The EV(M2+/M) value for copper is positive (+0.34V) which indicates that copper is more likely to be reduced than oxidized. This is due to its high ∆aHV (enthalpy of atomization) and low ∆hydHV (enthalpy of hydration).

1. High ∆aHV: The enthalpy of atomization is the energy required to convert a substance from its normal state into atoms. Copper has a high ∆aHV, which means it requires a lot of energy to be atomized. This makes it less likely to be oxidized (lose electrons), and more likely to be reduced (gain electrons).

2. Low ∆hydHV: The enthalpy of hydration is the energy released when a substance is dissolved in water. Copper has a low ∆hydHV, which means it releases less energy when it is hydrated. This also makes it less likely to be oxidized, and more likely to be reduced.

Therefore, the positive EV(M2+/M) value for copper is due to its high ∆aHV and low ∆hydHV, which make it more likely to be reduced than oxidized.

Question

The EV(M2+/M) value for copper is positive (+0.34V) which indicates that copper is more likely to be reduced than oxidized. This is due to its high ∆aHV (enthalpy of atomization) and low ∆hydHV (enthalpy of hydration).

1. High ∆aHV: The enthalpy of atomization is the energy required to convert a substance from its normal state into atoms. Copper has a high ∆aHV, which means it requires a lot of energy to be atomized. This makes it less likely to be oxidized (lose electrons), and more likely to be reduced (gain electrons).

2. Low ∆hydHV: The enthalpy of hydration is the energy released when a substance is dissolved in water. Copper has a low ∆hydHV, which means it releases less energy when it is hydrated. This also makes it less likely to be oxidized, and more likely to be reduced.

Therefore, the positive EV(M2+/M) value for copper is due to its high ∆aHV and low ∆hydHV, which make it more likely to be reduced than oxidized.

Knowee AI · Accepted Answer

The EV(M2+/M) value for copper is positive (+0.34V) which indicates that copper is more likely to be reduced than oxidized. This is due to its high ∆aHV (enthalpy of atomization) and low ∆hydHV (enthalpy of hydration).

1. High ∆aHV: The enthalpy of atomization is the energy required to convert a substance from its normal state into atoms. Copper has a high ∆aHV, which means it requires a lot of energy to be atomized. This makes it less likely to be oxidized (lose electrons), and more likely to be reduced (gain electrons).

2. Low ∆hydHV: The enthalpy of hydration is the energy released when a substance is dissolved in water. Copper has a low ∆hydHV, which means it releases less energy when it is hydrated. This also makes it less likely to be oxidized, and more likely to be reduced.

Therefore, the positive EV(M2+/M) value for copper is due to its high ∆aHV and low ∆hydHV, which make it more likely to be reduced than oxidized.

The EV(M2+/M) value for copper is positive (+0.34V). What is possiblereason for this? (Hint: consider its high ∆aHV and low ∆hydHV)

Question

The EV(M2+/M) value for copper is positive (+0.34V). What is possible reason for this?

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