Chemistry - Electrochemistry Question with Solution | TestHub

Explain the question: Find the cell with the highest EMF.

Concept: This question is based on the Nernst Equation and Equilibria.

Solution:

We calculate the cell potential for each option using $E_{cell}=E_{ca\mathrm{th}od}^{}-E_{anode}$ at 25°C.

1. Option 1:

Anode: $[A{g}^{+}]=K_{sp}/[C{l}^{-}]=10^{-10}/0.1=10^{-9}M$.

Cathode: $[A{g}^{+}]=10^{-3}M$.

Since $E_{cell}^{\circ }=0$, the cell potential is:

$$E=0.059\log \left(\frac{10^{-3}}{10^{-9}}\right)=0.059\times 6=0.354V$$

2. Option 2:

Anode: For a phosphate buffer, $pH=p{K}_a+\log \left(\frac{[base]}{[acid]}\right)$. Assuming $p{K}_a=12$ and concentrations are equal ($0.1/0.1$), $pH=12$.

So, $E_{anode}=-0.059\times 12=-0.708V$.

Cathode: $E_{cathode}=0.8+0.059\log (10^{-2})=0.8-0.118=0.682V$.

The cell potential is:

$$E_{cell}=E_{cathode}-E_{anode}=0.682-(-0.708)=1.39V$$

3. Option 3:

Anode: $pH=0.5(p{K}_a-\log C)=0.5(5+1)=3$.

Cathode: $pH=0.5(10+2)=6$.

The cell potential is:

$$E_{cell}=-0.059(6)-(-0.059(3))=-0.354+0.177=-0.177V$$

4. Option 4:

Anode: $[C{u}^{2+}]=K_{sp}/[O{H}^{-}{]}^2=10^{-20}/(0.1{)}^2=10^{-18}M$.

$E_{anode}=0.34+0.0295(-18)=0.34-0.531=-0.191V$.

Cathode: $pH=1⟹[H^{+}]=0.1M$.

$E_{cathode}=-0.059V$.

The cell potential is:

$$E_{cell}=E_{cathode}-E_{anode}=-0.059-(-0.191)=0.132V$$

Comparing these values, Option B provides the highest cell potential.