Question

$Zn+C{u}^{2+}\left(aq\right)\rightleftharpoons Cu+Z{n}^{2+}\left(aq\right)$
Reaction quotient is $Q=\frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}$. Variation of $E_{cell}$ with log Q is of the type with OA = 1.10 $V.E_{cell}$ will be $1.1591V$ 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

The correct option is B $\left[Z{n}^{2+}\right]/\left[C{u}^{2+}\right]=0.01$

$E_{cell}=E_{cell}^{\ominus }-\frac{0.0591}{2}log\frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}$
From line $OA=E_{cell}^{\ominus }=1.10V$
$\frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}={10}^{-2}$ M,
then $E_{cell}=1.1591V$