Question

Calculate the equilibrium constant at 25 ∘C for the reaction

Co_(s) + 2Ag⁺_(aq) → Co²⁺_(aq) + 2Ag_(s)

Standard Reduction Potentials at 25 ∘C

Co2+(aq)+2e−→Co(s)	E∘= −0.28 V
Ag+(aq)+e−→Ag(s)	E∘= 0.80 V

Express your answer using two significant figures.

K = ?

Answer 1

from data table:

Eo(Co2+/Co(s)) = -0.28 V

Eo(Ag+/Ag(s)) = 0.80 V

As per given reaction/cell notation,

cathode is (Ag+/Ag(s))

anode is (Co2+/Co(s))

Eocell = Eocathode - Eoanode

= (0.80) - (-0.28)

= 1.08 V

here, number of electrons being transferred, n = 2

Eo = (2.303*R*T)/(n*F) log Kc

At 25 oC or 298 K, R*T/F = 0.0592

So, Eo = (0.0592/n)*log Kc

1.08 = (0.0592/2)*log Kc

log Kc = 36.4865

Kc = 3.065*10^36

Answer: 3.06*10^36

Answer 2

To calculate the equilibrium constant ($K$) for the reaction:

$$\text{Co(s)} + 2\text{Ag}^+ (aq) \rightarrow \text{Co}^{2+} (aq) + 2\text{Ag}(s)$$

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Step 1: Identify half-reactions and their potentials

Oxidation half-reaction:

$$\text{Co(s)} \rightarrow \text{Co}^{2+}(aq) + 2e^- \quad E^\circ_{\text{oxidation}} = -(-0.28 \, \text{V}) = 0.28 \, \text{V}$$

Reduction half-reaction:

$$2\text{Ag}^+ (aq) + 2e^- \rightarrow 2\text{Ag}(s) \quad E^\circ_{\text{reduction}} = 0.80 \, \text{V}$$

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Step 2: Calculate the standard cell potential

$$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.80 - (-0.28) = 1.08 \, \text{V}$$

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Step 3: Calculate the equilibrium constant $K$

Use the Nernst equation at standard conditions:

$$E^\circ_{\text{cell}} = \frac{0.0257}{n} \ln K$$

Where:

• $E^\circ_{\text{cell}} = 1.08 \, \text{V}$

• $n = 2$ (since 2 electrons are transferred)

Rearrange to solve for $K$:

$$\ln K = \frac{n E^\circ_{\text{cell}}}{0.0257} = \frac{2 \times 1.08}{0.0257} \approx 84.03$$$$K = e^{84.03} \approx 4.0 \times 10^{36}$$

✅ Final answer:

$$K \approx 4.0 \times 10^{36}$$

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