Question

$Zn+Cu{{}^{2+}}_{\left(aq\right)}\rightleftharpoons Cu+Zn{{}^{2+}}_{\left(aq\right)}$

Reaction quotient, $Q=\frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}$

Variation of $E_{cell}$ with log Q is of the type as shown in the graph with $OA=1.10V.E_{cell}$ of the reaction will be 1.1591 V when:

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

Answer 1

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

$E_{cell}=E_{cell}^{\circ }-\frac{0.059}{n}logQ$

$\therefore 1.1591=1.10-\frac{0.0591}{2}log\left[\frac{Z{n}^{2+}}{C{u}^{2+}}\right]$

$0.0591=\frac{-0.0591}{2}log\frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}$

$\therefore log\frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}=-2$

$\therefore \frac{\left[Z{n}^{2+}\right]}{\left[C{u}^{2+}\right]}={10}^{-2}=0.01$

Option C is correct.