The standard potential for the reaction Zn + 2Ag⁺ $\rightarrow$ Zn²⁺ + 2Ag is 1.56 V. Given that the standard reduction potential for Ag⁺ + e^- $\rightarrow$ Ag is 0.80 V, determine the standard reduction potential for Zn²⁺ + 2e^- $\rightarrow$ Zn.

Question

Quizplus · Accepted Answer

We can use the equation:

∆G° = -nFE°

Where ∆G° is the standard Gibbs free energy change, n is the number of electrons transferred in the balanced chemical equation, F is the Faraday constant (96,485 C/mol), and E° is the standard reduction potential.

For the given reaction:

Zn + 2AgS → ZnS + 2Ag

n = 2 (since 2 electrons are transferred)
E° = 1.56 V (given)
F = 96,485 C/mol

From the standard Gibbs free energy equation:

∆G° = ∆H° - T∆S°

We know that at standard conditions:

∆G° = -RTln(K)

where R is the gas constant, T is the temperature, and K is the equilibrium constant.

So, combining the two equations and solving for E°:

E° = ∆G°/(-nF) = (-RTln(K))/(-nF)

E° = (RT/nF)ln(K)

At equilibrium, ∆G° = 0 and K = 1. Therefore:

0 = (RT/nF)ln(1)

0 = (RT/nF)(0)

So, at equilibrium:

E° = 0

With this information, we can set up the following cell:

Zn | Zn2+ || Ag+ | Ag

The cell potential can be found using the Nernst equation:

Ecell = E°cell - (RT/nF) ln(Q)

At equilibrium, the cell potential will be 0 V:

0 = E°cell - (RT/nF)ln(1)

So:

E°cell = (RT/nF)ln(1)

E°cell = 0

Using the known reduction potentials for Ag and Zn, we can write:

E°cell = E°red (cathode) - E°red (anode)

0 = E°red (Ag+) - E°red (Zn2+)

E°red (Zn2+) = E°red (Ag+)

E°red (Ag+) = 0.80 V (given)

Therefore:

E°red (Zn2+) = 0.80 V

However, we need to find the reduction potential for ZnS. We can use the following half-reactions to get there:

Zn2+ + 2e- → Zn (E° = E°red (Zn2+))

ZnS + 2e- → Zn + S2- (E° = ?)

S2- + 2e- → 2S2- (E° = -2.00 V)

Adding the last two half-reactions gives:

ZnS + 4e- → 2S2- (E° = -2.00 V)

Multiplying the first half-reaction by 2 and adding it to the above equation gives:

ZnS + 4e- + 2Zn2+ → 2S2- + 2Zn

E° = E°red (Zn2+) + E° for the above equation

E° = E°red (Zn2+) + (-2.00 V)

E° = 0.80 V - 2.00 V

E° = -1.20 V

Finally, we can use the following half-reactions to get the reduction potential for ZnS2-:

ZnS2- + 2e- → ZnS (E° = ?)

ZnS + 2e- → Zn + S2- (E° = -0.76 V)

Adding the two half-reactions gives:

ZnS2- + 4e- → Zn + 2S2- (E° = -0.76 V)

Multiplying the first equation by 2 and adding it to the above equation gives:

2ZnS2- + 8e- → 2Zn + 4S2-

E° = E°red (ZnS2-) + E° for the above equation

E° = E°red (ZnS2-) + (-0.76 V)

E° = -1.20 V + 0.76 V

E° = -0.44 V

However, the question asks for the standard reduction potential for ZnS2-, which is the negative of the value we just found:

E°red (ZnS2-) = +0.44 V

Therefore, the answer is (C) 0.76 V.

The Standard Potential for the Reaction Zn + 2Ag+ →\rightarrow→ Zn2+ + 2Ag Is 1

The Standard Potential for the Reaction Zn + 2Ag⁺ $\rightarrow$ Zn²⁺ + 2Ag Is 1