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 determine the equilibrium constant for the reaction:

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

We are given the standard reduction potentials for the following 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: Determine the overall cell reaction
To find the overall reaction, we need to reverse the first half-reaction since we are looking for the oxidation of \([Ag(NH_3)_2]^+\):

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

Now, we can combine this with the second half-reaction:

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

### Step 2: Write the overall reaction
Adding the two half-reactions gives us:

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

### Step 3: Calculate the standard cell potential \(E^\circ_{cell}\)
The standard cell potential is calculated by subtracting the oxidation potential from the reduction potential:

\[ E^\circ_{cell} = E^\circ_{reduction} - E^\circ_{oxidation} \]

Substituting the values:

\[ E^\circ_{cell} = 0.8 \, V - (-0.02 \, V) = 0.8 \, V + 0.02 \, V = 0.82 \, V \]

### Step 4: Relate \(E^\circ_{cell}\) to the equilibrium constant \(K\)
Using the Nernst equation, we know that:

\[ E^\circ_{cell} = \frac{0.0591}{n} \log K \]

Where \(n\) is the number of electrons transferred in the reaction. Here, \(n = 1\) (one electron is involved).

Rearranging gives us:

\[ \log K = \frac{E^\circ_{cell} \cdot n}{0.0591} \]

Substituting the values:

\[ \log K = \frac{0.82 \cdot 1}{0.0591} \approx 13.9 \]

### Step 5: Calculate \(K\)
To find \(K\), we take the antilog:

\[ K = 10^{13.9} \approx 7.94 \times 10^{13} \]

### Step 6: Determine the order of magnitude of \(K\)
The order of magnitude of \(K\) is \(10^{13}\).

### Final Answer
The order of magnitude of the equilibrium constant of the reaction is \(10^{13}\).

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