Question

Electra needs to answer this question to master this topic.
Help her calculate the equilibrium constant of the following reaction:
$Cu\left(s\right)+2A{g}^{+}\left(aq\right)\to C{u}^{2+}\left(aq\right)+2Ag\left(s\right)$
$E_{cell}^0=0.46V$

The options below are $logK$ values where $K$ is the equilibrium constant.

• A. 15.45
• B. 15.59
• C. 15.93
• D. 19.5

Answer 1

The correct option is B 15.59

The first step is to write down the half-cell reactions

Reduction: 2Ag⁺ (aq) + 2e^- → 2Ag(s)

Oxidation: Cu(s) → Cu2+ (aq) + 2e-

From this we get, n = 2

Let us first look at the general form of Nernst equation and try to derive our result.

E_cell = E^o - 2.303 $\frac{2.303RT}{nF}$log Q

But at equilibrium, E_cell goes to 0 and Q is replaced by K, which is the equilibrium constant.

So, we have
0 = E^o – 2.303 $\frac{2.303RT}{nF}$log K

Rearranging, E^o = 2.303 $\frac{2.303RT}{nF}$ log K

Now let us plug in the values given,
E⁰_cell = 0.46 V

R = 8.314 JK^-1mol^-1
T = 25^o C = 298 K and n = 2

We would end up with,

0.46 = $\frac{0.059}{2}$log K

log K = 15.59