Given that at `25^@C`
`[Ag(NH_(3))_(2)]^(+) +e^(0) to Ag_((s))+2NH_(3) E^(0)=0.02V and Ag^(+)+1e^(-) to Ag_((s))" " E^(0)=0.8V`
Hence the order of magnitude of equilibrium constant of the reaction
`[Ag(NH_(3))_(2)]^(+) Leftrightarrow Ag^(+) +2NH_(3)`, will be (antilog 0.22=1.66)

Options

A. `10^(-10)`
B. `10^(-6)`
C. `10^(-14)`
D. `10^(-18)`

Answer

To solve the problem, we need to find the order of magnitude of the equilibrium constant for the reaction:

\[ [Ag(NH_3)_2]^+ \leftrightarrow Ag^+ + 2NH_3 \]

We are given the standard reduction potentials for two half-reactions:

1. \([Ag(NH_3)_2]^+ + e^- \rightarrow Ag(s) + 2NH_3\) with \(E^\circ = 0.02 \, V\)
2. \(Ag^+ + e^- \rightarrow Ag(s)\) with \(E^\circ = 0.8 \, V\)

### Step 1: Write the half-reactions and their potentials

The half-reaction for the formation of silver from silver ion is already given. We can invert the second half-reaction to find the oxidation potential:

\[ Ag(s) \rightarrow Ag^+ + e^- \quad (E^\circ = -0.8 \, V) \]

### Step 2: Combine the half-reactions

Now, we can add the two half-reactions together. The first half-reaction is:

\[ [Ag(NH_3)_2]^+ + e^- \rightarrow Ag(s) + 2NH_3 \quad (E^\circ = 0.02 \, V) \]

The inverted second half-reaction is:

\[ Ag(s) \rightarrow Ag^+ + e^- \quad (E^\circ = -0.8 \, V) \]

Adding these reactions cancels out the electrons:

\[ [Ag(NH_3)_2]^+ \rightarrow Ag^+ + 2NH_3 \]

### Step 3: Calculate the overall standard cell potential

The overall standard cell potential \(E^\circ_{cell}\) is the sum of the two potentials:

\[ E^\circ_{cell} = E^\circ_{[Ag(NH_3)_2]^+} + E^\circ_{Ag^+} \]
\[ E^\circ_{cell} = 0.02 \, V - 0.8 \, V = -0.78 \, V \]

### Step 4: Use the Nernst equation

At equilibrium, the cell potential \(E_{cell} = 0\). We can use the Nernst equation to relate the cell potential to the equilibrium constant \(K\):

\[ E_{cell} = E^\circ_{cell} - \frac{0.059}{n} \log K \]

Where \(n\) is the number of electrons transferred (which is 1 in this case). Setting \(E_{cell} = 0\):

\[ 0 = -0.78 - \frac{0.059}{1} \log K \]

### Step 5: Solve for \(K\)

Rearranging gives:

\[ \frac{0.059}{1} \log K = -0.78 \]
\[ \log K = \frac{-0.78}{0.059} \]
\[ \log K \approx -13.22 \]

### Step 6: Calculate \(K\)

Taking the antilog to find \(K\):

\[ K \approx 10^{-13.22} \]

### Step 7: Order of magnitude of \(K\)

The order of magnitude of \(K\) is approximately \(10^{-13}\).

### Final Answer

Thus, the order of magnitude of the equilibrium constant for the reaction is:

\[ \text{Order of magnitude of } K \approx 10^{-13} \]