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What is the Nernst equation for a cell at 298 K with E° = 0.77 V and [Zn²⁺] = 0.
What is the Nernst equation for a cell at 298 K with E° = 0.77 V and [Zn²⁺] = 0.01 M?
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Practice Questions
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Q1
What is the Nernst equation for a cell at 298 K with E° = 0.77 V and [Zn²⁺] = 0.01 M?
E = 0.77 - 0.0591 log(0.01)
E = 0.77 + 0.0591 log(0.01)
E = 0.77 - 0.0591 log(100)
E = 0.77 + 0.0591 log(100)
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Using the Nernst equation: E = E° - (0.0591/n) log([Zn²⁺]), where n=2.
Questions & Step-by-step Solutions
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Q
Q: What is the Nernst equation for a cell at 298 K with E° = 0.77 V and [Zn²⁺] = 0.01 M?
Solution:
Using the Nernst equation: E = E° - (0.0591/n) log([Zn²⁺]), where n=2.
Steps: 9
Show Steps
Step 1: Identify the Nernst equation, which is E = E° - (0.0591/n) log(Q).
Step 2: Determine the standard cell potential (E°), which is given as 0.77 V.
Step 3: Identify the number of electrons transferred in the reaction (n). For zinc (Zn), n = 2.
Step 4: Calculate the reaction quotient (Q). Since we have [Zn²⁺] = 0.01 M, Q = [Zn²⁺] = 0.01.
Step 5: Plug the values into the Nernst equation: E = 0.77 V - (0.0591/2) log(0.01).
Step 6: Calculate log(0.01), which is -2.
Step 7: Substitute log(0.01) into the equation: E = 0.77 V - (0.0591/2) * (-2).
Step 8: Calculate (0.0591/2) * (-2) = -0.0591.
Step 9: Finally, calculate E = 0.77 V + 0.0591 = 0.8291 V.
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