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3 years ago

Given the chemical reaction: Hg2+(aq) + Cu(s) + Hg(s) + Cu2+(aq)

Chemistry

1 answer:

Pavel [41]3 years ago

5 0

Answer: When the reaction reaches equilibrium, the cell potential will be 0.00 V

Explanation:

Equilibrium state is the state when reactants and products are present but the concentrations does not change with time.

The equilibrium is dynamic in nature and the reactions are continuous in nature. Rate of forward reaction is equal to the rate of backward reaction.

The standard emf of a cell is related to Gibbs free energy by following relation:

$\Delta G=-nFE^0$

The Gibbs free energy is related to equilibrium constant by following relation:

$\Delta G=-2.303RTlog K$

For equilibrium $\Delta G=0$

Thus $0=-nFE^0$

$E^0=0$

Thus When the reaction reaches equilibrium, the cell potential will be 0.00 V

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Hope this helps

4 0

3 years ago

A chemist designs a galvanic cell that uses these two half-reactions: half-reaction standard reduction potential (s)(aq)(aq)(l)

miv72 [106K]

Answer:

Reduction (cathode): Cu²⁺(⁺aq) + 2 e⁻ → Cu(s)

Oxidation (anode): Zn(s) → Zn²⁺(⁺aq) + 2 e⁻

Cu²⁺(⁺aq) + Zn(s) → Cu(s) + Zn²⁺(⁺aq)

E°cell = 1.10 V

Explanation:

The half-reactions are missing, but I will propose some to show you the general procedure and then you can apply it to your equations.

Suppose we have the following half-reactions.

Cu²⁺(⁺aq) + 2 e⁻ → Cu(s) E°red = 0.34 V

Zn²⁺(⁺aq) + 2 e⁻ → Zn(s) E°red = -0.76 V

To identify how to make a spontaneous cell, we need to consider the standard reduction potentials (E°red). The half-reaction with the higher E°red will occur as a reduction (in the cathode), whereas the one with the lower E°red will occur as an oxidation (in the anode).

Reduction (cathode): Cu²⁺(⁺aq) + 2 e⁻ → Cu(s) E°red = 0.34 V

Oxidation (anode): Zn(s) → Zn²⁺(⁺aq) + 2 e⁻ E°red = -0.76 V

To get the overall equation we add both half-reactions.

Cu²⁺(⁺aq) + Zn(s) → Cu(s) + Zn²⁺(⁺aq)

The standard cell potential (E°cell) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E°cell = E°red, cat - E°red, an

E°cell = 0.34 V - (-0.76 V) = 1.10 V

Since E°cell > 0, the reaction is spontaneous.

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