The free energy change for the following reaction at 25 °C, when [Ni²+] = 1.15 M and [Cd²+] = 0.00360 M, is -43.8 kJ: Ni²+ (1.15 M) + Cd(s) → Ni(s) + Cd²+ (0.00360 M) AG = -43.8 kJ What is the cell potential for the reaction as written under these conditions? Ecell = Would this reaction be spontaneous in the forward or the reverse direction? O forward direction O reverse direction

Chemistry: Principles and Practice
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Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
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Chapter18: Electrochemistry
Section: Chapter Questions
Problem 18.43QE: For each of the reactions, calculate E from the table of standard potentials, and state whether the...
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The free energy change for the following reaction at 25 °C, when
[N₁²+] = 1.15 M and
Ni²+ (1.15 M) + Cd(s) → Ni(s) + Cd²+ (0.00360 M) AG = -43.8 kJ
What is the cell potential for the reaction as written under these conditions?
Ecell =
V
[Ca²+ ] =
=
Would this reaction be spontaneous in the forward or the reverse direction?
O forward direction
O reverse direction
0.00360 M, is -43.8 kJ:
Transcribed Image Text:The free energy change for the following reaction at 25 °C, when [N₁²+] = 1.15 M and Ni²+ (1.15 M) + Cd(s) → Ni(s) + Cd²+ (0.00360 M) AG = -43.8 kJ What is the cell potential for the reaction as written under these conditions? Ecell = V [Ca²+ ] = = Would this reaction be spontaneous in the forward or the reverse direction? O forward direction O reverse direction 0.00360 M, is -43.8 kJ:
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