- –→ 2 Cl¯(aq) E. red = +1.36 V Al3 + (aq) + 3 e Al(s) Eed = -1.66 V What is the standard cell potential? 3.02 V 7.70 V -3.02 V O -0.30 V
Given the balanced electrochemical reaction below (M and N represent generic metals), what is the [M3*], in moles per liter, required to generate a cell potential (Ecell) of 0.584 V for this reaction at 58.1 °C when [N*] = 1.246 M? Enter your answer to the thousandths place. Do not include units. M(s) + 3N*(aq) –→ M3+(aq) + 3N(s) E°cell = 0.572 V
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 =
17. Calculate the cell potential of the cell described below: 2+ Al(s) | Al³+ (aq) (1.00 M) || Pb²+ (aq) (1.00 M) | Pb(s)
Calculate Eu for the following cell: Pt(s) | H₂(g) | H *(aq) || Pb2+ (aq) | PbSO4(s) | Pb(s) given the following standard reduction potentials. 2 H(aq) + 2e →→ H₂(g) PbSO4(s) + 2e →→ Pb(s) + SO42 (aq) O a. -0.712 V O b. +0.356 V O c. -0.356 V O d. +0.712 V O e. -0.178 V E° = 0.000 V E = -0.356 V
Consider the electrolysis reactions cathode anode whose standard reduction potentials can be found in this table. Calculate the voltage needed to drive the net reaction if the current is negligible. Ecell = Cu²+ (aq, 0.177 M) +2e=Cu(s) H₂O(1) ⇒ 10₂(g, 1 bar) + 2 H+ (aq, 0.195 M) + 2 e¯ The cell has a resistance of 66.0 and a current of 7.17 mA when the cathode overpotential is 0.231 V and the anode overpotential is 0.050 V. Calculate the voltage needed to overcome these effects and drive the net reaction. Ecell = V V
Calculate the potential for each cell as written at 25 °C. Indicate whether the metal electrode in the half-reaction opposite the standard hydrogen electrode, S.H.E., would be the anode or the cathode if the cell was shorted and electrons were allowed to flow freely. S.H.E. || Zn²+ (aq, 0.0110 M)| Zn(s) Ecell = Pt(s) | V3+ (aq, 0.450 M), V²+ (aq, 0.00420 M) || S.H.E. Ecell = EZn²+ Zn Incorrect = -0.762 V V V E√3+√√²+ = -0.255 V
The standard cell potential for the reaction shown here is .653 V. Calculate the cell potential at 25°C with [Hg^2+] = 0.0500 M, [V^2+] = 0.100 M, [Hg2^2+] = 0.00500 M and [V^3+] = %3D 0.0200 M. [Hg?+] do.o500 M [V2+] = 0.100 M H0.0500 M [Hg* = 0.00500 M [V³*] = 0.0200 M 2 Hg²+ (aq) + 2 V²+ (aq) = Hg2²+ (aq) + 2 V³+ (aq)