**Consider the following cell:**\[ \text{Cr} \mid \text{Cr}^{3+}(1.00 \, M) \, \| \, \text{Ni}^{2+}(1.00 \, M) \mid \text{Ni} \]\[ E^\circ = 0.68 \, V \]**(a) How many milligrams of metallic nickel would be plated on the cathode if a current of 0.16 A is drawn for 1.3 hours?**- **Answer:** \( 2.3 \times 10^2 \, \text{mg} \) (Correct)**(b) What is the Cr\(^{3+}\) concentration when the Ni\(^{2+}\) concentration has dropped to \( 1.0 \times 10^{-4} \, M \)?**- **Answer:** \( 3.29 \times 10^{28} \, M \) (Incorrect)**What is the cell potential at this point?**- **Answer:** \( 8.45 \times 10^{7} \, V \) (Incorrect)

(b) Write the balanced chemical equation for the overall cell reaction and use the ICE table.The reduction half reaction is Ni2+(aq)+2e-→Ni(s) The oxidation half reaction is Cr(s)→Cr3+(aq)+3e- Multiply the reduction half reaction with 3 and oxidation half reaction with 2 and then add the resultant half reactions to arrive at the overall cell reaction. 3Ni2+(aq)+6e-→3Ni(s)2Cr(s)→2Cr3+(aq)+6e--------------2Cr(s)+3Ni2+(aq)→2Cr3+(aq)+3Ni(s) Overall cell reaction is 2Cr(s)+3Ni2+(aq)→2Cr3+(aq)+3Ni(s) Make ICE table for the overall reaction. 2Cr(s) + 3Ni2+(aq)→2Cr3+(aq)+3Ni(s)I - 1.00 1.00 -C - -3y +2y -E - 1.00-3y 1.00+2y - The equilibrium Ni2+ ion concentration is given as 1.0 x 10-4 M. So, [Ni2+]E =1.00-3y = 1.0×10-4 M y =1- 1.0×10-4 3=0.3333 The equilibrium concentration of Cr3+ is (1.00+2y) M [Cr3+] =(1.00+2y) M [Cr3+] =(1.00+2×0.3333) M =1.6666 M =1.67 M Therefore, answer for the question (b) is 1.67 M(C) Use the Nernst equation for the overall cell reaction to calculate the cell potential. E = Eo-0.05916 Vnlog[Cr3+]2[Ni2+]3 at 298.15 K n is the number of electrons involved in the overall reaction =6 E = Eo-0.05916 V6log[1.67]2[1.0×10-4]3 E = 0.68 V-0.05916 V6×12.445=0.68 V-0.1227 V =0.557 V =0.56 V The cell potential is 0.56 V Answer for (c) 0.56 V

Science

Chemistry

Consider the following cell. Cr | Cr³+ (1.00 M) || Ni²+(1.00 M) | Ni = 0.68 V (a) How many milligrams of metallic nickel would be plated on the cathode if a current of 0.16 A is drawn for 1.3 hrs? 4.0 2.3e2 mg (b) What is the Cr³+ concentration when the Ni²+ concentration has dropped to 1.0 x 10-4 M? 4.0 3.29e28 XM What is the cell potential at this point? x v 4.0 8.45e7

Consider the following cell. Cr | Cr³+ (1.00 M) || Ni²+(1.00 M) | Ni = 0.68 V (a) How many milligrams of metallic nickel would be plated on the cathode if a current of 0.16 A is drawn for 1.3 hrs? 4.0 2.3e2 mg (b) What is the Cr³+ concentration when the Ni²+ concentration has dropped to 1.0 x 10-4 M? 4.0 3.29e28 XM What is the cell potential at this point? x v 4.0 8.45e7

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

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**Consider the following cell:**

\[ \text{Cr} \mid \text{Cr}^{3+}(1.00 \, M) \, \| \, \text{Ni}^{2+}(1.00 \, M) \mid \text{Ni} \]

\[ E^\circ = 0.68 \, V \]

**(a) How many milligrams of metallic nickel would be plated on the cathode if a current of 0.16 A is drawn for 1.3 hours?**

- **Answer:** \( 2.3 \times 10^2 \, \text{mg} \) (Correct)

**(b) What is the Cr\(^{3+}\) concentration when the Ni\(^{2+}\) concentration has dropped to \( 1.0 \times 10^{-4} \, M \)?**

- **Answer:** \( 3.29 \times 10^{28} \, M \) (Incorrect)

**What is the cell potential at this point?**

- **Answer:** \( 8.45 \times 10^{7} \, V \) (Incorrect)

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