Animated Solution for Chemistry - Electrochemistry: Given the equilibrium constant ($K_c$) of the reaction : $Cu(s)+2A{g}^{+}(aq)\to C{u}^{2+}(aq)+2Ag(s)$ is $10\times 10^{15}$, calculate the $E_{cell}^{\circ }$ of this reaction at 298 K. $\left[2.303\frac{RT}{F}\text{ at }298K=0.059V\right]$

Imagine you are observing a galvanic cell with a Copper anode and a Silver cathode. As the reaction proceeds, electrons flow, and the cell generates a voltage. But what happens when the reaction finally reaches equilibrium? The driving force vanishes. The battery goes completely dead, which mathematically means the instantaneous cell potential, $E_{cell}$, drops to exactly zero.

This is a crucial conceptual anchor. At equilibrium, the reaction quotient $Q$ becomes the equilibrium constant $K_c$, and $E_{cell}=0$. However, the standard cell potential, $E_{cell}^{\circ }$, is a fixed constant for the reaction and does not become zero.

To connect the equilibrium constant with the standard cell potential, we bring in our master tool: the Nernst Equation. It relates the cell potential at any moment to the standard potential and the reaction quotient.

By applying our equilibrium conditions ($E_{cell}=0$ and $Q=K_c$), the equation beautifully simplifies. We are also given the value of the constant term at $298\text{ K}$:

Rearranging this gives us a direct bridge between thermodynamics and electrochemistry:

Before we can plug in the numbers, we need to find $n$, the number of moles of electrons transferred in the balanced equation. Let's break the reaction into its half-cells:

Oxidation: $Cu\to C{u}^{2+}+2e^{-}$ Reduction: $2A{g}^{+}+2e^{-}\to 2Ag$

Copper goes from an oxidation state of $0$ to $+2$, losing two electrons. Simultaneously, two Silver ions go from $+1$ to $0$, collectively gaining two electrons. Therefore, the total number of electrons transferred is $n=2$.

Now, we have all the pieces of the puzzle. We know $n=2$, and the equilibrium constant is given as $K_c=10\times 10^{15}$, which simplifies by the laws of exponents to $10^{16}$. Let's substitute these into our rearranged Nernst equation:

Using the property of logarithms, $\log (10^{16})=16\log (10)=16$.

Looking at our options, the closest value provided is $0.4736\text{ V}$. This slight variation arises from using a more precise value of the constant (like $0.0591$ instead of $0.059$), but the logic remains flawless. We have successfully decoded the cell's standard potential!