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Electrochemistry

Question

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If E° for copper electrode is + 0.34 V how will you calculate its emf value when the solution in contact with it is 0.1 M in copper ions? How does emf for copper electrode change when concentration of Cu²⁺ ions in the solution is decreased?

Solution

The emf of an electrode when dipped in different concentrated solution is given by Nernst equation.

E_{{Cu}^{2 +} / Cu} = E °_{{Cu}^{2 +} / Cu} + \frac{0.059}{2} \times \log {Cu}^{2 +}

Substituting the given values, we get

E_{{Cu}^{2 +} / Cu} = 0.34 V + [\frac{0.059}{2} \log 0.1] V = + 0.34 V - [0.0295 \log 10] V = + 0.34 V - 0.0295 V = + 0.3105 V .

When concentration of Cu²⁺ ion in the solution decreases the emf of the electrode decreases. In this case it has decreased from 0.34 V to 0.3105 V.

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