Question

$Zn\left(s\right)+C{u}^{2+}(aq.)\rightleftharpoons Cu\left(s\right)+Z{n}^{2+}(aq.)$
Reaction quotient $Q=\frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}$, variation $E_{cell}$ with $Q$ is of the type with $OA=1.10volt.E_{cell}=1.1591volt$ when:

• A. $\left[C{u}^{2+}\right]/\left[Z{n}^{2+}\right]=0.01$
• B. $\left[Z{n}^{2+}\right]/\left[C{u}^{2+}\right]=0.01$
• C. $\left[Z{n}^{2+}\right]/\left[C{u}^{2+}\right]=0.1$
• D. $\left[Z{n}^{2+}\right]/\left[C{u}^{2+}\right]=1$

Answer 1

Answer: (B)

$\left[Z{n}^{2+}\right]/\left[C{u}^{2+}\right]=0.01$

We have,

$E_{cell}=E_{cell}^0-\frac{0.0591}{2}log\frac{Z{n}^{2+}}{C{u}^{2+}}$

$E_{cell}^0=OA=1.1volt$

so, we have

$1.1591=1.1-0.0295log\frac{Z{n}^{2+}}{C{u}^{2+}}$

$log\frac{Z{n}^{2+}}{C{u}^{2+}}=2$

$\frac{Z{n}^{2+}}{C{u}^{2+}}=0.01$