Question

If a half cell $X|X^{2+}\left(0.1M\right)$ is connected to another half cell $Y|Y^{2+}\left(1.0M\right)$ by means of a salt bridge and an external circuit at ${25}^{\circ }C$, the cell voltage would be:

• A. $0.06V$
• B. $0.12V$
• C. $0.62V$
• D. $0.72V$

Answer 1

The correct option is C $0.62V$

When X is connected to SHE, electrons flow from X to SHE. This means that X is acting as anode and SHE as cathode and its oxidation potential is positive. Also, the reduction potential of Y is greater than the reduction potential of X
(as electrons flow X to Y).

$\Rightarrow Y^{2+}+2e^{-}\to Y\left(s\right)$

$E_{Y^{2+}|Y}^{\ominus }=-0.34V$;

$X^{2+}+2e^{-}\to X\left(s\right)$

$E_{X^{2+}|X}^{\ominus }=-0.25V$

Consider : $X\left(s\right)X^{2+}\left(0.1M\right)||Y^{2+}\left(1.0M\right)|Y\left(s\right)$

$E_{cell}^{\ominus }=E_{cell}^{\ominus }-\frac{0.059}{2}lo{g}_1\left[x^{2+}\right]\left[Y^{2+}\right]$

$=[0.34-(-0.25\left)\right]-\frac{0.059}{2}lo{g}_{10}\frac{0.1}{1}=0.62V$