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What is the value of the equilibrium constant at 25 oC for the reaction between the pair: Sn(s) and Pb2 (aq) to give Pb(s) and Sn2 (aq) Use the reduction potential values for Sn2 (aq) of -0.14 V and for Pb2 (aq) of -0.13 V Give your answer using E-notation with NO decimal places (e.g., 2 x 10-2 would be 2E-2; and 2.12 x 10-2 would also be 2E-2.). Do NOT include spaces, units, punctuation or anything else silly

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Answer:

The value of the equilibrium constant at 25°C is 2.177

Explanation:

Based on the half-reactions:

Sn(s) → Sn²⁺ + 2e⁻  E° = -(Reduction potential) = 0.14V

Pb²⁺ + 2e⁻ → Pb(s) E° = -0.13V

Sn(s) + Pb²⁺ → Pb(s) + Sn²⁺ E° = 0.14V - 0.13V = 0.01V

Using:

Eºcell = 0.0592/n (log K) At 25°C

Where E° cell = 0.01V

n are moles of electrons = 2

K is equilibrium constant

Replaing:

0.01V = 0.0592/2 * (logK)

0.3378 = log K

2.177 = K

The value of the equilibrium constant at 25°C is 2.177

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